JEE Electric Potential & Capacitance Questions

The electron enters symmetrically between the plates, meaning it starts at the midpoint with an initial vertical velocity of zero. The initial velocity is purely horizontal, denoted as $$u_x$$. The horizontal component of velocity remains constant because there is no force in the horizontal direction. After emerging, the horizontal component is given as $$10^6 \, \text{m/s}$$, so $$u_x = 10^6 \, \text{m/s}$$.

The length of each plate is $$L = 10 \, \text{cm} = 0.1 \, \text{m}$$. The time taken to cross the electric field region is calculated using the horizontal motion:

$$t = \frac{L}{u_x} = \frac{0.1}{10^6} = 10^{-7} \, \text{s}$$

The electric field magnitude is $$E = 9.1 \, \text{V/cm}$$. Convert to SI units: $$1 \, \text{V/cm} = 100 \, \text{V/m}$$, so $$E = 9.1 \times 100 = 910 \, \text{V/m}$$.

The force on the electron in the vertical direction is due to the electric field. The charge of the electron is $$q = -1.6 \times 10^{-19} \, \text{C}$$, but the magnitude of the force is $$|F| = eE$$, where $$e = 1.6 \times 10^{-19} \, \text{C}$$. The mass of the electron is $$m = 9.1 \times 10^{-31} \, \text{kg}$$.

The acceleration in the vertical direction is given by Newton's second law:

$$a_y = \frac{|F|}{m} = \frac{eE}{m}$$

Substitute the values:

$$a_y = \frac{(1.6 \times 10^{-19}) \times 910}{9.1 \times 10^{-31}}$$

First, compute the numerator: $$1.6 \times 10^{-19} \times 910 = 1.6 \times 10^{-19} \times 9.1 \times 10^2 = 14.56 \times 10^{-17} = 1.456 \times 10^{-16}$$

Now, divide by the denominator:

$$a_y = \frac{1.456 \times 10^{-16}}{9.1 \times 10^{-31}} = \frac{1.456}{9.1} \times 10^{-16 - (-31)} = 0.16 \times 10^{15} = 1.6 \times 10^{14} \, \text{m/s}^2$$

The initial vertical velocity is $$u_y = 0$$. The vertical component of velocity when the electron emerges is given by the equation of motion:

$$v_y = u_y + a_y t = 0 + (1.6 \times 10^{14}) \times (10^{-7}) = 1.6 \times 10^{14} \times 10^{-7} = 1.6 \times 10^{7} \, \text{m/s}$$

Simplify: $$1.6 \times 10^7 \, \text{m/s} = 16 \times 10^6 \, \text{m/s}$$.

The vertical component of velocity is $$16 \times 10^6 \, \text{m/s}$$, which corresponds to option C.

Final Answer: C. $$16 \times 10^{6} \, \text{m/s}$$

Three infinitely long wires with linear charge density $$\lambda$$ are placed along the x-axis, y-axis, and z-axis respectively. We need to find the equation of an equipotential surface.

The electric potential at a perpendicular distance $$r$$ from an infinitely long line charge with linear charge density $$\lambda$$ is:

$$V = -\frac{\lambda}{2\pi\epsilon_0} \ln r + C$$

For a point $$(x, y, z)$$:

- Distance from the x-axis: $$r_x = \sqrt{y^2 + z^2}$$

- Distance from the y-axis: $$r_y = \sqrt{x^2 + z^2}$$

- Distance from the z-axis: $$r_z = \sqrt{x^2 + y^2}$$

$$V_{total} = -\frac{\lambda}{2\pi\epsilon_0}[\ln\sqrt{y^2+z^2} + \ln\sqrt{x^2+z^2} + \ln\sqrt{x^2+y^2}] + C'$$

$$= -\frac{\lambda}{4\pi\epsilon_0}\ln[(y^2+z^2)(x^2+z^2)(x^2+y^2)] + C'$$

For an equipotential surface, $$V_{total} = \text{constant}$$, which requires:

$$(y^2 + z^2)(x^2 + z^2)(x^2 + y^2) = \text{constant}$$

This can be rewritten as:

$$(x^2 + y^2)(y^2 + z^2)(z^2 + x^2) = \text{constant}$$

The correct answer is Option 3: $$(x^2 + y^2)(y^2 + z^2)(z^2 + x^2) = \text{constant}$$.

Both capacitors are in parallel ⇒ same voltage at all times

$$q=CV\Rightarrow q\propto C$$

From the graph:

At final (steady state), $$q_2>q_1$$

⇒ $$C_2>C_1$$

Energy stored:

$$U=\frac{1}{2}CV^2\Rightarrow U\propto C$$

(since V is same)

So:

$$U_2>U_1$$

U = (1 / 4πϵ₀) · (q₁q₂ / r)

given:

q₁ = 7 μC = 7 × 10⁻⁶ C
q₂ = −4 μC = −4 × 10⁻⁶ C

distance between charges:

positions are at −7 cm and +7 cm
so separation r = 14 cm = 0.14 m

step 1: substitute

U = 9 × 10⁹ × (7×10⁻⁶)(−4×10⁻⁶) / 0.14

step 2: simplify

(7×−4) = −28

U = 9 × 10⁹ × (−28 × 10⁻¹²) / 0.14

= 9 × (−28) × 10⁻³ / 0.14

= (−252 × 10⁻³) / 0.14

step 3: final calculation

−0.252 / 0.14 = −1.8

dipole is along +x direction.

point A is on the axial line at distance r
point B is on the equatorial line at distance 2r (on y-axis)

at A (axial point):

for a dipole,

V ∝ p cosθ / r²

here θ = 0° ⇒ cosθ = 1

so,

V₀ = (1/4πϵ₀)(p/r²)

electric field on axial line:

E₀ = (1/4πϵ₀)(2p/r³)

at B (equatorial point):

here θ = 90° ⇒ cosθ = 0

so potential:

V_B = 0

(this is because contributions from +q and −q cancel in potential)

electric field at equatorial point:

fields from +q and −q don’t cancel completely — their components add opposite to dipole moment direction

magnitude:

E_B = (1/4πϵ₀)(p/(distance)³)

distance = 2r

so:

E_B = (1/4πϵ₀)(p/(2r)³)
= (1/4πϵ₀)(p/8r³)

now compare with E₀:

E₀ = (1/4πϵ₀)(2p/r³)

so,

E_B / E₀ = (p/8r³) / (2p/r³) = 1/16

First configuration

Two dielectrics fill half the thickness each, so they are in series.

Each slab has thickness

$$\frac{d}{2}$$

and full area AAA.

Capacitances:

$$C_a=\frac{\varepsilon_1A}{d/2}$$

$$=\frac{2\varepsilon_1A}{d}$$

$$C_b=\frac{\varepsilon_2A}{d/2}=\frac{2\varepsilon_2A}{d}$$

Series combination:

$$\frac{1}{C_1}=\frac{1}{C_a}+\frac{1}{C_b}$$$$\frac{1}{C_1}=\frac{d}{2A}\left(\frac{1}{\varepsilon_1}+\frac{1}{\varepsilon_2}\right)$$

Thus

$$C_1=\frac{2A\varepsilon_1\varepsilon_2}{d(\varepsilon_1+\varepsilon_2)}$$

Second configuration

Each dielectric occupies half area

$$\frac{A}{2}$$

and full thickness ddd, so they are in parallel.

Capacitances:

$$C'_1=\frac{\varepsilon_1(A/2)}{d}$$

$$C'_2=\frac{\varepsilon_2(A/2)}{d}$$

Hence

$$C_2=C'_1+C'_2$$$$C_2=\frac{A}{2d}(\varepsilon_1+\varepsilon_2)$$

Now ratio:

$$\frac{C_1}{C_2}=\frac{\frac{2A\varepsilon_1\varepsilon_2}{d(\varepsilon_1+\varepsilon_2)}}{\frac{A}{2d}(\varepsilon_1+\varepsilon_2)}$$

Simplifying,

$$\frac{C_1}{C_2}=\frac{4\varepsilon_1\varepsilon_2}{(\varepsilon_1+\varepsilon_2)^2}$$

So correct answer is

$$\frac{4\varepsilon_1\varepsilon_2}{(\varepsilon_1+\varepsilon_2)^2}$$

The first capacitor has $$C_1 = 100\,$$pF and is initially charged to $$V_{1i}=60\,$$V.

Initial charge on this capacitor:
$$Q_{\text{initial}} = C_1\,V_{1i} = 100\,\text{pF}\times 60\,\text{V} = 6000\,\text{pC}$$

After disconnecting the battery, this charge cannot escape; it is conserved inside the closed system formed by the two capacitors.

An uncharged capacitor of capacitance $$C_2$$ is then connected in parallel. In a parallel connection the final voltage on both capacitors is the same. The problem states that this common voltage is $$V_f = 20\,$$V.

Let $$Q_{1f}$$ and $$Q_{2f}$$ be the final charges on $$C_1$$ and $$C_2$$ respectively.
Using $$Q = C\,V$$:
$$Q_{1f} = C_1\,V_f$$
$$Q_{2f} = C_2\,V_f$$

Because total charge is conserved (battery is removed),
$$Q_{\text{initial}} = Q_{1f} + Q_{2f}$$
$$\Rightarrow 6000 = C_1\,V_f + C_2\,V_f$$

Divide both sides by $$V_f$$ to isolate $$C_2$$:
$$\frac{6000}{V_f} = C_1 + C_2$$
$$\Rightarrow C_2 = \frac{6000}{20} - C_1$$

Substitute the known values:
$$C_2 = 300 - 100 = 200\,\text{pF}$$

Therefore, the capacitance of the second capacitor is $$200\,$$pF, which corresponds to Option B.

Between parallel plates, electric field is uniform.

If plate separation is 10cm and potential difference is V, field is

$$E=\frac{V}{10}​(volts/cm)$$

We need potential difference between A and B.

From figure:

AC=3cm

AB=5 cm

Triangle ACB is right angled, so

$$CB=\sqrt{\ 5^2-3^2}$$

Now electric field is horizontal (between vertical plates), so only horizontal separation matters for potential difference.

Vertical displacement does not affect potential.

Thus only CB=4 cm contributes.

So

$$V_{AB}=E(4)$$

$$=\frac{V}{10}\times4=$$

$$=\frac{2V}{5}$$

Initially

$$C_1=C_2=C_3=5\mu F$$

Dielectric constant 4 is inserted in $$C_1$$​, so

$$C_1'=4(5)=20\mu F$$

From circuit:

• Top branch: $$C_1\ ​and\ C_2$$ are in series
• Bottom branch:$$C_3$$ alone

These two branches are in parallel.

Top branch equivalent

$$C_{top}=\frac{C_1'C_2}{C_1'+C_2}$$

$$=\frac{20\cdot5}{20+5}$$

$$\frac{100}{25}=4\mu F$$

Total equivalent

Parallel combination:

$$C_{eq}=C_{top}+C_3$$

$$=4+5$$

$$=9\mu F$$

step 1: initial condition

only $$C_1$$​ is charged:

$$q_{\text{initial}}=C_1V=6\mu F\times5=30\mu C$$

$$C_2$$​ is uncharged

step 2: after switch is closed

the two capacitors are connected in parallel (like plates joined)

so:

• total charge is conserved = 30 μC
• final voltage across both is same

step 3: find final voltage

$$C_{\text{eq}}=C_1+C_2=6+12=18\mu F$$

$$V_f=\frac{Q_{\text{total}}}{C_{\text{eq}}}=\frac{30}{18}=\frac{5}{3}\text{ V}$$

step 4: final charges

$$q_1=C_1V_f=6\times\frac{5}{3}=10\mu C$$

$$q_2=C_2V_f=12\times\frac{5}{3}=20\mu C$$

The shell is a thin, uniformly charged spherical shell of radius $$R = 10 \text{ cm} = 0.10 \text{ m}$$.
The given electrostatic potential on its surface is $$V(R)=120 \text{ V}$$.

Key electrostatics results for a uniformly charged thin spherical shell
1. For any point inside the shell ($$r \lt R$$): the electric field is zero, $$E=0$$, and the electrostatic potential is the same everywhere as on the surface. Hence $$V(r)=V(R)$$.
2. For any point outside the shell ($$r \gt R$$): the shell behaves as if all its charge were concentrated at its centre. Therefore $$V(r)=\dfrac{kQ}{r}$$, where $$Q$$ is the total charge and $$k=\dfrac{1}{4\pi\varepsilon_0}$$.

We use these two facts to find the required potentials.

Case 1: At the centre of the shell ( $$r = 0$$ )
Since $$r \lt R$$, the potential equals the surface potential: $$V(0)=V(R)=120 \text{ V}$$.

Case 2: At an interior point $$r = 5 \text{ cm} = 0.05 \text{ m}$$
Again $$r \lt R$$, so $$V(5\text{ cm})=V(R)=120 \text{ V}$$.

Case 3: At an exterior point $$r = 15 \text{ cm} = 0.15 \text{ m}$$
Here $$r \gt R$$, so use the point-charge formula.

The charge $$Q$$ is not given directly, but we can eliminate it using the known surface potential:
$$V(R)=\dfrac{kQ}{R}=120 \text{ V} \quad -(1)$$

For $$r = 0.15 \text{ m}$$, $$V(15\text{ cm})=\dfrac{kQ}{r}$$

Divide this by equation $$(1)$$ to remove $$Q$$:
$$\dfrac{V(15\text{ cm})}{120}=\dfrac{\dfrac{kQ}{r}}{\dfrac{kQ}{R}}=\dfrac{R}{r}$$

Hence
$$V(15\text{ cm})=120 \times \dfrac{R}{r}=120 \times \dfrac{0.10}{0.15}=120 \times \dfrac{2}{3}=80 \text{ V}$$.

Summary of potentials
Centre (0 cm): $$120 \text{ V}$$
Inside at 5 cm: $$120 \text{ V}$$
Outside at 15 cm: $$80 \text{ V}$$

These values match Option A: 120 V, 120 V, 80 V.

The original capacitor has $$l = 3$$ cm, $$b = 1$$ cm, $$d = 3$$ $$\mu$$m.

Capacitance formula: $$C = \frac{\epsilon_0 A}{d} = \frac{\epsilon_0 \cdot l \cdot b}{d}$$.

Original: $$C_0 = \frac{\epsilon_0 \cdot 3 \cdot 1}{3} = \epsilon_0$$ (in relative units).

We need $$C = 10C_0$$, i.e., $$\frac{l \cdot b}{d} = 10 \times \frac{3 \cdot 1}{3} = 10$$.

A: $$l = 30, b = 1, d = 1$$: $$\frac{30 \times 1}{1} = 30 \neq 10$$. No.

B: $$l = 3, b = 1, d = 30$$: $$\frac{3 \times 1}{30} = 0.1 \neq 10$$. No.

C: $$l = 6, b = 5, d = 3$$: $$\frac{6 \times 5}{3} = 10$$. Yes!

D: $$l = 1, b = 1, d = 10$$: $$\frac{1 \times 1}{10} = 0.1 \neq 10$$. No.

E: $$l = 5, b = 2, d = 1$$: $$\frac{5 \times 2}{1} = 10$$. Yes!

Options C and E give a capacitance 10 times the original.

The correct answer is Option D: C and E only.

We need to find the energy density between the plates of a parallel plate capacitor.

The energy stored per unit volume (energy density) in an electric field is:

$$ u = \frac{1}{2}\epsilon_0 E^2 $$

This formula comes from the total energy stored in a capacitor $$U = \frac{1}{2}CV^2$$ divided by the volume between the plates.

For a parallel plate capacitor, the electric field is uniform and given by:

$$ E = \frac{V}{d} $$

where $$V = 20$$ V is the potential difference and $$d = 1\,\mu m = 1 \times 10^{-6}$$ m is the plate separation.

$$ E = \frac{20}{1 \times 10^{-6}} = 2 \times 10^7 \text{ V/m} $$

Using $$\epsilon_0 = 8.85 \times 10^{-12}$$ F/m (permittivity of free space):

$$ u = \frac{1}{2} \times 8.85 \times 10^{-12} \times (2 \times 10^7)^2 $$

$$ = \frac{1}{2} \times 8.85 \times 10^{-12} \times 4 \times 10^{14} $$

$$ = \frac{1}{2} \times 8.85 \times 4 \times 10^{2} $$

$$ = \frac{1}{2} \times 35.4 \times 10^{2} $$

$$ = 17.7 \times 10^{2} \approx 1.77 \times 10^3 \approx 1.8 \times 10^3 \text{ J/m}^3 $$

The correct answer is Option C: $$1.8 \times 10^3$$ J/m$$^3$$.

To express the potential energy stored in a parallel plate capacitor in terms of the electric field $$E$$, plate area $$A$$, separation $$d$$, and permittivity $$\varepsilon_0$$, recall that the general formula for the energy is$$U = \frac{1}{2}CV^2$$where $$C$$ is the capacitance and $$V$$ the potential difference across the plates. For a parallel plate capacitor, the capacitance is given by$$C = \frac{\varepsilon_0 A}{d}$$and the uniform electric field relates to the potential difference via$$V = Ed\,. $$Substituting these expressions into the energy formula leads to$$U = \frac{1}{2}\,\frac{\varepsilon_0 A}{d}\,(Ed)^2 = \frac{1}{2}\varepsilon_0 E^2 A d\,. $$

An alternative derivation begins with the energy density of an electric field,$$u = \frac{1}{2}\varepsilon_0 E^2\,, $$and notes that the volume between the plates is $$A\,d$$. Multiplying gives$$U = u\times(A d) = \frac{1}{2}\varepsilon_0 E^2 A d\,, $$which agrees with the previous result.

Thus, the energy stored in the parallel plate capacitor can be written as$$\frac{1}{2}\varepsilon_0 E^2 A d\,. $$

A parallel-plate capacitor of capacitance $$40\mu F$$ is connected to a 100 V supply. A dielectric of K=2 is inserted while the supply remains connected.

Initial capacitance: $$C_0 = 40 \mu F$$

$$Q_0 = C_0 V = 40 \times 10^{-6} \times 100 = 4 \times 10^{-3} \text{ C} = 4 \text{ mC}$$

$$U_0 = \frac{1}{2}C_0 V^2 = \frac{1}{2} \times 40 \times 10^{-6} \times (100)^2 = 0.2 \text{ J}$$

New capacitance: $$C = KC_0 = 2 \times 40 = 80 \mu F$$

$$Q = CV = 80 \times 10^{-6} \times 100 = 8 \times 10^{-3} \text{ C} = 8 \text{ mC}$$

$$U = \frac{1}{2}CV^2 = \frac{1}{2} \times 80 \times 10^{-6} \times (100)^2 = 0.4 \text{ J}$$

Extra charge: $$\Delta Q = Q - Q_0 = 8 - 4 = 4 \text{ mC}$$

Change in energy: $$\Delta U = U - U_0 = 0.4 - 0.2 = 0.2 \text{ J}$$

The correct answer is Option 1: 4 mC and 0.2 J.

A. no current through resistor
wrong. at the instant of closing, the capacitor behaves like a short circuit, so the circuit is complete. current flows through the resistor and is not zero.

B. maximum current in wires
correct. initially the capacitor has zero voltage across it, so the entire battery voltage drives current through the circuit. hence current is maximum at that instant.

C. potential difference between plates A and B is minimum
correct. since the capacitor is uncharged at the beginning, the voltage across its plates is zero, which is the minimum possible value.

D. charge on capacitor plates is minimum
correct. at the instant of closing, no charge has yet accumulated on the capacitor, so the charge is zero, i.e., minimum.

Let the free charge on the plates of the capacitor be $$Q_f = 5 \times 10^{-6}\, \text{C}$$.

When a dielectric slab is inserted, the molecules of the slab polarise and create two layers of induced (bound) charge. The magnitude of the induced charge on either face of the slab is given as $$Q_{\text{ind}} = 4 \times 10^{-6}\, \text{C}$$.

For a parallel-plate capacitor completely (or almost completely) filled with a dielectric of relative permittivity $$K$$, the relation between the free surface charge density $$\sigma_f$$ and the bound surface charge density $$\sigma_b$$ is derived as follows:

• The electric displacement vector $$\mathbf{D}$$ inside the dielectric is $$\mathbf{D} = \sigma_f \hat{n}$$ (since $$\mathbf{D}$$ originates from free charge only).
• Electric field inside the dielectric: $$E = \frac{\sigma_f}{K \varepsilon_0}$$.
• Polarisation $$P$$ of the dielectric: $$P = \varepsilon_0 \chi_e E = \varepsilon_0 (K-1) E$$, where $$\chi_e = K-1$$.
• Surface bound charge density: $$\sigma_b = P = \varepsilon_0 (K-1) \frac{\sigma_f}{K \varepsilon_0} = \frac{K-1}{K} \, \sigma_f$$.

Therefore the ratio of the magnitudes of bound to free charge is

$$\frac{Q_{\text{ind}}}{Q_f} = \frac{\sigma_b}{\sigma_f} = \frac{K-1}{K}\;.$$

Substitute the given values:

$$\frac{4 \times 10^{-6}}{5 \times 10^{-6}} = \frac{K-1}{K}$$

$$\frac{4}{5} = \frac{K-1}{K}$$

Cross-multiplying:

$$5(K-1) = 4K \quad\Rightarrow\quad 5K - 5 = 4K$$

$$K = 5$$

Hence, the dielectric constant of the slab is $$5$$.

A parallel plate capacitor with C = 2.5 μF produces a displacement current of 0.25 mA. Find the rate of change of voltage.

We use the displacement current formula: $$I_d = C\frac{dV}{dt}$$

Solving for the rate of change of voltage gives $$\frac{dV}{dt} = \frac{I_d}{C} = \frac{0.25 \times 10^{-3}}{2.5 \times 10^{-6}} = \frac{0.25}{2.5} \times 10^3 = 100$$ V/s

The answer is 100.

For a parallel plate capacitor, the capacitance is given by $$C = \frac{\varepsilon_0 A}{d}$$.

The charging current satisfies $$I = C\frac{dV}{dt}$$, so $$C = \frac{I}{dV/dt}$$. Equating this with the capacitance expression yields

$$d = \frac{\varepsilon_0 A}{C} = \frac{\varepsilon_0 A \cdot (dV/dt)}{I}$$.

Here, $$\varepsilon_0 = 9 \times 10^{-12}\text{ F/m}$$, the plate radius is $$r = 10\text{ cm} = 0.1\text{ m}$$ so $$A = \pi r^2 = \pi(0.01)$$, the current is $$I = 0.15\text{ A}$$, and $$\frac{dV}{dt} = 7 \times 10^8\text{ V/s}\,$$.

Substituting these values into the expression for $$d$$ gives

$$d = \frac{9 \times 10^{-12} \times \frac{22}{7} \times 0.01 \times 7 \times 10^8}{0.15}$$

which simplifies to

$$= \frac{9 \times 10^{-12} \times 22 \times 0.01 \times 10^8}{0.15}$$

$$= \frac{9 \times 22 \times 10^{-12+8-2}}{0.15} = \frac{198 \times 10^{-6}}{0.15}$$

$$= 1320 \times 10^{-6} \text{ m} = 1320 \,\mu\text{m}$$

The answer is 1320 μm.

Now the given Circuit / figure can be redrawn as

So here all 4 are in parellel

If capacitances are parellel then effective capacitance $$C_{eff}$$ will be

$$C_{eff}=C_1+C_2+C_3+C_4$$

as Capacitancce values are equal to $$C=16\mu\ F$$

$$C_{eff}=4\left(C\right)$$

$$C_{eff}=4\left(16\mu\ \right)F$$

$$C_{eff}=64\mu\ F$$

At steady state capacitor behaves as open circuit.

It is connected in parallel with the $$10Ω$$ resistor, so voltage across capacitor = voltage across $$10Ω$$.

Current in series circuit:

$$I=\frac{5}{10+15}$$

$$=\frac{5}{25}=0.2A$$

Voltage across $$10Ω$$:

V=IR

$$=0.2\times10=2V$$

Charge on capacitor:

Q=CV

$$8\mu F\times2$$

$$=16μC$$

A bulb and capacitor are in series across an AC supply. When a dielectric is placed between the capacitor plates:

The capacitance increases ($$C' = \kappa C$$ where $$\kappa > 1$$).

The capacitive reactance decreases: $$X_C = \frac{1}{\omega C}$$, so larger C means smaller $$X_C$$.

The total impedance of the series circuit decreases: $$Z = \sqrt{R^2 + X_C^2}$$ (R is the resistance of the bulb).

Since $$Z$$ decreases, the current increases: $$I = \frac{V}{Z}$$.

More current through the bulb means the glow increases.

The correct answer is Option 1: increases.

A point charge $$q = 10^{-6}\mu C = 10^{-12}$$ C is placed at the origin, and we consider the points P $$(\sqrt{3},\sqrt{3})$$ and Q $$(\sqrt{6},0)$$.

The distance of point P from the origin is $$r_P = \sqrt{3+3} = \sqrt{6}$$ m, while the distance of point Q is $$r_Q = \sqrt{6}$$ m.

The electric potential due to a point charge is given by $$V = \frac{kq}{r}$$. Since $$r_P = r_Q = \sqrt{6}$$, it follows that

$$V_P = V_Q = \frac{kq}{\sqrt{6}}$$

Consequently, the potential difference between P and Q is

$$V_P - V_Q = 0$$

The correct answer is Option 3: 0 V.

We need to evaluate two statements about equipotential surfaces and electric fields.

Assertion (A): Work done by electric field on moving a positive charge on an equipotential surface is always zero. This statement is true. The work done by the electric field in moving a charge $$q$$ from point 1 to point 2 is: $$W = q(V_1 - V_2)$$. On an equipotential surface, all points have the same potential, so $$V_1 = V_2$$, which gives: $$W = q(V_1 - V_2) = q \times 0 = 0$$. This holds true regardless of the path taken along the equipotential surface.

Reason (R): Electric lines of forces are always perpendicular to equipotential surfaces. This statement is true. The electric field $$\vec{E}$$ is related to the potential by $$\vec{E} = -\nabla V$$, which means the electric field points in the direction of the steepest decrease in potential. Since the potential is constant along an equipotential surface ($$dV = 0$$), the gradient of $$V$$ (and hence $$\vec{E}$$) must be perpendicular to the surface.

Because the electric field lines are perpendicular to the equipotential surface, the component of the electric force along the surface is zero. Therefore, when a charge moves along the surface, the force and displacement are perpendicular, and the work done ($$W = \vec{F} \cdot \vec{ds} = F \cdot ds \cdot \cos 90° = 0$$) is zero. The perpendicularity directly explains why the work is zero, so (R) is the correct explanation of (A).

The correct answer is Option 4: Both (A) and (R) are correct and (R) is the correct explanation of (A).

Find the dielectric constant of a slab inserted in a parallel plate capacitor.

Plate area $$A = 12$$ cm$$^2$$, separation $$d = 0.6$$ cm, and the original capacitance (air) is $$C_0 = \frac{\varepsilon_0 A}{d}$$.

When a dielectric slab of thickness $$t = 0.6$$ cm fills the gap and one plate is moved by 0.2 cm to maintain the same capacitance, the new separation becomes 0.6 + 0.2 = 0.8 cm. The dielectric slab still has thickness 0.6 cm, and the remaining 0.2 cm is air.

The capacitance with the dielectric and air gap is $$C = \frac{\varepsilon_0 A}{\frac{t}{\kappa} + d_{\text{air}}} = \frac{\varepsilon_0 A}{\frac{0.6}{\kappa} + 0.2}$$.

Setting $$C = C_0$$ gives $$\frac{\varepsilon_0 A}{\frac{0.6}{\kappa} + 0.2} = \frac{\varepsilon_0 A}{0.6}$$, so $$\frac{0.6}{\kappa} + 0.2 = 0.6 \implies \frac{0.6}{\kappa} = 0.4 \implies \kappa = \frac{0.6}{0.4} = 1.5$$.

The correct answer is Option (4): 1.50.

step 1: understand geometry

the bottom plate is flat (area A)
the top plate has 3 steps, each of area A/3

their separations from bottom plate are:

• lowest step → distance = d
• middle step → distance = 2d
• top step → distance = 3d

step 2: treat as capacitors

each step forms a capacitor with the bottom plate

all are connected in parallel (same two plates → same potential difference)

step 3: capacitance of each

$$C_1=\frac{\varepsilon_0(A/3)}{d}$$

$$C_2=\frac{\varepsilon_0(A/3)}{2d}$$

$$C_3=\frac{\varepsilon_0(A/3)}{3d}$$

step 4: total capacitance (parallel)

$$C=C_1+C_2+C_3$$

$$=\frac{\varepsilon_0A}{3d}\left(1+\frac{1}{2}+\frac{1}{3}\right)$$

$$=\frac{\varepsilon_0A}{3d}\cdot\frac{11}{6}$$

step 1: steady state idea

capacitors act as open circuits
so only resistors + galvanometer (2 Ω) matter

step 2: find equivalent resistance

the network reduces to:

4 Ω + 2 Ω + 6 Ω = 12 Ω

step 3: current

$$I=\frac{6}{12}=0.5\text{ A}$$

step 4: potential distribution

drop across 4 Ω:

$$V=0.5\times4=2\text{ V}$$

drop across 2 Ω:

$$V=0.5\times2=1\text{ V}$$

drop across 6 Ω:

$$V=0.5\times6=3\text{ V}$$

step 5: capacitor voltages

from node potentials:

• voltage across $$C_1$$​ = 3 V
• voltage across $$C_2$$ = 4 V

step 6: charges

$$q_2=C_2V_2=6\mu F\times4=24\mu C$$

$$q_1=C_1V_1=4\mu F\times3=12\mu C$$

Let the capacitance of each capacitor be $$C$$. Capacitor 1 is charged to potential $$V$$ and capacitor 2 is charged to potential $$2V$$.

Charge on capacitor 1: $$Q_1 = C \times V = CV$$.
Charge on capacitor 2: $$Q_2 = C \times 2V = 2CV$$.

Initial energy of capacitor 1: $$U_1 = \frac12 C V^2$$.
Initial energy of capacitor 2: $$U_2 = \frac12 C (2V)^2 = \frac12 C \times 4V^2 = 2CV^2$$.

Total initial energy, $$U_i = U_1 + U_2 = \frac12 C V^2 + 2CV^2 = \frac{1}{2}CV^2 + 2CV^2 = \frac{5}{2}CV^2$$. $$-(1)$$

After connecting the negative ends together and then the positive ends, the total charge on the combined positive terminal is:
$$Q_{\text{total}} = Q_1 + Q_2 = CV + 2CV = 3CV$$.

Equivalent capacitance of two identical capacitors in parallel is $$C_{\text{eq}} = C + C = 2C$$.

Final common potential $$V_f$$ is given by the charge‐capacitance relation:
$$V_f = \frac{Q_{\text{total}}}{C_{\text{eq}}} = \frac{3CV}{2C} = \frac{3V}{2}$$. $$-(2)$$

Final energy of the system:
$$U_f = \frac12 C_{\text{eq}} V_f^2 = \frac12 \times 2C \times \Bigl(\frac{3V}{2}\Bigr)^2 = C \times \frac{9V^2}{4} = \frac{9}{4}CV^2$$.

Decrease in energy:
$$\Delta U = U_i - U_f = \frac{5}{2}CV^2 - \frac{9}{4}CV^2 = \frac{10}{4}CV^2 - \frac{9}{4}CV^2 = \frac{1}{4}CV^2$$.

Therefore, the decrease in energy of the combined system is $$\frac{1}{4}CV^2$$, which corresponds to Option A.

An LCR circuit at resonance has its resistance halved. Find the new current amplitude.

At resonance in an LCR circuit, the inductive reactance equals the capacitive reactance: $$X_L = X_C$$. These cancel out, so the impedance equals just the resistance:

$$Z = R \quad \text{(at resonance)}$$

$$I = \frac{V}{Z} = \frac{V}{R}$$

The resonance condition depends on $$L$$ and $$C$$ (not $$R$$), so the circuit remains at resonance. With $$R' = R/2$$:

$$I' = \frac{V}{R'} = \frac{V}{R/2} = \frac{2V}{R} = 2I$$

The current amplitude doubles.

The correct answer is Option (4): double.

The magnetic field inside a solenoid is along its axis. An electron moving along the axis has velocity parallel to the magnetic field.

The magnetic force on a charged particle is: $$\vec{F} = q\vec{v} \times \vec{B}$$

Since $$\vec{v}$$ is parallel to $$\vec{B}$$, the cross product $$\vec{v} \times \vec{B} = 0$$.

Therefore, no magnetic force acts on the electron, and it continues to move with uniform velocity along the axis.

The correct answer is Option 1.

Resonance: f = 1/(2π√(LC)). For same f with 4C: L_new = L/4. Reduction = L - L/4 = 3L/4.

The correct answer is Option 3: reduced by 3L/4.

The given values are the amplitude $$V_0 = 40$$ V, the frequency $$f = 4$$ kHz = $$4000$$ Hz, and the capacitance $$C = 12 \mu$$F $$= 12 \times 10^{-6}$$ F.

The displacement current through a capacitor equals the conduction current, so

$$ i = C\frac{dV}{dt}$$

For $$V = V_0 \sin(\omega t)$$, the maximum current is

$$ i_{max} = C \omega V_0 = C \times 2\pi f \times V_0$$

$$ i_{max} = 12 \times 10^{-6} \times 2\pi \times 4000 \times 40$$

$$ i_{max} = 12 \times 10^{-6} \times 8\pi \times 10^3 \times 40$$

$$ = 12 \times 8\pi \times 40 \times 10^{-3}$$

$$ = 3840\pi \times 10^{-3}$$

$$ = 3.84\pi \approx 3.84 \times 3.14 \approx 12.06 \text{ A}$$

Thus the maximum displacement current is nearly $$12$$ A.

The correct answer is Option (2): 12 A.

Three capacitors are given as $$C_1 = 25\mu$$F, $$C_2 = 30\mu$$F, $$C_3 = 45\mu$$F, with a supply voltage $$V = 100$$ V.

When these capacitors are connected in parallel, the equivalent capacitance is

$$ C_p = 25 + 30 + 45 = 100 \mu\text{F} $$

and the energy stored in this configuration is

$$ E = \frac{1}{2}C_p V^2 = \frac{1}{2} \times 100 \times 10^{-6} \times 10000 = 0.5 \text{ J} $$

In the case of series connection, the equivalent capacitance satisfies

$$ \frac{1}{C_s} = \frac{1}{25} + \frac{1}{30} + \frac{1}{45} $$

LCM of 25, 30, 45 = 450:

$$ \frac{1}{C_s} = \frac{18 + 15 + 10}{450} = \frac{43}{450} $$

$$ C_s = \frac{450}{43} \mu\text{F} $$

Thus the energy stored in series is

$$ E_s = \frac{1}{2}C_s V^2 = \frac{1}{2} \times \frac{450}{43} \times 10^{-6} \times 10000 = \frac{450}{43} \times 5 \times 10^{-3} $$

Finally, the ratio of the energy stored in series to that in parallel is

$$ \frac{E_s}{E} = \frac{C_s}{C_p} = \frac{450/43}{100} = \frac{450}{4300} = \frac{9}{86} $$

Therefore, $$E_s = \frac{9}{86}E$$, which implies that $$x = 86\,$$.

The energy stored in a capacitor is given by: $$E = \frac{1}{2}CV^2$$.

Initially, for the first capacitor, $$E_1 = \frac{1}{2}CV^2 = E$$ (given), so $$CV^2 = 2E$$. For the second capacitor, $$E_2 = \frac{1}{2}(2C)(2V)^2 = \frac{1}{2} \times 2C \times 4V^2 = 4CV^2 = 8E$$. Hence the total initial energy is $$E_i = E + 8E = 9E$$.

After connecting the capacitors, charge redistributes while the total charge remains conserved: $$Q_{total} = CV + 2C(2V) = CV + 4CV = 5CV$$. The total capacitance becomes $$C_{total} = C + 2C = 3C$$, so the common potential is $$V_f = \frac{Q_{total}}{C_{total}} = \frac{5CV}{3C} = \frac{5V}{3}$$.

The final energy stored is $$E_f = \frac{1}{2}(3C)\left(\frac{5V}{3}\right)^2 = \frac{1}{2} \times 3C \times \frac{25V^2}{9} = \frac{25CV^2}{6} = \frac{25 \times 2E}{6} = \frac{25E}{3}$$.

The energy loss is given by $$\Delta E = E_i - E_f = 9E - \frac{25E}{3} = \frac{27E - 25E}{3} = \frac{2E}{3} = \frac{x}{3}E$$, which yields $$x = 2$$.

Initial capacitance of the parallel-plate capacitor
$$C = 12.5\ \text{pF} = 12.5 \times 10^{-12}\ \text{F}$$

The plates are first charged by a battery to a potential difference
$$V = 12.0\ \text{V}$$

Because the battery is disconnected before inserting the dielectric, the charge $$Q$$ on the plates remains constant.

Step 1 : Calculate the initial charge
Using $$Q = C V$$,

$$Q = \left(12.5 \times 10^{-12}\right)\!\text{F}\;\left(12\ \text{V}\right) = 150 \times 10^{-12}\ \text{C}$$

Step 2 : Initial electrostatic energy
For a charged capacitor, the energy stored is
$$U = \frac{1}{2} C V^{2}$$

$$U_i = \tfrac{1}{2}\left(12.5 \times 10^{-12}\right)\!\text{F}\;\left(12\ \text{V}\right)^{2}$$
$$U_i = \tfrac{1}{2}\left(12.5 \times 10^{-12}\right)\!\left(144\right)$$
$$U_i = 9.0 \times 10^{-10}\ \text{J}$$

Step 3 : Capacitance after inserting the dielectric
Relative permittivity of the slab $${\epsilon_r} = 6$$. New capacitance
$$C' = \epsilon_r\,C = 6 \times 12.5\ \text{pF} = 75\ \text{pF}$$
$$C' = 75 \times 10^{-12}\ \text{F}$$

Step 4 : Final electrostatic energy
Charge stays the same, so use $$U = \dfrac{Q^{2}}{2C}$$:

$$U_f = \frac{\left(150 \times 10^{-12}\ \text{C}\right)^{2}}{2 \left(75 \times 10^{-12}\ \text{F}\right)}$$
$$U_f = \frac{2.25 \times 10^{-20}}{150 \times 10^{-12}}$$
$$U_f = 1.5 \times 10^{-10}\ \text{J}$$

Step 5 : Change in potential energy
$$\Delta U = U_f - U_i = 1.5 \times 10^{-10}\ \text{J} \;-\; 9.0 \times 10^{-10}\ \text{J}$$
$$\Delta U = -7.5 \times 10^{-10}\ \text{J}$$

The negative sign shows that the energy decreases. Expressing the magnitude in units of $$10^{-12}\ \text{J}$$:

$$|\Delta U| = 7.5 \times 10^{-10}\ \text{J} = 750 \times 10^{-12}\ \text{J}$$

Hence, the change in potential energy is $$\mathbf{750} \times 10^{-12}\ \text{J}$$ (decrease).

A parallel plate capacitor with plate separation $$d = 5$$ mm draws 25% more charge when a dielectric sheet of thickness $$t = 2$$ mm is introduced (with the battery connected). We need to find the dielectric constant $$K$$.

We start by writing the original capacitance.

$$ C_0 = \frac{\epsilon_0 A}{d} = \frac{\epsilon_0 A}{5} $$

(where the separation is in mm for now; the units will cancel)

Next, when a dielectric of thickness $$t$$ and dielectric constant $$K$$ is inserted, with the remaining gap being air, the effective capacitance is:

$$ C = \frac{\epsilon_0 A}{d - t + \frac{t}{K}} = \frac{\epsilon_0 A}{5 - 2 + \frac{2}{K}} = \frac{\epsilon_0 A}{3 + \frac{2}{K}} $$

This formula comes from treating the system as two capacitors in series (dielectric region and air region) and simplifying.

Since the battery remains connected, the voltage is constant and the charge follows $$Q = CV$$, so a 25% increase in charge means:

$$ C = 1.25 \times C_0 $$

$$ \frac{\epsilon_0 A}{3 + \frac{2}{K}} = 1.25 \times \frac{\epsilon_0 A}{5} $$

Cancelling $$\epsilon_0 A$$:

$$ \frac{1}{3 + \frac{2}{K}} = \frac{1.25}{5} = \frac{1}{4} $$

$$ 3 + \frac{2}{K} = 4 $$

$$ \frac{2}{K} = 1 $$

$$ K = 2 $$

The dielectric constant is $$\boxed{2}$$.

For a half ring of radius $$R$$ with linear charge density $$\lambda$$, the total charge is $$Q = \lambda \pi R$$. Since every element is at distance $$R$$ from the centre, the potential is:

$$V = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{R} = \frac{1}{4\pi\epsilon_0} \cdot \frac{\lambda \pi R}{R} = k\lambda\pi$$

where $$k = 9 \times 10^9$$ N m$$^2$$/C$$^2$$.

The linear charge density is $$\lambda = 4$$ nC/m = $$4 \times 10^{-9}$$ C/m.

Substituting these values gives

$$V = 9 \times 10^9 \times 4 \times 10^{-9} \times \pi = 36\pi \text{ V}$$

Hence, $$x = \boxed{36}$$.

step 1: positions of charges

On the x-axis:

• +2q at −2l
• −q at −l
• +q at +l
• −2q at +2l

step 2: net charge

$$2q-q+q-2q=0$$

So monopole term = 0

step 3: dipole moment

Take dipole moment about O:

$$p=\sum_{ }^{ }q_ix_i$$

$$=(2q)(-2l)+(-q)(-l)+(q)(l)+(-2q)(2l)$$

So,

$$p=-6ql$$

(direction along negative x-axis)

step 4: potential at point P

For dipole:

$$V=\frac{1}{4\pi\epsilon_0}\frac{p\cos\theta}{r^2}$$

Here $$\theta=60^{\circ}$$

$$\cos60^{\circ}=\frac{1}{2}$$

$$V=9\times10^9\cdot\frac{-6ql\cdot(1/2)}{r^2}$$

$$=9\times10^9\cdot\frac{-3ql}{r^2}$$

The capacitor ($$C = 1.5\mu$$F) discharges through a wire of resistance $$R$$ and the electric field drops to $$\frac{1}{3}$$ of its initial value in $$6.6\mu$$s.

Because the electric field is proportional to the voltage across the capacitor, $$E \propto V$$, the voltage during discharge follows

$$ V = V_0 e^{-t/RC} $$

When $$V = V_0/3$$, we have

$$ \frac{1}{3} = e^{-t/RC} $$

$$ \ln 3 = \frac{t}{RC} $$

$$ RC = \frac{t}{\ln 3} $$

Substituting $$t = 6.6 \mu$$s and $$\ln 3 = \log_e 3 = 1.1$$ gives

$$ RC = \frac{6.6 \times 10^{-6}}{1.1} = 6 \times 10^{-6} \text{ s} $$

$$ R = \frac{6 \times 10^{-6}}{1.5 \times 10^{-6}} = 4 \text{ }\Omega $$

Hence, the answer is 4.

At steady state (DC), the capacitor branch carries no current. So the 5 Ω branch is open.

step 1: reduce circuit

Only active loop:

battery (10 V) → 4 Ω → 6 Ω → back

Total resistance:

R=4+6=10 Ω

step 2: current in circuit

$$I=\frac{10}{10}=1A$$

step 3: voltage drop

Voltage drop across 4 Ω:

$$V_4=1\times4=4V$$

Voltage drop across 6 Ω:

$$V_6=6V$$

step 4: capacitor voltage

Capacitor is connected between left node and right node (same as ends of 6 Ω)

So voltage across capacitor = voltage across 6 Ω = 6 V

step 5: charge

$$Q=CV=10μF\times6=10\times10^{-6}\times6=60μC$$

step 1: steady state
capacitor is open → no current through AB

So the diamond reduces to two parallel branches between top and bottom nodes

• right branch: 6Ω + 4Ω = 10Ω
• left branch: 1Ω + 2Ω = 3Ω

step 2: equivalent of parallel

$$R_{eq}=10∥3=\frac{30}{13}Ω$$

Total current from 10 V source:

$$I=\frac{10}{30/13}=\frac{130}{30}=\frac{13}{3}\text{ A}$$

step 3: current division

current in left branch (1Ω + 2Ω):

$$I_L=I\cdot\frac{10}{10+3}=\frac{13}{3}\cdot\frac{10}{13}=\frac{10}{3}\text{ A}$$

current in right branch:

IR=133−103=1 AI_R = \frac{13}{3} - \frac{10}{3} = 1 \text{ A}IR​=313​−310​=1 A

step 4: find potentials

take bottom node = 0 V, top = 10 V

Point A (on left branch):

drop across 1Ω:

$$V_A=10-\frac{10}{3}=\frac{20}{3}$$

Point B (on right branch):

drop across 6Ω:

$$V_B=10-6(1)=4$$

step 5: capacitor voltage

$$V_{AB}=\frac{20}{3}-4=\frac{8}{3}$$

step 6: charge

$$Q=CV=150\times\frac{8}{3}=400μC$$

We need to find the RMS value of current when an alternating EMF $$E = 110\sqrt{2} \sin 100t$$ volts is applied to a capacitor of $$2 \, \mu F$$. Since the general form of an AC voltage is $$E = E_0 \sin \omega t$$, comparing with the given expression yields the peak EMF $$E_0 = 110\sqrt{2}$$ V and the angular frequency $$\omega = 100$$ rad/s.

Substituting the peak EMF into the RMS relation gives the RMS voltage. From the equation $$E_{rms} = \frac{E_0}{\sqrt{2}} = \frac{110\sqrt{2}}{\sqrt{2}} = 110 \text{ V}$$, we obtain an RMS voltage of 110 V.

Next, the capacitive reactance is given by $$X_C = \frac{1}{\omega C}$$. Substituting $$\omega = 100$$ rad/s and $$C = 2 \, \mu F = 2 \times 10^{-6}$$ F into this expression, we find $$X_C = \frac{1}{100 \times 2 \times 10^{-6}} = \frac{1}{2 \times 10^{-4}} = 5000 \, \Omega$$.

Then the RMS current in a purely capacitive circuit follows from $$I_{rms} = \frac{E_{rms}}{X_C} = \frac{110}{5000} = 0.022 \text{ A} = 22 \text{ mA}$$. Therefore, the current is 22 mA.

The answer is 22 mA.

At resonance in series LCR,

$$f=\frac{1}{2\pi\sqrt{LC}}$$

Given,

$$L=100mH=0.1H$$

$$C=2.5\text{ nF}=2.5\times10^{-9}F$$

So,

$$f=\frac{1}{2\pi\sqrt{(0.1)(2.5\times10^{-9})}}$$

$$=\frac{1}{2\pi\sqrt{2.5\times10^{-10}}}$$

$$\sqrt{2.5\times10^{-10}}=\frac{5\times10^{-5}}{\sqrt{10}}$$

Thus,

$$f=\frac{1}{2\pi\cdot\frac{5\times10^{-5}}{\sqrt{10}}}$$

Given

$$\pi^2=10\Rightarrow\sqrt{10}=\pi$$

Hence,

$$f=\frac{1}{2\pi\cdot\frac{5\times10^{-5}}{\pi}}$$

$$=\frac{1}{10\times10^{-5}}$$

$$=10^4\text{ Hz}$$

$$=10\times10^3\text{ Hz}$$

When A and B are shorted, they become the same node.

step 1: identify final steady state

In DC steady state, capacitors act as open circuits.
So no current flows through capacitors, but they can hold charge.

step 2: reduce the circuit

Now A and B are same node.

So:

• top node = 9 V
• bottom node = 0 V (ground)
• A = B is some intermediate node

Between top and A:

6 Ω

Between A and bottom:

3 Ω

So it becomes a simple voltage divider.

step 3: find voltage at node A (= B)

$$V_A=\frac{3}{6+3}\times9=\frac{3}{9}\times9=3\text{ V}$$

step 4: voltage across 6 μF capacitor

Top plate is at 9 V, bottom plate at node B = 3 V

$$V_C=9-3=6\text{ V}$$

step 5: charge

$$Q=CV=6\mu F\times6=36\mu C$$

The four metal sheets $$S_1,S_2,S_3,S_4$$ are identical squares of area $$A=a^{2}$$ kept successively at equal separations $$d\,(d\ll a)$$.
Between any two adjacent sheets the capacitance is

$$C_0=\frac{\varepsilon_0A}{d}\quad -(1)$$

Thus the stack contains three individual capacitors:
$$S_1\!-\!S_2: C_{12}=C_0,\; S_2\!-\!S_3: C_{23}=C_0,\; S_3\!-\!S_4: C_{34}=C_0$$.

Required: capacitance between $$S_1$$ and $$S_4$$ with $$S_2,S_3$$ left isolated (not connected).
The three capacitors are then in series:

$$\frac{1}{C_{\text{eq}}}= \frac{1}{C_{12}}+\frac{1}{C_{23}}+\frac{1}{C_{34}} =\frac{1}{C_0}+\frac{1}{C_0}+\frac{1}{C_0}= \frac{3}{C_0}$$

$$\Rightarrow\; C_{\text{eq}}=\frac{C_0}{3}$$

So (P) corresponds to value $$\mathbf{C_0/3}$$, i.e. List-II (3).

Required: capacitance between $$S_1$$ and $$S_4$$ with $$S_2$$ shorted to $$S_3$$ (equipotential).
Because $$S_2$$ and $$S_3$$ form one common conductor, the region between them is at a single potential and contributes no electric field. The stack now looks like two capacitors in series:

$$S_1\;{\large\|}\;[S_2\!+\!S_3]\;{\large\|}\;S_4$$

Each of the two gaps still has capacitance $$C_0$$, hence

$$C_{\text{eq}}=\frac{C_0 \times C_0}{C_0+C_0}= \frac{C_0}{2}$$

Thus (Q) gives $$\mathbf{C_0/2}$$, that is List-II (2).

Required: capacitance between $$S_1$$ and $$S_3$$ when $$S_2$$ is shorted to $$S_4$$.
Define the three electrical nodes:

Node A: $$S_1$$ (one terminal)
Node B: $$S_3$$ (other terminal)
Node C: $$S_2$$ shorted to $$S_4$$ (floating internal conductor)

Network of capacitors:

$$S_1\!\!-\!\!S_2 : C_{AC}=C_0,\quad S_2\!\!-\!\!S_3 : C_{CB}=C_0,\quad S_3\!\!-\!\!S_4 : C_{CB}'=C_0$$

Capacitances between the same pair of nodes add, so between nodes B and C we have

$$C_{BC}=C_{CB}+C_{CB}'=2C_0$$

Now A-C and C-B are in series. The equivalent between A and B is therefore

$$C_{\text{eq}}=\frac{C_{AC}\,C_{BC}}{C_{AC}+C_{BC}} =\frac{C_0\,(2C_0)}{C_0+2C_0} =\frac{2C_0}{3}$$

Hence (R) gives $$\mathbf{2C_0/3}$$, i.e. List-II (4).

Required: capacitance between $$S_1$$ and $$S_2$$ when $$S_3$$ is shorted to $$S_1$$ and $$S_2$$ is shorted to $$S_4$$.
Thus we have two final nodes:

Node A: $$S_1$$ and $$S_3$$ (same potential)
Node B: $$S_2$$ and $$S_4$$ (same potential)

All three original capacitors now connect directly between these two nodes:

$$C_{12}=C_0,\; C_{23}=C_0,\; C_{34}=C_0$$

They are in parallel, so

$$C_{\text{eq}}=C_{12}+C_{23}+C_{34}=3C_0$$

Therefore (S) corresponds to $$\mathbf{3C_0}$$, which is List-II (1).

Collecting the four matches:

P → 3,  Q → 2,  R → 4,  S → 1

The option that lists this combination is Option C.

Parallel plate capacitor: plate area $$A = 40 \text{ cm}^2 = 40 \times 10^{-4} \text{ m}^2$$, plate separation $$d = 2 \text{ mm}$$.

Dielectric: thickness $$d_1 = 1 \text{ mm}$$, dielectric constant $$K = 5$$. Remaining air gap: $$d_2 = 1 \text{ mm}$$.

This is equivalent to two capacitors in series:

$$C_1 = \frac{K\varepsilon_0 A}{d_1} = \frac{5 \times \varepsilon_0 \times 40 \times 10^{-4}}{1 \times 10^{-3}} = 20\varepsilon_0 \text{ F}$$

$$C_2 = \frac{\varepsilon_0 A}{d_2} = \frac{\varepsilon_0 \times 40 \times 10^{-4}}{1 \times 10^{-3}} = 4\varepsilon_0 \text{ F}$$

For series combination:

$$\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{1}{20\varepsilon_0} + \frac{1}{4\varepsilon_0} = \frac{1 + 5}{20\varepsilon_0} = \frac{6}{20\varepsilon_0} = \frac{3}{10\varepsilon_0}$$

$$C = \frac{10\varepsilon_0}{3} \text{ F}$$

The answer is Option C: $$\frac{10}{3}\varepsilon_0$$ F.

Assertion A: Two metallic spheres are charged to the same potential. One is hollow and another is solid, both with the same radii. The solid sphere will have lower charge than the hollow one.

Reason R: Capacitance of metallic spheres depends on the radii of the spheres.

Analysis of Assertion A:

For a conducting sphere (whether solid or hollow), charge resides entirely on the outer surface. The potential of a sphere of radius $$R$$ carrying charge $$Q$$ is:

$$V = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{R}$$

The capacitance is $$C = 4\pi\epsilon_0 R$$, which depends only on the radius $$R$$.

Since both spheres have the same radius $$R$$ and are charged to the same potential $$V$$:

$$Q = CV = 4\pi\epsilon_0 R \cdot V$$

Both spheres carry the same charge. Therefore, the assertion that the solid sphere has lower charge is false.

Analysis of Reason R:

The capacitance of a metallic sphere is $$C = 4\pi\epsilon_0 R$$, which indeed depends only on the radius. This statement is true.

Conclusion: A is false but R is true.

The answer is Option A: A is false but R is true.

We need to find the ratio $$\frac{C_2}{C_1}$$ when a metal sheet of thickness $$\frac{2d}{3}$$ is introduced between the plates of a parallel plate capacitor with plate separation $$d$$.

We begin by noting that for a parallel plate capacitor with air as the medium, the capacitance without the metal sheet is given by $$C_1 = \frac{\varepsilon_0 A}{d}$$, where $$\varepsilon_0$$ is the permittivity of free space, $$A$$ is the area of the plates, and $$d$$ is the distance between the plates.

Next, we recall that a metal (conductor) is a perfect conductor, which means the electric field inside the metal is zero. When a metal sheet of thickness $$t$$ is placed between the plates of a capacitor, it effectively reduces the gap over which the electric field exists. The metal sheet acts as an equipotential region with no field inside it, so the effective distance over which the electric field acts is the total gap minus the thickness of the metal sheet.

This effective distance is given by
$$d_{eff} = d - t$$.

Substituting $$t = \frac{2d}{3}$$ into this expression yields
$$d_{eff} = d - \frac{2d}{3} = \frac{3d - 2d}{3} = \frac{d}{3}$$.

Therefore, the new capacitance becomes
$$C_2 = \frac{\varepsilon_0 A}{d_{eff}} = \frac{\varepsilon_0 A}{d/3} = \frac{3\varepsilon_0 A}{d} = 3C_1$$.

Finally, taking the ratio yields
$$\frac{C_2}{C_1} = \frac{3C_1}{C_1} = 3$$, so that $$C_2 : C_1 = 3 : 1$$.

Hence, the correct answer is Option 1: 3 : 1.

We have a capacitor of capacitance $$C = 2$$ F initially charged to potential $$V$$.

The initial energy is:

$$E_1 = \frac{1}{2}CV^2 = \frac{1}{2}(2)V^2 = V^2$$

Now, when this capacitor is connected to an identical uncharged capacitor in parallel, the total capacitance becomes $$C_{total} = 2 + 2 = 4$$ F. By conservation of charge, $$Q = CV = 2V$$, so the new voltage across the combination is:

$$V' = \frac{Q}{C_{total}} = \frac{2V}{4} = \frac{V}{2}$$

The energy of the combination is:

$$E_2 = \frac{1}{2}C_{total}(V')^2 = \frac{1}{2}(4)\left(\frac{V}{2}\right)^2 = \frac{1}{2}(4)\frac{V^2}{4} = \frac{V^2}{2}$$

So the ratio is:

$$\frac{E_2}{E_1} = \frac{V^2/2}{V^2} = \frac{1}{2}$$

Hence, $$E_2 : E_1 = 1 : 2$$.

We have a point charge $$q = 5 \times 10^{-9}$$ C, potential at P is $$V = 50$$ V, and $$\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9$$ N m$$^2$$ C$$^{-2}$$.

The electric potential due to a point charge is:

$$V = \frac{1}{4\pi\varepsilon_0} \cdot \frac{q}{r}$$

Solving for $$r$$:

$$r = \frac{1}{4\pi\varepsilon_0} \cdot \frac{q}{V}$$

$$r = 9 \times 10^9 \times \frac{5 \times 10^{-9}}{50}$$

$$r = 9 \times 10^9 \times 10^{-10}$$

$$r = 0.9 \text{ m} = 90 \text{ cm}$$

So, the distance of P from the point charge is 90 cm. Hence, the correct answer is Option 4.

For a uniformly charged thin spherical shell of radius R, electric potential varies with distance r from the centre as follows:

For r≤R (inside and on the shell)

Electric field inside a conducting shell is zero:

E=0

Since

$$E=-\frac{dV}{dr}$$

zero electric field means potential is constant throughout the interior.

Its value equals the surface potential:

$$V=\frac{1}{4\pi\varepsilon_0}\frac{Q}{R}$$

So for all

0≤r≤R

potential remains constant and maximum.

For r>R (outside the shell)

The shell behaves like a point charge concentrated at the centre:

$$V=\frac{1}{4\pi\varepsilon_0}\frac{Q}{r}$$

Thus

$$V\propto\frac{1}{r}$$

and decreases with distance.

Graphical variation:

• From r=0 to r=R: horizontal straight line (constant potential)
• At r=R: potential is continuous,

$$V=\frac{Q}{4\pi\varepsilon_0R}$$

• For r>R: a decreasing $$\frac{1}{r}$$ curve approaching zero as

$$r\longrightarrow \infty$$

So the graph is a flat line inside the shell followed by a falling hyperbolic curve outside.

Potential due to a charged arc at its centre is

$$V=\frac{1}{4\pi\varepsilon_0}\int\frac{dq}{R}$$

Since RRR is constant for an arc,

$$V=\frac{1}{4\pi\varepsilon_0}\frac{Q}{R}$$

For a half ring of radius R, length is

$$\pi R$$

So charge on it is

$$\left(Q=\lambda(\pi R\right)$$

Thus potential due to one half ring:

$$V=\frac{1}{4\pi\varepsilon_0}\frac{\lambda\pi R}{R}$$

$$=\frac{\lambda\pi}{4\pi\varepsilon_0}$$​

$$=\frac{\lambda}{4\varepsilon_0}$$

Interesting result: it is independent of radius.

So each half ring contributes

$$\frac{\lambda}{4\varepsilon_0}​$$

For two concentric half rings, potentials add:

$$V=\frac{\lambda}{4\varepsilon_0}+\frac{\lambda}{4\varepsilon_0}$$

$$V=\frac{\lambda}{2\varepsilon_0}$$

There are four capacitors each of capacitance C.

Notice the two vertical capacitors in the middle are short-circuited.

Why?

Their top plates are connected to the top wire, and bottom plates to the bottom wire — but those top and bottom wires are already directly joined together at the right side by a conducting wire.

So both vertical capacitors have zero potential difference across them:

V=0

They contribute nothing.

Only two capacitors remain:

• top-left capacitor C
• bottom-left capacitor C

Both are connected directly between the left terminal and right terminal.

So they are in parallel.

Equivalent capacitance:

$$C_{eq}=C+C$$

=2C

Two metallic spheres of radii $$R$$ and $$2R$$ with same surface charge density $$\sigma$$ are connected by a wire.

$$Q_1 = \sigma \cdot 4\pi R^2$$

$$Q_2 = \sigma \cdot 4\pi(2R)^2 = 16\sigma\pi R^2$$

Total charge: $$Q = Q_1 + Q_2 = 4\sigma\pi R^2 + 16\sigma\pi R^2 = 20\sigma\pi R^2$$

$$\frac{Q_1'}{4\pi\epsilon_0 R} = \frac{Q_2'}{4\pi\epsilon_0 \cdot 2R}$$

$$\frac{Q_1'}{R} = \frac{Q_2'}{2R} \implies Q_2' = 2Q_1'$$

$$Q_1' + Q_2' = 20\sigma\pi R^2$$

$$Q_1' + 2Q_1' = 20\sigma\pi R^2$$

$$Q_1' = \frac{20\sigma\pi R^2}{3}, \quad Q_2' = \frac{40\sigma\pi R^2}{3}$$

$$\sigma' = \frac{Q_2'}{4\pi(2R)^2} = \frac{40\sigma\pi R^2}{3 \times 16\pi R^2} = \frac{40\sigma}{48} = \frac{5\sigma}{6}$$

$$\frac{\sigma'}{\sigma} = \frac{5}{6}$$

The correct answer is Option 4: $$\frac{5}{6}$$.

For a charged spherical conductor:

• For

r<R

electric field inside conductor is zero, so potential remains constant:

$$V=\frac{1}{4\pi\varepsilon_0}\frac{Q}{R}$$

Thus graph is a horizontal line from centre to surface.

• For

r>R

potential varies as

$$V=\frac{1}{4\pi\varepsilon_0}\frac{Q}{r}$$

so it falls as a 1/r curve.

• At

r=R

potential is continuous.

So correct graph is:

flat inside, then decreasing hyperbola outside.

The two conducting walls act as parallel-plate capacitor plates.
Plate area: $$A = 50\ \text{cm} \times 50\ \text{cm} = 0.50\ \text{m} \times 0.50\ \text{m} = 0.25\ \text{m}^2$$

Plate separation (distance between the conducting walls): $$d = 5\ \text{cm} = 0.05\ \text{m}$$

The container is filled at a uniform rate of $$250\ \text{cm}^3\ \text{s}^{-1}$$.
In $$10\ \text{s}$$ the liquid volume is
$$V = 250 \times 10 = 2500\ \text{cm}^3 = 2.5 \times 10^{-3}\ \text{m}^3$$

Cross-sectional base area of the container (same as the overlap area of plates in the vertical cross-section):
$$A_{\text{base}} = 50\ \text{cm} \times 5\ \text{cm} = 0.50\ \text{m} \times 0.05\ \text{m} = 0.025\ \text{m}^2$$

Hence the height of the liquid column after 10 s is
$$h = \frac{V}{A_{\text{base}}} = \frac{2.5 \times 10^{-3}}{0.025} = 0.10\ \text{m} = 10\ \text{cm}$$

The plates therefore have two regions:
• Bottom region (height $$h = 0.10\ \text{m}$$) filled with liquid of dielectric constant $$k = 3$$.
• Top region (height $$50\ \text{cm} - 10\ \text{cm} = 40\ \text{cm} = 0.40\ \text{m}$$) filled with air (dielectric constant $$1$$).

Areas of these two regions on each plate:
Bottom (with dielectric): $$A_1 = 0.50\ \text{m} \times 0.10\ \text{m} = 0.05\ \text{m}^2$$
Top (with air): $$A_2 = 0.50\ \text{m} \times 0.40\ \text{m} = 0.20\ \text{m}^2$$

Since both regions experience the same potential difference, the two parts behave like capacitors in parallel. The equivalent capacitance is
$$C = \frac{\epsilon_0}{d}\,\bigl(k A_1 + A_2\bigr)$$

Substituting the values:
$$C = \frac{9 \times 10^{-12}}{0.05} \Bigl(3 \times 0.05 + 0.20\Bigr)$$
$$C = (9 \times 10^{-12} \times 20)\,(0.15 + 0.20)$$
$$C = 1.8 \times 10^{-10} \times 0.35 = 6.3 \times 10^{-11}\ \text{F}$$

Convert to picofarads: $$6.3 \times 10^{-11}\ \text{F} = 63\ \text{pF}$$

Therefore, the capacitance of the container after 10 s is
Option B which is: 63 pF

A 900 μF capacitor initially charged to 100 V is connected in parallel to an identical uncharged capacitor. The initial energy stored in the first capacitor is $$U_i = \frac{1}{2}CV^2 = \frac{1}{2} \times 900 \times 10^{-6} \times (100)^2 = 4.5$$ J.

After connection, the total charge is $$Q = CV = 900 \times 10^{-6} \times 100 = 0.09$$ C and the combined capacitance is $$C_{total} = 2C = 1800$$ μF. The final voltage across each capacitor is $$V_f = \frac{Q}{C_{total}} = \frac{0.09}{1800 \times 10^{-6}} = 50$$ V, resulting in a final energy of $$U_f = \frac{1}{2}C_{total}V_f^2 = \frac{1}{2} \times 1800 \times 10^{-6} \times 2500 = 2.25$$ J.

The energy loss during the redistribution of charge is $$\Delta U = U_i - U_f = 4.5 - 2.25 = 2.25$$ J $$= 225 \times 10^{-2}$$ J, hence $$x = \boxed{225}$$.

We have two identical capacitors $$C_1 = C_2 = 600$$ pF, with $$C_1$$ initially charged to $$V = 200$$ V.

The initial energy stored in $$C_1$$ is:

$$U_i = \frac{1}{2}C_1V^2 = \frac{1}{2} \times 600 \times 10^{-12} \times (200)^2 = 12 \times 10^{-6} \text{ J} = 12 \text{ } \mu\text{J}$$

When $$C_1$$ is connected to the uncharged $$C_2$$, charge is conserved. The common voltage becomes:

$$V' = \frac{C_1 V}{C_1 + C_2} = \frac{600 \times 200}{600 + 600} = \frac{120000}{1200} = 100 \text{ V}$$

Now, the final energy of the combination is:

$$U_f = \frac{1}{2}(C_1 + C_2)V'^2 = \frac{1}{2} \times 1200 \times 10^{-12} \times (100)^2 = 6 \times 10^{-6} \text{ J} = 6 \text{ } \mu\text{J}$$

So the energy lost is:

$$\Delta U = U_i - U_f = 12 - 6 = 6 \text{ } \mu\text{J}$$

Hence, the answer is $$6$$ $$\mu$$J.

Initial capacitance with air:

$$C_0=5μF$$

When a dielectric slab of thickness $$\frac{d}{2}$$​ and dielectric constant K=1.5 is inserted, the system behaves like two capacitors in series:

• Air layer thickness $$\frac{d}{2}$$
• Dielectric layer thickness $$\frac{d}{2}$$

Effective separation:

$$d_{\text{eff}}=\frac{d}{2}+\frac{1}{K}\cdot\frac{d}{2}$$

Substitute K=1.5:

$$d_{\text{eff}}=\frac{d}{2}+\frac{d}{3}$$

$$d_{\text{eff}}=\frac{5d}{6}$$

Since capacitance is inversely proportional to effective distance,

$$C=\frac{\varepsilon_0A}{d_{\text{eff}}}$$

$$C=\frac{d}{5d/6}C_0$$

$$C=\frac{6}{5}\times5$$

C=6 μF

We have a parallel plate capacitor with plate area $$A$$ and plate separation $$d$$, containing a dielectric of constant $$K = 4$$ and thickness $$x$$.

Treating it as two capacitors in series (one with dielectric and one without), the capacitance is

$$C = \frac{\varepsilon_0 A}{(d - x) + \frac{x}{K}} = \frac{\varepsilon_0 A}{d - x + \frac{x}{4}} = \frac{\varepsilon_0 A}{d - \frac{3x}{4}}$$

Case 1: For $$x = \frac{d}{3}$$,

$$C_1 = \frac{\varepsilon_0 A}{d - \frac{3}{4} \cdot \frac{d}{3}} = \frac{\varepsilon_0 A}{d - \frac{d}{4}} = \frac{\varepsilon_0 A}{\frac{3d}{4}} = \frac{4\varepsilon_0 A}{3d} = 2\;\mu\text{F}$$

Case 2: For $$x = \frac{2d}{3}$$,

$$C_2 = \frac{\varepsilon_0 A}{d - \frac{3}{4} \cdot \frac{2d}{3}} = \frac{\varepsilon_0 A}{d - \frac{d}{2}} = \frac{\varepsilon_0 A}{\frac{d}{2}} = \frac{2\varepsilon_0 A}{d}$$

Now taking the ratio,

$$\frac{C_2}{C_1} = \frac{\frac{2\varepsilon_0 A}{d}}{\frac{4\varepsilon_0 A}{3d}} = \frac{2 \times 3}{4} = \frac{3}{2}$$

So $$C_2 = \frac{3}{2} \times C_1 = \frac{3}{2} \times 2 = 3\;\mu\text{F}$$.

Hence, the answer is $$3$$ $$\mu$$F. So, the answer is $$3$$.

So

$$\frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_2}$$

with

$$C_1=\frac{\varepsilon_0A}{a}$$

$$C_2=\frac{\varepsilon_0A}{b}$$

Thus

$$\frac{1}{C_{eq}}=\frac{a}{\varepsilon_0A}+\frac{b}{\varepsilon_0A}$$

$$\frac{1}{C_{eq}}=\frac{a+b}{\varepsilon_0A}$$

So

$$C_{eq}=\frac{\varepsilon_0A}{a+b}$$

Now from geometry,

$$a+b=d-c$$

Given

d=5 mm,c=1 mm so

a+b=4 mm=$$4\times\ 10^{-3}\ m$$ 

Area:

A=200 $$cm^2$$

$$=2\times10^{-2}m$$

$$=2\times10^{-2}m^2$$

Hence

$$C_{eq}=\frac{\varepsilon_0(2\times10^{-2})}{4\times10^{-3}}$$

$$=5\varepsilon_0$$

Comparing with

$$C_{eq}=x\varepsilon_0$$

we get

x=5

Given:

C1=2,  C2=0.2,  C3=2,  C4=4,  C5=2,  C6=2 μFC

Battery:

10V

Need charge on $$C_4$$​.

Let top-left node be 10V, bottom-left be 0V.

Let central top node = $$V_a$$

central bottom node = $$V_b$$

top-right node = $$V_c$$

bottom-right node = $$V_d$$

For floating nodes, net charge = 0.

At $$V_a$$:

$$2(V_a-10)+0.2(V_a-V_b)+2(V_a-V_c)=0$$

$$4.2V_a-0.2V_b-2V_c=20$$

At $$V_b$$​:

$$2(V_b-0)+0.2(V_b-V_a)+2(V_b-V_d)=0$$

$$-0.2V_a+4.2V_b-2V_d=0$$

At $$V_c$$​:

$$2(V_c-V_a)+4(V_c-V_d)=0$$

$$-2V_a+6V_c-4V_d=0$$

At $$V_d$$​:

$$2(V_d-V_b)+4(V_d-V_c)=0$$

$$-2V_b-4V_c+6V_d=0$$

Solving gives

$$V_c-V_d=1V$$

Charge on $$C_4$$:

$$Q_4=C_4(V_c-V_d)$$

$$=4\times1=4$$

We have a parallel plate capacitor with plate separation $$d$$ and original capacitance $$C_0 = \frac{\varepsilon_0 A}{d}$$. A dielectric slab of constant $$K$$ and thickness $$\frac{3d}{4}$$ is inserted between the plates.

When a dielectric slab is inserted, the system can be treated as two capacitors in series: one with the dielectric (thickness $$\frac{3d}{4}$$) and one with air (thickness $$d - \frac{3d}{4} = \frac{d}{4}$$).

The capacitance of the dielectric portion is $$C_1 = \frac{K\varepsilon_0 A}{3d/4} = \frac{4K\varepsilon_0 A}{3d}$$.

The capacitance of the air portion is $$C_2 = \frac{\varepsilon_0 A}{d/4} = \frac{4\varepsilon_0 A}{d}$$.

For capacitors in series, $$\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{3d}{4K\varepsilon_0 A} + \frac{d}{4\varepsilon_0 A} = \frac{d}{4\varepsilon_0 A}\left(\frac{3}{K} + 1\right) = \frac{d}{4\varepsilon_0 A} \cdot \frac{3+K}{K}$$.

So $$C = \frac{4K\varepsilon_0 A}{d(3+K)}$$. Since $$C_0 = \frac{\varepsilon_0 A}{d}$$, we can write $$C = \frac{4KC_0}{3+K}$$.

Hence, the correct answer is Option A.

Given:
Capacitors in series: 10 μF,  15 μF,  20 μF
Potential difference V=13 V

Step 1: Equivalent Capacitance (Series)

$$\ \frac{\ 1}{C_{eq}}\ =\ \frac{\ 1}{10}\ +\ \ \frac{1\ }{15}\ +\ \ \frac{1\ }{20}\ =\ \ \frac{\ 6\ +\ 4\ +\ 3}{60}$$
$$C_{eq}\ =\ \ \frac{\ 60}{13}μF$$

Step 2: Charge in Series Combination

$$Q=Ceq​⋅V=\ \frac{\ 60}{13}\times\ 13=60μC$$

Step 3: Charge on Each Capacitor
In series, charge is same on all capacitors.

Final Answer:

Q=60 μC

Four capacitors $$C_1 = 1\,\mu F$$, $$C_2 = 2\,\mu F$$, $$C_3 = 4\,\mu F$$, and $$C_4 = 3\,\mu F$$ are connected in parallel across a $$20 \text{ V}$$ battery. We need to find the total charge on the system.

For capacitors in parallel, the equivalent capacitance is the sum of individual capacitances:

$$C_{eq} = C_1 + C_2 + C_3 + C_4 = 1 + 2 + 4 + 3 = 10\,\mu F$$

Using $$Q = C_{eq} \times V$$:

$$Q = 10\,\mu F \times 20 \text{ V} = 200\,\mu C$$

The correct answer is Option A: $$200\,\mu C$$.

Two capacitors of 40 μF each are connected in series. One capacitor is filled with a dielectric of constant K, and the equivalent capacitance becomes 24 μF.

When a dielectric of constant $$K$$ is inserted in one capacitor:

$$C_1 = 40K \; \mu F$$ and $$C_2 = 40 \; \mu F$$

$$\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}$$

$$\frac{1}{24} = \frac{1}{40K} + \frac{1}{40}$$

$$\frac{1}{24} = \frac{1}{40}\left(\frac{1}{K} + 1\right)$$

$$\frac{40}{24} = \frac{1}{K} + 1$$

$$\frac{5}{3} = \frac{1}{K} + 1$$

$$\frac{1}{K} = \frac{5}{3} - 1 = \frac{2}{3}$$

$$K = \frac{3}{2} = 1.5$$

Hence, the correct answer is Option A: 1.5.

Given:
Plate separation d=2 m, original capacitance C=4 μF, dielectric constant K=3

Step 1: Divide the capacitor into two regions
Each region has thickness d/2, so they act as series capacitors.

Step 2: Capacitance of each part

• Without dielectric:

$$C_1=\frac{\varepsilon_0A}{d/2}=\frac{2\varepsilon_0A}{d}=2C$$

• With dielectric K=3K:

$$C_2=\frac{K\varepsilon_0A}{d/2}=\frac{2K\varepsilon_0A}{d}=2KC=6C$$

Step 3: Equivalent capacitance (series)

$$C_{\text{new}}=\frac{C_1C_2}{C_1+C_2}$$

$$=\frac{(2C)(6C)}{2C+6C}$$

$$=\frac{12C^2}{8C}$$

$$Cnew=\frac{3}{2}C​$$

Step 4: Substitute value

$$C_{new}​=\ \frac{\ 3}{2}​\times4=6μF$$

Final Answer:

6 μF

Initially switch KKK is closed, so both identical capacitors C are in parallel across voltage V.

Equivalent capacitance:

$$C_{eq}=2C$$

Initial energy:

$$E_1=\frac{1}{2}(2C)V^2$$

Now switch is opened and dielectric constant

k=5

is inserted.

Left capacitor remains connected to battery, so voltage remains constant.

New capacitance:

$$C_1'=5C$$

Energy in left capacitor:

$$E_L=\frac{1}{2}(5C)V^2$$

$$=\frac{5}{2}CV^2$$

Right capacitor becomes isolated, so charge remains constant.

Initial charge on it was

Q=CV

After dielectric insertion,

$$C_2'=5C$$

Energy:

$$=\frac{1}{10}CV^2$$

Total final energy:

$$E_2=\frac{5}{2}CV^2+\frac{1}{10}CV^2$$

$$=\frac{26}{10}CV^2$$

$$=\frac{13}{5}CV^2$$

Therefore

$$=\frac{5}{13}$$

We need to find the ratio $$\frac{R_2}{R_1}$$ when the capacitance of an isolated sphere becomes $$n$$ times after enclosing it with a grounded concentric sphere.

Write the capacitance of an isolated sphere.

$$C_1 = 4\pi\varepsilon_0 R_1$$

Write the capacitance of the spherical capacitor.

When the inner sphere of radius $$R_1$$ is enclosed by a grounded outer sphere of radius $$R_2$$, the capacitance is:

$$C_2 = \frac{4\pi\varepsilon_0 R_1 R_2}{R_2 - R_1}$$

Apply the condition $$C_2 = nC_1$$.

$$\frac{4\pi\varepsilon_0 R_1 R_2}{R_2 - R_1} = n \times 4\pi\varepsilon_0 R_1$$

$$\frac{R_2}{R_2 - R_1} = n$$

$$R_2 = n(R_2 - R_1) = nR_2 - nR_1$$

$$nR_1 = nR_2 - R_2 = R_2(n - 1)$$

$$\frac{R_2}{R_1} = \frac{n}{n - 1}$$

The correct answer is Option A: $$\dfrac{n}{n-1}$$.

We are given: area $$A = 30\pi$$ cm$$^2 = 30\pi \times 10^{-4}$$ m$$^2$$, separation $$d = 1$$ mm $$= 10^{-3}$$ m, dielectric strength $$E_{max} = 3.6 \times 10^7$$ V/m, maximum charge $$Q = 7 \times 10^{-6}$$ C, $$\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9$$ N m$$^2$$ C$$^{-2}$$.

From the relation $$\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9$$, it follows that

$$ \varepsilon_0 = \frac{1}{4\pi \times 9 \times 10^9} $$

The capacitance with dielectric is given by

$$ C = \frac{\kappa \varepsilon_0 A}{d} $$

The maximum voltage before dielectric breakdown is

$$ V_{max} = E_{max} \times d = 3.6 \times 10^7 \times 10^{-3} = 3.6 \times 10^4 \text{ V} $$

Using $$Q = CV$$ yields

$$ Q = C \times V_{max} = \frac{\kappa \varepsilon_0 A}{d} \times E_{max} \times d = \kappa \varepsilon_0 A \times E_{max} $$

$$ \kappa = \frac{Q}{\varepsilon_0 \times A \times E_{max}} $$

Substituting the given values:

$$ \kappa = \frac{7 \times 10^{-6}}{\frac{1}{4\pi \times 9 \times 10^9} \times 30\pi \times 10^{-4} \times 3.6 \times 10^7} $$

Computing the denominator step by step:

$$ \varepsilon_0 \times A = \frac{30\pi \times 10^{-4}}{4\pi \times 9 \times 10^9} = \frac{30 \times 10^{-4}}{4 \times 9 \times 10^9} = \frac{30 \times 10^{-4}}{36 \times 10^9} = \frac{10^{-4}}{1.2 \times 10^9} = \frac{1}{1.2} \times 10^{-13} $$

$$ \varepsilon_0 \times A \times E_{max} = \frac{1}{1.2} \times 10^{-13} \times 3.6 \times 10^7 = \frac{3.6}{1.2} \times 10^{-6} = 3 \times 10^{-6} $$

$$ \kappa = \frac{7 \times 10^{-6}}{3 \times 10^{-6}} = \frac{7}{3} \approx 2.33 $$

Therefore, the correct answer is Option D.

We consider each statement about conductors in electrostatic equilibrium.

Statement I: In electrostatic equilibrium, the electric field inside a conductor is zero. Since $$\vec{E} = -\nabla V$$, a zero electric field throughout the conductor means the potential does not change — it is constant within and on the surface of the conductor. This statement is correct.

Statement II: For a charged conductor in equilibrium, any net charge resides on the surface. The electric field just outside the surface must be perpendicular to the surface at every point. If there were a tangential component, it would cause surface charges to move, contradicting the assumption of electrostatic equilibrium. Therefore, the field is purely normal (perpendicular) to the surface. This statement is correct.

Since both statements are correct, hence the correct answer is Option A.

Given: $$V = 3x^2$$ volt. Find the electric field at $$(1, 0, 3)$$.

$$\vec{E} = -\nabla V = -\left(\frac{\partial V}{\partial x}\hat{i} + \frac{\partial V}{\partial y}\hat{j} + \frac{\partial V}{\partial z}\hat{k}\right)$$

$$\frac{\partial V}{\partial x} = 6x, \quad \frac{\partial V}{\partial y} = 0, \quad \frac{\partial V}{\partial z} = 0$$

$$\vec{E} = -6x\hat{i}$$

$$\vec{E} = -6(1)\hat{i} = -6\hat{i} \text{ Vm}^{-1}$$

The magnitude is 6 Vm$$^{-1}$$, directed along the negative $$x$$-axis.

Hence, the correct answer is Option C: 6 Vm$$^{-1}$$, directed along negative $$x$$-axis.

A capacitor discharges through resistance $$R$$. In time $$t_1$$, energy reduces to half. In time $$t_2$$, charge reduces to one-eighth. Find $$t_1/t_2$$.

$$q = q_0 e^{-t/RC}$$

Energy $$U = \frac{q^2}{2C} = \frac{q_0^2}{2C}e^{-2t/RC}$$

So energy decays as $$U = U_0 e^{-2t/RC}$$.

Energy reduces to half: $$\frac{1}{2} = e^{-2t_1/RC}$$

$$\ln 2 = \frac{2t_1}{RC}$$

$$t_1 = \frac{RC \ln 2}{2}$$

Charge reduces to $$\frac{1}{8}$$: $$\frac{1}{8} = e^{-t_2/RC}$$

$$\ln 8 = \frac{t_2}{RC}$$

$$t_2 = RC \ln 8 = 3RC \ln 2$$

$$\frac{t_1}{t_2} = \frac{RC\ln 2/2}{3RC\ln 2} = \frac{1}{6}$$

Hence, the correct answer is Option D: $$\dfrac{1}{6}$$.

Let the original charge on the capacitor be $$Q$$ (in Coulombs) and the capacitance be $$C$$.

Write the energy stored in a capacitor: The energy stored in a capacitor is:

$$E = \frac{Q^2}{2C}$$

Set up the equation for the given condition: When the charge is increased by $$2$$ C, the new charge is $$(Q + 2)$$ C.

The new energy is:

$$E' = \frac{(Q + 2)^2}{2C}$$

The energy increases by $$44\%$$, so:

$$E' = 1.44 \times E$$

Substitute and simplify: $$\frac{(Q + 2)^2}{2C} = 1.44 \times \frac{Q^2}{2C}$$

Cancel $$\frac{1}{2C}$$ from both sides:

$$(Q + 2)^2 = 1.44 \, Q^2$$

Expand and solve: $$Q^2 + 4Q + 4 = 1.44 \, Q^2$$

$$0.44 \, Q^2 - 4Q - 4 = 0$$

Multiply through by 100 to clear decimals:

$$44Q^2 - 400Q - 400 = 0$$

Divide by 4:

$$11Q^2 - 100Q - 100 = 0$$

Solve the quadratic equation: Using the quadratic formula:

$$Q = \frac{100 \pm \sqrt{10000 + 4400}}{22} = \frac{100 \pm \sqrt{14400}}{22} = \frac{100 \pm 120}{22}$$

Taking the positive root (charge must be positive):

$$Q = \frac{100 + 120}{22} = \frac{220}{22} = 10 \text{ C}$$

The original charge on the capacitor is $$10$$ C.

The correct answer is Option A.

A capacitor of 50 pF is charged by a 100 V source, then connected to another uncharged identical capacitor.

The initial energy stored in the capacitor is $$U_i = \frac{1}{2}CV^2 = \frac{1}{2} \times 50 \times 10^{-12} \times (100)^2$$. Substituting the values gives $$U_i = \frac{1}{2} \times 50 \times 10^{-12} \times 10000 = 250 \times 10^{-9} \text{ J} = 250 \text{ nJ}$$.

When this capacitor is connected to an identical uncharged one, the total charge is shared equally and the final voltage across each capacitor is $$V_f = \frac{Q_{total}}{C_{total}} = \frac{CV}{2C} = \frac{V}{2} = 50 \text{ V}$$.

The total energy stored in the two capacitors after connection is $$U_f = \frac{1}{2}(2C)V_f^2 = \frac{1}{2} \times 2 \times 50 \times 10^{-12} \times (50)^2$$, which gives $$U_f = 50 \times 10^{-12} \times 2500 = 125 \times 10^{-9} \text{ J} = 125 \text{ nJ}$$.

The energy loss is the difference between the initial and final energies: $$\Delta U = U_i - U_f = 250 - 125 = 125 \text{ nJ}$$.

The electrostatic energy loss is 125 nJ.

Treat it as two capacitors in series with dielectric slabs separated by conducting foil F

$$\varepsilon_{r1}=3\ and\ t_1\ =\ 0.5mm$$

$$\varepsilon_{r2}=4\ and\ t_2\ =\ 1mm$$

V =100V

For dielectrics in series, same charge appears on each, so electric displacement is same, and voltage division is proportional to $$\frac{t}{\varepsilon}$$

so,

$$V_1:V_2\ =\ \frac{t_1}{\varepsilon_{r1}}:\frac{t_2}{\varepsilon_{r2}}$$

Substitute:

$$V_1:V_2=\frac{0.5}{3}:\frac{1}{4}$$

$$=\frac{1}{6}:\frac{1}{4}$$

Multiply by 12:

$$2:3$$

Total voltage is

$$V_1+V_2=100$$

So

$$V_1=40V,\quad V_2=60V$$

The conducting foil is between the two dielectric sections. Its potential (with lower plate as reference zero) equals the voltage across lower dielectric, i.e.

V=60V

Concept:
All sections are connected between the same two conductors, hence they have the same potential difference ⇒ they are in parallel.

Why distances are b,3b,5b:
Each step increases separation by b:

• First gap = b
• Next gap = b+b+b=3b
• Last gap = b+b+b+b+b=5b

Step 1: Individual capacitances

$$C=\frac{\varepsilon_0A}{d}​$$

$$C_1=\frac{\varepsilon_0A}{b},\quad$$

$$C_2=\frac{\varepsilon_0A}{3b},\quad$$

$$C_3=\frac{\varepsilon_0A}{5b}$$

Step 2: Equivalent capacitance (parallel)

$$C_{\text{eq}}=C_1+C_2+C_3$$

= $$\varepsilon_0A\left(\frac{1}{b}+\frac{1}{3b}+\frac{1}{5b}\right)$$

= $$\frac{\varepsilon_0A}{b}\cdot\frac{23}{15}$$

Step 3: Compare with given form

$$C_{\text{eq}}=\frac{x}{15}\cdot\frac{\varepsilon_0A}{b}$$

$$\Rightarrow x=23$$

Final Answer:

23

Two capacitors $$C$$ and $$3C$$ are connected in parallel and charged to $$18 \text{ V}$$.

Total initial charge:

$$Q = (C + 3C) \times 18 = 72C$$

The battery is disconnected, so charge is conserved.

When a dielectric of constant $$K = 9$$ is inserted in the capacitor of capacity $$C$$, its new capacitance becomes:

$$C' = KC = 9C$$

The new total capacitance of the parallel combination is:

$$C_{total} = 9C + 3C = 12C$$

Since charge is conserved, the new potential difference is:

$$V = \dfrac{Q}{C_{total}} = \dfrac{72C}{12C} = 6 \text{ V}$$

Therefore, the final potential difference is $$\boxed{6}$$ V.

Number of identical drops: $$n = 27$$

Potential of each small drop: $$V = 22$$ V

Let the radius of each small drop be $$r$$. When 27 drops combine, the total volume is conserved:

$$\frac{4}{3}\pi R^3 = 27 \times \frac{4}{3}\pi r^3$$

$$R^3 = 27r^3$$

$$R = 3r$$

The potential of each small drop: $$V = \frac{kq}{r}$$, so $$q = \frac{Vr}{k}$$.

Total charge on the bigger drop:

$$Q = 27q = \frac{27Vr}{k}$$

$$V' = \frac{kQ}{R} = \frac{k \times 27q}{3r} = \frac{27kq}{3r} = 9 \times \frac{kq}{r} = 9V$$

$$V' = 9 \times 22 = 198 \text{ V}$$

Hence, the potential of the bigger drop is 198 V.

Let the area of each plate be $$A$$ (very large) and let the battery keep the potential difference between the plates fixed at $$V$$ throughout the entire process.

Initial configuration - Figure (a)
• Separation between plates = $$d$$.
• The whole region between the plates is filled with a dielectric of constant $$K$$.
Hence

Capacitance: $$C_i = \dfrac{\varepsilon_0 K A}{d}$$ $$-(1)$$

Electric field inside the dielectric:
Since the potential difference is $$V$$ and the field is uniform, $$E_i = \dfrac{V}{d}$$ $$-(2)$$

Final configuration - Figure (b)
• Each plate is pulled away from its original position by $$\dfrac{d}{2}$$, so the new plate separation is $$d + \dfrac{d}{2} + \dfrac{d}{2} = 2d$$.
• The dielectric slab remains where it was; its thickness is still $$d$$ and it now occupies the middle part of the gap.
Thus the space between the plates consists of three layers placed one after another (in series):
  1. Vacuum of thickness $$\dfrac{d}{2}$$,
  2. Dielectric of thickness $$d$$ and constant $$K$$,
  3. Vacuum of thickness $$\dfrac{d}{2}$$.

Equivalent capacitance in the final set-up
For layers in series, the capacitance per unit area is found from $$\dfrac{1}{C_f/A} = \dfrac{d_1}{\varepsilon_0} + \dfrac{d_2}{\varepsilon_0 K} + \dfrac{d_3}{\varepsilon_0}$$

Here $$d_1 = d_3 = \dfrac{d}{2},\qquad d_2 = d$$.
Therefore

$$\dfrac{1}{C_f/A} = \dfrac{d/2}{\varepsilon_0} + \dfrac{d}{\varepsilon_0 K} + \dfrac{d/2}{\varepsilon_0} = \dfrac{d}{\varepsilon_0} + \dfrac{d}{\varepsilon_0 K} = \dfrac{d}{\varepsilon_0}\Bigl(1 + \dfrac{1}{K}\Bigr) = \dfrac{d}{\varepsilon_0}\,\dfrac{K+1}{K}$$

Hence

$$C_f = \dfrac{\varepsilon_0 A K}{d\,(K+1)}$$ $$-(3)$$

Comparison of capacitances
Using $$(1)$$ and $$(3)$$,

$$\dfrac{C_f}{C_i} = \dfrac{\varepsilon_0 A K}{d\,(K+1)} \bigg/ \dfrac{\varepsilon_0 K A}{d} = \dfrac{1}{K+1}$$

Thus the capacitance decreases by the factor $$\dfrac{1}{K+1}$$. Statement B is correct.

Electric fields in the final configuration
Let $$E_0$$ be the field in either vacuum gap and $$E_d$$ the field in the dielectric. Because there is no free charge inside, the displacement field is continuous: $$\varepsilon_0 E_0 = \varepsilon_0 K E_d \;\Longrightarrow\; E_0 = K E_d$$ $$-(4)$$

The total potential difference is still $$V$$ (battery connected):

$$V = E_0\Bigl(\dfrac{d}{2} + \dfrac{d}{2}\Bigr) + E_d (d) = dE_0 + dE_d = d\bigl(E_0 + E_d\bigr)$$

Using $$(4)$$, $$E_0 + E_d = K E_d + E_d = E_d (K+1)$$. Hence

$$E_d = \dfrac{V}{d\,(K+1)},\qquad E_0 = \dfrac{VK}{d\,(K+1)}$$

Comparing $$E_d$$ with the initial field $$E_i = \dfrac{V}{d}$$ (from $$(2)$$), we get $$\dfrac{E_d}{E_i} = \dfrac{1}{K+1}$$, not $$\dfrac{1}{2K}$$. Therefore Statement A is false.

Voltage between the plates
It remains fixed at $$V$$, so it is certainly not multiplied by $$(K+1)$$. Statement C is false.

Work done
The energy stored initially: $$U_i = \dfrac{1}{2} C_i V^2$$.
The energy stored finally: $$U_f = \dfrac{1}{2} C_f V^2$$.
Because $$C_f \neq C_i$$ and depends on $$K$$, the work (equal to $$U_i - U_f$$, any difference being supplied by the battery) does depend on the dielectric. Statement D is false.

Only Statement B is correct.
Option B which is: The capacitance is decreased by a factor of $$\dfrac{1}{K+1}$$.

We are given a parallel plate capacitor with width = 4 cm = 0.04 m, length = 8 cm = 0.08 m, and separation = 4 mm = 0.004 m, connected to a battery of voltage V = 20 V.

A dielectric slab of dielectric constant $$\kappa = 5$$, length = 1 cm, width = 4 cm, and thickness = 4 mm is inserted into the capacitor.

Since the dielectric slab occupies the same width and thickness as the capacitor but spans only 1 cm of the total 8 cm length, the system can be modeled as two capacitors in parallel: one region containing the dielectric of length 1 cm = 0.01 m (denoted $$C_1$$) and another region without the dielectric of length 7 cm = 0.07 m (denoted $$C_2$$).

Substituting into the standard expression for a parallel‐plate capacitor, we find

$$C_1 = \frac{\kappa \epsilon_0 \times \text{width} \times \text{length}_1}{\text{separation}} = \frac{5\epsilon_0 \times 0.04 \times 0.01}{0.004} = \frac{5\epsilon_0 \times 4 \times 10^{-4}}{4 \times 10^{-3}} = 0.5\epsilon_0$$

and

$$C_2 = \frac{\epsilon_0 \times \text{width} \times \text{length}_2}{\text{separation}} = \frac{\epsilon_0 \times 0.04 \times 0.07}{0.004} = \frac{\epsilon_0 \times 28 \times 10^{-4}}{4 \times 10^{-3}} = 0.7\epsilon_0$$

Therefore, the total capacitance is $$C = C_1 + C_2 = 0.5\epsilon_0 + 0.7\epsilon_0 = 1.2\epsilon_0\,.$$

From this capacitance and the applied voltage, the electrostatic energy stored is $$U = \tfrac{1}{2} C V^2 = \tfrac{1}{2} \times 1.2\epsilon_0 \times (20)^2 = \tfrac{1}{2} \times 1.2\epsilon_0 \times 400 = 240\epsilon_0 \text{ J}\,.$$

Thus, the answer is 240.

initially capacitance =

$$\left(\frac{A\varepsilon_0}{d}\right)=C$$

when sheet of thickness $$t=\left(\frac{d}{2}\right)$$ is introduced then capacitance be C'. 

and $$C'=\frac{A\varepsilon_0}{d-t}$$

$$\left(\frac{C'}{C}\right)=\frac{\left(d-t\right)}{d}=\frac{\left(d-(d/2)\right)}{d}\ $$

$$\frac{C'}{C}=\frac{2}{1}$$

ratio of capacitance is 2 : 1

For a parallel-plate capacitor we assume the plates have free charge $$+Q$$ and $$-Q$$. Throughout the dielectric the displacement field is uniform, so by Gauss’s law we have

$$D=\frac{Q}{A}\;.$$

The local permittivity is a function of the distance $$x$$ measured from one plate. Therefore the electric field at that point is obtained from the relation $$D=\varepsilon_x\,E(x)$$, giving

$$E(x)=\frac{D}{\varepsilon_x}= \frac{Q}{A\,\varepsilon_x}\;.$$

The total potential difference between the plates equals the integral of the electric field across the whole separation,

$$V=\int_0^{d}E(x)\,dx=\frac{Q}{A}\int_0^{d}\frac{dx}{\varepsilon_x}\;.$$

The capacitance is defined by $$C=\dfrac{Q}{V}$$, hence

$$C=\frac{A}{\displaystyle\int_{0}^{d}\frac{dx}{\varepsilon_x}}\;.$$

Now we substitute the given piece-wise form of the permittivity.

For $$0<x\le\dfrac{d}{2}$$ we have $$\varepsilon_x=\varepsilon_0+kx$$, and for $$\dfrac{d}{2}\le x\le d$$ we have $$\varepsilon_x=\varepsilon_0+k(d-x)$$. Thus the integral in the denominator splits into two parts:

$$\int_{0}^{d}\frac{dx}{\varepsilon_x}= \int_{0}^{d/2}\frac{dx}{\varepsilon_0+kx}+ \int_{d/2}^{d}\frac{dx}{\varepsilon_0+k(d-x)}\;.$$

In the second integral we set $$u=d-x\; (du=-dx)$$. When $$x=d/2$$, $$u=d/2$$ and when $$x=d$$, $$u=0$$. Therefore

$$\int_{d/2}^{d}\frac{dx}{\varepsilon_0+k(d-x)}=\int_{d/2}^{0}\frac{-du}{\varepsilon_0+ku}= \int_{0}^{d/2}\frac{du}{\varepsilon_0+ku}\;.$$

Both integrals are now identical, so the total becomes

$$\int_{0}^{d}\frac{dx}{\varepsilon_x}=2\int_{0}^{d/2}\frac{dx}{\varepsilon_0+kx}\;.$$

We evaluate the remaining logarithmic integral. Using the standard result

$$\int\frac{dx}{a+bx}=\frac{1}{b}\ln(a+bx)\;,$$

we get

$$\int_{0}^{d/2}\frac{dx}{\varepsilon_0+kx}= \left[\frac{1}{k}\ln\bigl(\varepsilon_0+kx\bigr)\right]_{0}^{d/2}= \frac{1}{k}\Bigl[\ln\!\left(\varepsilon_0+\frac{kd}{2}\right)-\ln(\varepsilon_0)\Bigr] =\frac{1}{k}\ln\!\left(\frac{\varepsilon_0+\dfrac{kd}{2}}{\varepsilon_0}\right).$$

Multiplying by 2 gives

$$\int_{0}^{d}\frac{dx}{\varepsilon_x}= \frac{2}{k}\ln\!\left(\frac{\varepsilon_0+\dfrac{kd}{2}}{\varepsilon_0}\right) =\frac{2}{k}\ln\!\left(\frac{2\varepsilon_0+kd}{2\varepsilon_0}\right).$$

Substituting this result into the capacitance formula we finally obtain

$$C=\frac{A}{\dfrac{2}{k}\ln\!\left(\dfrac{2\varepsilon_0+kd}{2\varepsilon_0}\right)} =\frac{kA}{2\,\ln\!\left(\dfrac{2\varepsilon_0+kd}{2\varepsilon_0}\right)}\;.$$

Hence, the correct answer is Option B.

The original capacitance of the parallel plate capacitor is $$C_0 = \frac{\varepsilon_0 A}{d}$$, where $$A$$ is the plate area and $$d$$ is the separation between the plates.

When a dielectric slab of thickness $$t = \frac{3}{4}d$$ and dielectric constant $$K$$ is inserted, the capacitor can be thought of as two capacitors in series: one with the dielectric (thickness $$t$$) and one with air (thickness $$d - t$$).

The effective capacitance formula for a partially filled capacitor is $$C' = \frac{\varepsilon_0 A}{(d - t) + \frac{t}{K}}$$.

Substituting $$t = \frac{3d}{4}$$: $$C' = \frac{\varepsilon_0 A}{\left(d - \frac{3d}{4}\right) + \frac{3d}{4K}} = \frac{\varepsilon_0 A}{\frac{d}{4} + \frac{3d}{4K}}$$.

Taking the common denominator in the denominator: $$\frac{d}{4} + \frac{3d}{4K} = \frac{dK + 3d}{4K} = \frac{d(K + 3)}{4K}$$.

Therefore, $$C' = \frac{\varepsilon_0 A}{\frac{d(K+3)}{4K}} = \frac{4K \varepsilon_0 A}{d(K+3)} = \frac{4K}{K+3} \cdot \frac{\varepsilon_0 A}{d} = \frac{4K}{K+3} \, C_0$$.

We have two identically sized, thin, coaxial rings of radius $$a$$ separated along their common axis by a distance $$s$$. The upper ring (call it Ring 1) carries charge $$+Q$$, while the lower ring (Ring 2) carries charge $$-Q$$. We wish to find the potential difference between the centres of the two rings; let these centres be the points $$O_1$$ (on Ring 1) and $$O_2$$ (on Ring 2).

First, we recall the standard result for the electric potential on the axis of a uniformly charged ring. If a ring of total charge $$q$$ and radius $$a$$ is observed from a point on its axis at a distance $$x$$ from its centre, the potential there is

$$V=\frac{1}{4\pi\varepsilon_0}\frac{q}{\sqrt{a^2+x^2}}\;.$$

Now we evaluate the potentials at the two centres one by one, adding the contributions from both rings in each case.

Potential at point $$O_1$$ (centre of Ring 1):

• Contribution of Ring 1 itself: here the axial distance is $$x=0$$, so

$$V_{(1\to O_1)}=\frac{1}{4\pi\varepsilon_0}\frac{+Q}{\sqrt{a^2+0^2}}=\frac{1}{4\pi\varepsilon_0}\frac{Q}{a}\;.$$

• Contribution of Ring 2: its charge is $$-Q$$ and the distance from $$O_1$$ to the centre of Ring 2 is $$x=s$$, hence

$$V_{(2\to O_1)}=\frac{1}{4\pi\varepsilon_0}\frac{-Q}{\sqrt{a^2+s^2}}\;.$$

Adding these two gives the net potential at $$O_1$$:

$$$\begin{aligned} V_1&=V_{(1\to O_1)}+V_{(2\to O_1)}\\[4pt] &=\frac{1}{4\pi\varepsilon_0}\left(\frac{Q}{a}-\frac{Q}{\sqrt{a^2+s^2}}\right). \end{aligned}$$$

Potential at point $$O_2$$ (centre of Ring 2):

• Contribution of Ring 2 itself: here $$x=0$$ and the charge is $$-Q$$, so

$$V_{(2\to O_2)}=\frac{1}{4\pi\varepsilon_0}\frac{-Q}{a}\;.$$

• Contribution of Ring 1: its charge is $$+Q$$ and the distance from $$O_2$$ to Ring 1 is again $$x=s$$, giving

$$V_{(1\to O_2)}=\frac{1}{4\pi\varepsilon_0}\frac{+Q}{\sqrt{a^2+s^2}}\;.$$

Thus, the net potential at $$O_2$$ is

$$$\begin{aligned} V_2&=V_{(2\to O_2)}+V_{(1\to O_2)}\\[4pt] &=\frac{1}{4\pi\varepsilon_0}\left(-\frac{Q}{a}+\frac{Q}{\sqrt{a^2+s^2}}\right). \end{aligned}$$$

We are asked for the potential difference between the two centres. Choosing the order $$V_1-V_2$$ (the magnitude will be the same even if the sign is reversed), we obtain

$$$\begin{aligned} V_1-V_2 &=\frac{1}{4\pi\varepsilon_0}\left(\frac{Q}{a}-\frac{Q}{\sqrt{a^2+s^2}}\right) -\frac{1}{4\pi\varepsilon_0}\left(-\frac{Q}{a}+\frac{Q}{\sqrt{a^2+s^2}}\right)\\[6pt] &=\frac{1}{4\pi\varepsilon_0}\left[\frac{Q}{a}-\frac{Q}{\sqrt{a^2+s^2}}+\frac{Q}{a}-\frac{Q}{\sqrt{a^2+s^2}}\right]\\[6pt] &=\frac{1}{4\pi\varepsilon_0}\left(\frac{2Q}{a}-\frac{2Q}{\sqrt{a^2+s^2}}\right). \end{aligned}$$$

Extracting the common factor $$2Q$$ and dividing by $$4$$ in the denominator gives

$$V_1-V_2=\frac{Q}{2\pi\varepsilon_0}\left[\frac{1}{a}-\frac{1}{\sqrt{a^2+s^2}}\right].$$

This expression matches Option A exactly. Hence, the correct answer is Option A.

For a series combination of a resistor $$R$$ and a capacitor $$C$$ connected to a constant supply voltage $$V$$, the rise of voltage across the capacitor with time is governed by the charging equation

$$V_C(t)=V\bigl(1-e^{-t/RC}\bigr).$$

Here we have a battery of $$V=20\ \text{V}$$, a resistance $$R=10\ \Omega$$ and an observed capacitor voltage $$V_C=2\ \text{V}$$ after a time interval $$t=1\ \mu\text{s}=1\times10^{-6}\ \text{s}.$$ Substituting these numbers into the formula gives

$$2 = 20\left(1-e^{-t/RC}\right).$$

Dividing both sides by $$20$$, we obtain

$$\frac{2}{20}=1-e^{-t/RC}.$$

This simplifies to

$$0.1 = 1-e^{-t/RC}.$$

Re-arranging, we isolate the exponential term:

$$e^{-t/RC}=1-0.1=0.9.$$

Taking the natural logarithm of both sides, we write

$$-\,\frac{t}{RC}=\ln(0.9).$$

Since $$\ln(0.9)=-\ln\!\left(\dfrac{10}{9}\right)$$, we have

$$-\,\frac{t}{RC}= -\,\ln\!\left(\dfrac{10}{9}\right).$$

Removing the minus signs on both sides gives

$$\frac{t}{RC}= \ln\!\left(\dfrac{10}{9}\right).$$

We now substitute the known numerical values: $$t = 1\times10^{-6}\ \text{s}$$, $$R = 10\ \Omega$$ and, from the question, $$\ln\!\left(\dfrac{10}{9}\right)=0.105.$$ Thus,

$$\frac{1\times10^{-6}}{10\,C}=0.105.$$

Multiplying both sides by $$10C$$, we obtain

$$1\times10^{-6}=1.05\,C.$$

Dividing by $$1.05$$ isolates the capacitance:

$$C=\frac{1\times10^{-6}}{1.05}.$$

Carrying out the division,

$$C\approx0.952\times10^{-6}\ \text{F}.$$

Since $$1\times10^{-6}\ \text{F}=1\ \mu\text{F}$$, we convert the answer to microfarads:

$$C\approx0.952\ \mu\text{F}\,.$$

Rounding to two significant figures, this is $$0.95\ \mu\text{F}$$.

Hence, the correct answer is Option A.

We begin by recalling the fundamental relation that connects the electric displacement vector $$\mathbf D$$ with the free surface charge density $$\sigma_f$$ on a dielectric-filled capacitor plate. By definition,

$$\mathbf D \cdot \hat n \;=\; \sigma_f.$$

Here $$\hat n$$ is the outward normal to the surface and $$\sigma_f$$ is linked to the total free charge $$q_f$$ on the plate of area $$A$$ through

$$\sigma_f \;=\; \dfrac{q_f}{A}.$$

Next, we write the constitutive relation between $$\mathbf D$$, the electric field $$\mathbf E$$ and the polarisation $$\mathbf P$$ inside a linear isotropic dielectric:

$$\mathbf D \;=\; \varepsilon_0 \mathbf E \;+\; \mathbf P.$$

The dielectric constant $$k$$ (relative permittivity $$\varepsilon_r$$) is defined by

$$k = \varepsilon_r \;=\; 1 \;+\; \chi_e,$$

where $$\chi_e$$ is the electric susceptibility. Therefore

$$\mathbf P \;=\; \varepsilon_0 \chi_e \mathbf E \;=\; \varepsilon_0 (k-1)\mathbf E.$$

The bound surface charge density $$\sigma_b$$ appears wherever the polarisation vector $$\mathbf P$$ terminates on a surface, and it is given by the formula

$$\sigma_b \;=\; \mathbf P \cdot \hat n.$$

Substituting the expression of $$\mathbf P$$ we obtain

$$\sigma_b \;=\; \varepsilon_0 (k-1) \mathbf E \cdot \hat n \;=\; \varepsilon_0 (k-1) E,$$

because $$\mathbf E$$ is perpendicular to the plate and points along $$\hat n$$ in a parallel-plate capacitor.

In precisely the same geometry the displacement vector is purely normal, giving

$$\sigma_f \;=\; \mathbf D \cdot \hat n \;=\; \varepsilon_0 k E.$$

Now we take the ratio of the bound to the free surface charge densities:

$$\dfrac{\sigma_b}{\sigma_f} \;=\; \dfrac{\varepsilon_0 (k-1) E}{\varepsilon_0 k E} \;=\; \dfrac{k-1}{k} \;=\; 1 - \dfrac{1}{k}.$$

Since both surface charge densities refer to the same plate area $$A$$, multiplying by $$A$$ converts them to the corresponding total charges:

$$q_b \;=\; \sigma_b A, \qquad q_f \;=\; \sigma_f A.$$

Therefore the ratio of the total charges is identical to the ratio of the surface charge densities:

$$\dfrac{q_b}{q_f} \;=\; 1 - \dfrac{1}{k}.$$

Solving for the bound charge $$q_b$$ we finally obtain

$$q_b \;=\; q_f\left(1 - \dfrac{1}{k}\right).$$

Hence, the correct answer is Option B.

Let each capacitor have capacitance $$C$$.

When two equal capacitors are connected in series, the equivalent capacitance is $$C_s = \frac{C \times C}{C + C} = \frac{C}{2}$$.

When the same two capacitors are connected in parallel, the equivalent capacitance is $$C_p = C + C = 2C$$.

The ratio of the equivalent capacitances is $$\frac{C_s}{C_p} = \frac{C/2}{2C} = \frac{1}{4}$$.

So the ratio is $$1 : 4$$.

Hence, the correct answer is Option B.

Initially, the two capacitors are joined in parallel and connected to a battery of potential difference $$V$$. For any capacitor we recall the basic formula

$$Q = C\,V,$$

where $$Q$$ is the charge stored, $$C$$ is the capacitance and $$V$$ is the potential across the plates.

Because the capacitors are in parallel, the same potential $$V$$ appears across each of them while the battery is connected. Hence, using the above formula, we find the individual charges:

For the $$2C$$ capacitor, the charge is

$$Q_1 = (2C)\,V = 2CV.$$

For the $$C$$ capacitor, the charge is

$$Q_2 = C\,V = CV.$$

Adding these two charges gives the total charge present on the combination just before the battery is detached:

$$Q_{\text{total}} = Q_1 + Q_2 = 2CV + CV = 3CV.$$

After the battery is removed, the wires still connect the two capacitors, so they must continue to share a common potential. At this stage the capacitor of initial capacitance $$C$$ is completely filled with a dielectric of constant $$K$$. The formula for a parallel-plate capacitor with a dielectric tells us that its new capacitance becomes

$$C_{\text{new}} = K\,C.$$

The other capacitor keeps its original value $$2C$$. Let the common potential difference after equilibrium be $$V_f$$. Using $$Q = C\,V$$ once more, we write the new charges:

On the $$2C$$ capacitor the charge is

$$Q_1' = (2C)\,V_f = 2C\,V_f.$$

On the capacitor now of value $$KC$$ the charge is

$$Q_2' = (KC)\,V_f = KC\,V_f.$$

No charge has any external path to escape because the entire system is isolated, so the total charge after inserting the dielectric must equal the total charge that was present before. Therefore we impose charge conservation:

$$Q_1' + Q_2' = Q_{\text{total}}.$$

Substituting the expressions gives

$$2C\,V_f + KC\,V_f = 3CV.$$

We now take $$C$$ common on the left-hand side:

$$C\,(2 + K)\,V_f = 3C\,V.$$

Dividing both sides by $$C(2 + K)$$ yields the final potential difference:

$$V_f = \frac{3V}{K + 2}.$$

Thus, after the dielectric is inserted, the potential across both capacitors becomes $$\displaystyle\frac{3V}{K+2}$$.

Hence, the correct answer is Option C.

We have a parallel-plate capacitor whose initial capacitance is given as $$C_0 = 200\;\mu\text{F} = 200 \times 10^{-6}\,\text{F}.$$

This capacitor is connected across a constant battery of potential difference $$V = 200\;\text{V},$$ and the battery remains connected throughout the process. Because the battery is always attached, the voltage across the plates stays fixed at 200 V even after any change in the capacitor.

First, let us determine the energy stored in the capacitor before inserting the dielectric. The expression for electrostatic energy of a capacitor at constant voltage is stated as $$U = \tfrac12 C V^{2}.$$ Substituting the initial values, we get $$U_i = \tfrac12 \, C_0 \, V^{2} = \tfrac12 \left(200 \times 10^{-6}\,\text{F}\right) \left(200\,\text{V}\right)^{2}.$$

Now, $$\left(200\,\text{V}\right)^{2} = 200^{2} = 40\,000.$$ So, $$U_i = \tfrac12 \left(200 \times 10^{-6}\right) \left(40\,000\right) = \tfrac12 \left(200 \times 40\,000 \times 10^{-6}\right).$$

Multiplying inside the parenthesis: $$200 \times 40\,000 = 8\,000\,000.$$ Thus, $$U_i = \tfrac12 \left(8\,000\,000 \times 10^{-6}\right) = \tfrac12 \left(8\right) = 4\;\text{J}.$$

Next, a dielectric slab of dielectric constant $$k = 2$$ is inserted completely between the plates. When a dielectric fills the entire space, the new capacitance is given by the known relation $$C_f = k\,C_0.$$ Hence, $$C_f = 2 \times 200 \times 10^{-6}\,\text{F} = 400 \times 10^{-6}\,\text{F}.$$

Because the battery is still connected, the voltage remains $$V = 200\;\text{V}.$$ We again use the same energy formula for the final state: $$U_f = \tfrac12 C_f V^{2} = \tfrac12 \left(400 \times 10^{-6}\,\text{F}\right) \left(200\,\text{V}\right)^{2}.$$

We already evaluated $$V^{2} = 40\,000.$$ Hence, $$U_f = \tfrac12 \left(400 \times 10^{-6} \times 40\,000\right) = \tfrac12 \left(400 \times 40\,000 \times 10^{-6}\right).$$

Compute the product inside: $$400 \times 40\,000 = 16\,000\,000.$$ Therefore, $$U_f = \tfrac12 \left(16\,000\,000 \times 10^{-6}\right) = \tfrac12 \left(16\right) = 8\;\text{J}.$$

The change in electrostatic energy is the difference between the final and initial energies: $$\Delta U = U_f - U_i = 8\;\text{J} - 4\;\text{J} = 4\;\text{J}.$$

So, the answer is $$4\;\text{J}$$.

We have 27 identical mercury drops, each maintained at a potential of 10 V. Let the radius of each small drop be $$r$$ and charge be $$q$$.

Since the volume is conserved when the drops merge, $$27 \times \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3$$, which gives $$R = 3r$$.

The total charge on the big drop is $$Q = 27q$$.

The potential energy (self-energy) of a charged sphere is $$U = \frac{kQ^2}{2R}$$.

For a small drop: $$U_s = \frac{kq^2}{2r}$$.

For the big drop: $$U_B = \frac{k(27q)^2}{2(3r)} = \frac{729 \, kq^2}{6r} = \frac{243 \, kq^2}{2r}$$.

Therefore, $$\frac{U_B}{U_s} = \frac{243 \, kq^2 / (2r)}{kq^2 / (2r)} = 243$$.

The potential energy of the bigger drop is $$243$$ times that of a smaller drop.

We have 512 identical mercury drops, each charged to a potential of 2 V. Let each small drop have radius $$r$$ and charge $$q$$. The potential of each small drop is $$V = \frac{kq}{r} = 2\,\text{V}$$.

When 512 drops merge into one large drop, charge is conserved: $$Q = 512q$$. Volume is also conserved: $$\frac{4}{3}\pi R^3 = 512 \times \frac{4}{3}\pi r^3$$, which gives $$R = 8r$$ (since $$512^{1/3} = 8$$).

The potential of the large drop is $$V' = \frac{kQ}{R} = \frac{k(512q)}{8r} = 64 \times \frac{kq}{r} = 64 \times 2 = 128\,\text{V}$$.

Therefore, the potential of the combined drop is $$128$$ V.

We have a parallel plate capacitor with capacitance $$C = 14$$ pF charged to a potential difference $$V = 12$$ V. After the battery is disconnected, a porcelain plate with dielectric constant $$k = 7$$ is inserted between the plates. We need to find the constant mechanical energy with which the dielectric plate oscillates.

The initial energy stored in the capacitor (before inserting the dielectric) is $$U_i = \frac{1}{2}CV^2 = \frac{1}{2} \times 14 \times 12^2 = \frac{1}{2} \times 14 \times 144 = 1008$$ pJ.

Since the battery is disconnected, the charge on the capacitor remains constant: $$Q = CV = 14 \times 12 = 168$$ pC. After the dielectric is fully inserted, the new capacitance becomes $$C' = kC = 7 \times 14 = 98$$ pF.

The final energy stored with the dielectric fully inside is $$U_f = \frac{Q^2}{2C'} = \frac{Q^2}{2kC} = \frac{U_i}{k} = \frac{1008}{7} = 144$$ pJ.

By conservation of energy, the decrease in electrical energy converts to mechanical energy (kinetic energy of the oscillating plate). The mechanical energy is $$E = U_i - U_f = 1008 - 144 = 864$$ pJ.

Since there is no friction, this mechanical energy remains constant throughout the oscillation, and the dielectric plate oscillates back and forth with a constant mechanical energy of $$\boxed{864}$$ pJ.

For a parallel-plate capacitor we first recall the fundamental relation for its capacitance.

Formula: $$C=\varepsilon_r\varepsilon_0\frac{A}{d}$$ where $$\varepsilon_r$$ is the relative permittivity, $$\varepsilon_0=8.854\times10^{-12}\,\text{F\,m}^{-1}$$ is the permittivity of free space, $$A$$ is the plate area and $$d$$ is the separation.

Re-arranging, the geometry factor becomes

$$\frac{A}{d}=\frac{C}{\varepsilon_r\varepsilon_0}$$.

Now we write the electrical resistance offered by the same slab of dielectric. For a material of resistivity $$\rho$$ placed between parallel plates,

Formula: $$R=\rho\frac{d}{A}$$.

We can eliminate the unknown geometry by substituting $$d/A$$ from the reciprocal of the earlier expression:

$$R=\rho\frac{d}{A}=\rho\left(\frac{1}{A/d}\right)=\rho\left(\frac{1}{C/(\varepsilon_r\varepsilon_0)}\right)=\rho\varepsilon_r\varepsilon_0\,\frac{1}{C}.$$

Substituting the numerical values $$\rho=200\,\Omega\text{ m},\; \varepsilon_r=50,\; C=2\;\text{pF}=2\times10^{-12}\,\text{F}$$, we get

$$R=200\times50\times8.854\times10^{-12}\,\frac{1}{2\times10^{-12}}$$ $$\;=10000\times8.854\times10^{-12}\,\frac{1}{2\times10^{-12}}$$ $$\;=8.854\times10^{-8}\,\frac{1}{2\times10^{-12}}$$ $$\;=\frac{8.854}{2}\times10^{(-8)-(-12)}$$ $$\;=4.427\times10^{4}\;\Omega$$ $$\;=4.427\times10^{4}\;\Omega \;(\text{about }44.3\;\text{k}\Omega).$$

The leakage current is obtained from Ohm’s law $$I=\dfrac{V}{R}$$, with the applied potential difference $$V=40\;\text{V}$$:

$$I=\frac{40}{4.427\times10^{4}}\;\text{A}$$ $$\;=0.904\times10^{-3}\;\text{A}$$ $$\;=9.04\times10^{-4}\;\text{A}$$ $$\;=0.9\;\text{mA}\;(\text{approximately}).$$

Hence, the correct answer is Option A.

We have a first capacitor of capacitance $$C_1 = 10\;\mu\text{F}$$ that is initially charged to a potential difference $$V_1 = 50\;\text{V}$$.

According to the basic relation for a capacitor,
$$Q = C\,V,$$
where $$Q$$ is the charge stored, $$C$$ is the capacitance and $$V$$ is the potential difference across the plates.

Using this formula, the initial charge on the first capacitor is

$$Q_1 = C_1\,V_1 = (10\;\mu\text{F})\,(50\;\text{V}) = 10 \times 10^{-6}\,\text{F}\;\times\;50\;\text{V} = 500 \times 10^{-6}\,\text{C} = 5 \times 10^{-4}\,\text{C}.$$

Now the battery is removed, so this charge $$Q_1$$ is trapped on the capacitor plates. The charged capacitor is then connected in parallel to another, previously uncharged, capacitor of capacitance $$C_2$$ (the value we must find).

Because the two capacitors are connected in parallel, after connection they share the same final potential difference. Let us denote this common final voltage by $$V_f$$. The problem states that this final voltage is

$$V_f = 20\;\text{V}.$$

No charge is lost to the external circuit (we assume ideal wires with no resistance and no radiation), so total charge is conserved. Therefore, the total charge after connection must still be equal to the initial charge $$Q_1$$. After connection, the total charge is the sum of the charges on each capacitor:

$$Q_{\text{total after}} = Q_1' + Q_2' = C_1\,V_f + C_2\,V_f = (C_1 + C_2)\,V_f.$$

By conservation of charge,

$$Q_1 = (C_1 + C_2)\,V_f.$$

We already know $$Q_1 = C_1 V_1$$, so substituting this into the conservation equation gives

$$C_1 V_1 = (C_1 + C_2)\,V_f.$$

Our aim is to solve this equation for $$C_2$$. First, divide both sides by $$V_f$$ to make $$C_2$$ appear alone:

$$\frac{C_1 V_1}{V_f} = C_1 + C_2.$$

Next, subtract $$C_1$$ from both sides:

$$C_2 = \frac{C_1 V_1}{V_f} - C_1.$$

Now substitute the known numerical values $$C_1 = 10\;\mu\text{F}$$, $$V_1 = 50\;\text{V}$$, and $$V_f = 20\;\text{V}$$:

$$C_2 = \frac{(10\;\mu\text{F})(50\;\text{V})}{20\;\text{V}} - 10\;\mu\text{F} = (10\;\mu\text{F})\left(\frac{50}{20}\right) - 10\;\mu\text{F}.$$

Simplify the fraction $$\frac{50}{20} = 2.5$$, so

$$C_2 = (10\;\mu\text{F})(2.5) - 10\;\mu\text{F} = 25\;\mu\text{F} - 10\;\mu\text{F} = 15\;\mu\text{F}.$$

Hence, the correct answer is Option A.

For electrostatic forces, the work done in moving a charge depends only on the initial and final positions because the electric field is conservative. Therefore, the change in energy of the moving charge equals its charge multiplied by the change in electric potential experienced during the displacement.

We have two fixed charges on the $$x$$-axis:

$$4q \text{ at } x=-\frac{d}{2},\qquad -q \text{ at } x=\frac{d}{2}.$$

A third charge $$q$$ is moved from the origin $$(0,0)$$ to the point $$(d,0)$$. Since potential is a state function, we only need the potentials at these two points.

The potential $$V$$ at a point due to a single point charge $$Q$$ located a distance $$r$$ away is given by the well-known formula

$$V=\frac{1}{4\pi\varepsilon_0}\,\frac{Q}{r}.$$

1. Potential at the origin, $$V_{\text{initial}}$$.

Distances from the origin to the two fixed charges:

$$r_1=\left|0-\!\Bigl(-\frac{d}{2}\Bigr)\right|=\frac{d}{2},\qquad r_2=\left|0-\!\Bigl(\frac{d}{2}\Bigr)\right|=\frac{d}{2}.$$

Hence

$$$ V_{\text{initial}}=\frac{1}{4\pi\varepsilon_0}\!\left( \frac{4q}{d/2}+\frac{-q}{d/2} \right) =\frac{1}{4\pi\varepsilon_0}\!\left( \frac{4q- q}{d/2} \right) =\frac{1}{4\pi\varepsilon_0}\,\frac{3q}{d/2} =\frac{1}{4\pi\varepsilon_0}\,\frac{6q}{d}. $$$

Thus

$$V_{\text{initial}}=\frac{6q}{4\pi\varepsilon_0 d}.$$

2. Potential at the final point $$(d,0)$$, $$V_{\text{final}}$$.

Distances from $$(d,0)$$ to the two fixed charges:

$$ R_1 = \left|d -\!\Bigl(-\frac{d}{2}\Bigr)\right| = \frac{3d}{2},\qquad R_2 = \left|d -\!\Bigl(\frac{d}{2}\Bigr)\right| = \frac{d}{2}. $$

Therefore

$$$ V_{\text{final}}=\frac{1}{4\pi\varepsilon_0}\!\left( \frac{4q}{3d/2}+\frac{-q}{d/2} \right) =\frac{1}{4\pi\varepsilon_0}\!\left( \frac{8q}{3d}-\frac{2q}{d} \right). $$$

We bring both fractions to a common denominator $$3d$$:

$$$ \frac{8q}{3d}-\frac{2q}{d} =\frac{8q}{3d}-\frac{6q}{3d} =\frac{2q}{3d}. $$$

Hence

$$V_{\text{final}}=\frac{1}{4\pi\varepsilon_0}\,\frac{2q}{3d} =\frac{q}{6\pi\varepsilon_0 d}.$$

3. Change in potential experienced by the moving charge.

$$\Delta V = V_{\text{final}}-V_{\text{initial}} =\frac{q}{6\pi\varepsilon_0 d}-\frac{6q}{4\pi\varepsilon_0 d}.$$

Writing both terms over the common denominator $$6\pi\varepsilon_0 d$$:

$$$ \Delta V =\frac{q}{6\pi\varepsilon_0 d}-\frac{9q}{6\pi\varepsilon_0 d} =-\frac{8q}{6\pi\varepsilon_0 d} =-\frac{4q}{3\pi\varepsilon_0 d}. $$$

4. Change in potential energy of the charge $$q$$.

The change in energy is simply $$q$$ times the change in potential:

$$$ \Delta U = q\,\Delta V =q\left(-\frac{4q}{3\pi\varepsilon_0 d}\right) =-\frac{4q^{2}}{3\pi\varepsilon_0 d}. $$$

The negative sign tells us that the energy decreases, and the magnitude of the decrease is $$\dfrac{4q^2}{3\pi\varepsilon_0 d}$$.

Hence, the correct answer is Option D.

We have two concentric metallic hollow spheres. The inner sphere has radius $$R$$ and carries charge $$Q_1$$. The outer sphere has radius $$4R$$ and carries charge $$Q_2$$.

Because the spheres are metallic and the charges reside only on their outer surfaces, the surface charge density of each sphere is given by the usual relation

$$\sigma = \dfrac{\text{total charge}}{\text{surface area}}.$$

For the inner sphere, the surface charge density is

$$\sigma_1=\dfrac{Q_1}{4\pi R^{2}}.$$

For the outer sphere, the surface charge density is

$$\sigma_2=\dfrac{Q_2}{4\pi (4R)^{2}}=\dfrac{Q_2}{4\pi \,16R^{2}}=\dfrac{Q_2}{64\pi R^{2}}.$$

The problem states that the two surface charge densities are equal, so

$$\sigma_1=\sigma_2\quad\Longrightarrow\quad \dfrac{Q_1}{4\pi R^{2}}=\dfrac{Q_2}{64\pi R^{2}}.$$

The factors $$4\pi R^{2}$$ cancel on both sides, leaving

$$Q_1=\dfrac{Q_2}{16}\quad\Longrightarrow\quad Q_2=16Q_1.$$

Now we evaluate the electric potential at two points: on the surface of the inner sphere (at radial distance $$r=R$$) and on the surface of the outer sphere (at $$r=4R$$). For any spherical shell, the standard electrostatic result is:

• Outside the shell, the potential behaves as if all charge were concentrated at the centre, i.e. $$V=\dfrac{kQ}{r}$$ with $$k=\dfrac{1}{4\pi\varepsilon_0}.$$
• Inside the shell, the potential is constant everywhere and equal to the value at the surface, i.e. $$V=kQ/R_{\text{shell}}.$$

Potential at $$r=R$$ (surface of the inner sphere)

1. Contribution of the inner sphere itself (point on its surface): $$V_{1\to R}=k\dfrac{Q_1}{R}.$$
2. Contribution of the outer sphere (point lies inside that shell, so potential is constant and equals the surface value of the outer shell): $$V_{2\to R}=k\dfrac{Q_2}{4R}.$$

Adding them,

$$V(R)=k\left(\dfrac{Q_1}{R}+\dfrac{Q_2}{4R}\right).$$

Potential at $$r=4R$$ (surface of the outer sphere)

1. Contribution of the inner sphere (point is outside the inner shell): $$V_{1\to 4R}=k\dfrac{Q_1}{4R}.$$
2. Contribution of the outer sphere itself (point on its surface): $$V_{2\to 4R}=k\dfrac{Q_2}{4R}.$$

Thus,

$$V(4R)=k\left(\dfrac{Q_1}{4R}+\dfrac{Q_2}{4R}\right) =k\dfrac{Q_1+Q_2}{4R}.$$

Required potential difference

Subtracting the two results,

$$\begin{aligned} V(R)-V(4R) & = k\left(\dfrac{Q_1}{R}+\dfrac{Q_2}{4R}\right) \;-\;k\left(\dfrac{Q_1+Q_2}{4R}\right) \\[4pt] & = k\left[\dfrac{Q_1}{R}-\dfrac{Q_1}{4R} +\dfrac{Q_2}{4R}-\dfrac{Q_2}{4R}\right] \\[6pt] & = k\left[\dfrac{4Q_1-Q_1}{4R}\right] \\[4pt] & = k\dfrac{3Q_1}{4R}. \end{aligned}$$

Finally, substituting $$k=\dfrac{1}{4\pi\varepsilon_0}$$, we obtain

$$V(R)-V(4R)=\dfrac{1}{4\pi\varepsilon_0}\,\dfrac{3Q_1}{4R} =\dfrac{3Q_1}{16\pi\varepsilon_0 R}.$$

This expression matches Option C.

Hence, the correct answer is Option C.

For an isolated charged metallic sphere of radius $$R$$ carrying charge $$Q$$, the electrostatic potential on its surface (with respect to infinity) is given by the well-known formula

$$V \;=\; \dfrac{1}{4\pi\varepsilon_0}\,\dfrac{Q}{R}.$$

The electric field just outside the surface of the same sphere is obtained from Coulomb’s law and is

$$E \;=\; \dfrac{1}{4\pi\varepsilon_0}\,\dfrac{Q}{R^{2}}.$$

Dividing the second expression by the first, we eliminate the common constant $$\dfrac{1}{4\pi\varepsilon_0}$$ and the charge $$Q$$:

$$\dfrac{E}{V} \;=\; \dfrac{\dfrac{Q}{R^{2}}}{\dfrac{Q}{R}} \;=\; \dfrac{1}{R}.$$

Re-arranging this simple relation, we obtain a very useful connection between the electric field and the potential of a spherical conductor:

$$E \;=\; \dfrac{V}{R}\quad\text{or equivalently}\quad V = E\,R.$$

Now we apply this result separately to the two spheres $$S_1$$ and $$S_2$$. For sphere $$S_1$$ we have $$V_1 = E_1\,R_1$$, and for sphere $$S_2$$ we have $$V_2 = E_2\,R_2$$. Taking the ratio of the two potentials, we write

$$\dfrac{V_1}{V_2} \;=\; \dfrac{E_1\,R_1}{E_2\,R_2} \;=\; \left(\dfrac{E_1}{E_2}\right)\!\left(\dfrac{R_1}{R_2}\right).$$

The problem statement tells us that the electric fields on the two spheres satisfy

$$\dfrac{E_1}{E_2} \;=\; \dfrac{R_1}{R_2}.$$

Substituting this given ratio into the previous expression, we have

$$\dfrac{V_1}{V_2} \;=\; \left(\dfrac{R_1}{R_2}\right)\!\left(\dfrac{R_1}{R_2}\right) \;=\; \left(\dfrac{R_1}{R_2}\right)^{2}.$$

Therefore, the ratio of the electrostatic potentials on the two spheres is the square of the ratio of their radii.

Hence, the correct answer is Option B.

We know that the electrostatic energy stored in a capacitor having capacitance $$C$$ and potential difference $$V$$ is given by the well-known formula

$$U=\frac{1}{2}CV^{2}\,.$$

Initially only one capacitor of capacitance $$C$$ is present and it is fully charged to a voltage $$V_0$$. Applying the above formula, its initial energy is

$$U_{\text{initial}}=\frac{1}{2}C{V_0}^{2}.$$

The corresponding initial charge on this capacitor is obtained from $$Q=CV$$. Hence,

$$Q_{\text{initial}}=C\,V_0.$$

Now the battery is removed and this charged capacitor is connected in parallel with another capacitor whose capacitance is $$\dfrac{C}{2}$$ and which is completely uncharged. Because the circuit is isolated, the total charge remains conserved while the two capacitors share the charge until they reach a common final voltage, say $$V_f$$.

In the parallel combination, the equivalent capacitance becomes

$$C_{\text{eq}}=C+\frac{C}{2}=\frac{3C}{2}.$$

Using charge conservation we write

$$Q_{\text{initial}}=Q_{\text{final}}.$$

But $$Q_{\text{final}}=C_{\text{eq}}\,V_f,$$ so

$$C\,V_0=\frac{3C}{2}\,V_f.$$

Solving this for the final voltage gives

$$V_f=\frac{2}{3}\,V_0.$$

Next we compute the total energy stored after equilibrium is reached. Using the same energy formula for the equivalent capacitance at the final common voltage, we have

$$U_{\text{final}}=\frac{1}{2}\,C_{\text{eq}}\,V_f^{2}=\frac{1}{2}\left(\frac{3C}{2}\right)\left(\frac{2}{3}V_0\right)^{2}.$$

Simplifying step by step:

$$U_{\text{final}}=\frac{1}{2}\cdot\frac{3C}{2}\cdot\frac{4}{9}\,V_0^{2} =\frac{3C}{4}\cdot\frac{4}{9}\,V_0^{2} =\frac{3\times4}{4\times9}\,C\,V_0^{2} =\frac{1}{3}C\,V_0^{2}.$$

The energy lost in the process is the difference between the initial and final energies:

$$\Delta U=U_{\text{initial}}-U_{\text{final}} =\frac{1}{2}C\,V_0^{2}-\frac{1}{3}C\,V_0^{2} =\left(\frac{3}{6}-\frac{2}{6}\right)C\,V_0^{2} =\frac{1}{6}C\,V_0^{2}.$$

Hence, the correct answer is Option D.

We start with a parallel-plate capacitor whose plates have length $$l$$, width $$w$$ and are separated by a distance $$d$$. The total plate area is therefore $$A = lw$$. While the capacitor is connected to a battery of constant emf $$V$$, its voltage remains fixed at $$V$$ throughout the entire process.

For any capacitor of capacitance $$C$$ kept at a fixed potential difference $$V$$, the electrostatic energy stored is given by the basic formula

$$U = \dfrac12\,C\,V^2.$$

Initial situation (no dielectric)
The initial capacitance is that of an empty parallel-plate capacitor,

$$C_i \;=\; \dfrac{\varepsilon_0 A}{d} \;=\; \dfrac{\varepsilon_0\,lw}{d}.$$

Therefore the initial energy is

$$U_i \;=\; \dfrac12\,C_i\,V^2 \;=\; \dfrac12\left(\dfrac{\varepsilon_0\,lw}{d}\right)V^2.$$

Situation after inserting the slab
A dielectric slab of the same thickness $$d$$ and dielectric constant $$K = 4$$ is now pushed in from one side. Let the length of slab already inside be $$x$$ (so the slab covers an area $$A_d = w\,x$$), while the remaining portion of the plates, having length $$l - x$$, is still filled with air. Because both regions share the same pair of plates and the same voltage $$V$$, the two regions act like two capacitors connected in parallel.

• Capacitance of the dielectric-filled region:

$$C_{\text{dielectric}} \;=\; \dfrac{K\,\varepsilon_0 A_d}{d} \;=\; \dfrac{4\,\varepsilon_0\,w\,x}{d}.$$

• Capacitance of the air-filled region:

$$C_{\text{air}} \;=\; \dfrac{\varepsilon_0 A_{\text{air}}}{d} \;=\; \dfrac{\varepsilon_0\,w\,(l - x)}{d}.$$

Because they are in parallel, their capacitances simply add:

$$C_f \;=\; C_{\text{air}} + C_{\text{dielectric}} \;=\; \dfrac{\varepsilon_0 w}{d}\,\bigl[(l - x) + 4x\bigr] \;=\; \dfrac{\varepsilon_0 w}{d}\,\bigl[l + 3x\bigr].$$

The new energy stored is therefore

$$U_f \;=\; \dfrac12\,C_f\,V^2 \;=\; \dfrac12\left[\dfrac{\varepsilon_0 w}{d}(l + 3x)\right]V^2.$$

Condition for doubling the energy
The problem asks for the slab length $$x$$ that makes the stored energy twice the initial energy, i.e.

$$U_f \;=\; 2\,U_i.$$

Substituting the expressions for $$U_f$$ and $$U_i$$ we have

$$\dfrac12\left[\dfrac{\varepsilon_0 w}{d}(l + 3x)\right]V^2 \;=\; 2\left[\dfrac12\left(\dfrac{\varepsilon_0\,lw}{d}\right)V^2\right].$$

The common factors $$\dfrac12,\,\varepsilon_0,\,w/d,\,V^2$$ cancel out immediately, leaving

$$l + 3x \;=\; 2l.$$

Now we isolate $$x$$ step by step:

$$3x \;=\; 2l - l \;=\; l,$$

$$x \;=\; \dfrac{l}{3}.$$

Thus the dielectric slab must be inserted to a length $$\dfrac{l}{3}$$ inside the plates in order for the energy stored in the capacitor to become double its original value.

Hence, the correct answer is Option B.

We are told that two capacitors $$C_1$$ and $$C_2$$ are first connected in parallel and their equivalent (effective) capacitance is measured to be $$10\;\mu\text{F}$$.

For capacitors joined in parallel, the equivalent capacitance is simply the sum of the individual capacitances. Therefore we write the relation first:

$$C_{\text{parallel}} \;=\; C_1 + C_2$$

Substituting the given value $$C_{\text{parallel}} = 10\;\mu\text{F}$$, we have

$$C_1 + C_2 = 10\;\mu\text{F} \quad -(1)$$

Next, each capacitor is separately connected to a voltage source of $$1\;\text{V}$$. The energy stored in a capacitor is given by the basic formula

$$U = \dfrac{1}{2}\,C\,V^{2}$$

Applying this formula to the first capacitor, which has capacitance $$C_1$$ and voltage $$V = 1\;\text{V}$$, we get

$$U_1 = \dfrac{1}{2}\,C_1\,(1)^2 = \dfrac{1}{2}C_1$$

In the same way, for the second capacitor $$C_2$$ we obtain

$$U_2 = \dfrac{1}{2}\,C_2\,(1)^2 = \dfrac{1}{2}C_2$$

The problem states that the energy stored in $$C_2$$ is four times the energy stored in $$C_1$$. Hence we set up the proportionality condition

$$U_2 = 4\,U_1$$

Substituting the expressions for $$U_1$$ and $$U_2$$ found above, we write

$$\dfrac{1}{2}C_2 = 4 \times \left(\dfrac{1}{2}C_1\right)$$

The factor $$\dfrac{1}{2}$$ is common on both sides, allowing it to be cancelled, giving us the direct relation between the capacitances:

$$C_2 = 4C_1 \quad -(2)$$

Now we possess two linear equations: equation (1) $$C_1 + C_2 = 10$$ and equation (2) $$C_2 = 4C_1$$. We substitute equation (2) into equation (1) to solve for $$C_1$$:

$$C_1 + 4C_1 = 10$$

Simplifying the left-hand side, we find

$$5C_1 = 10$$

Dividing both sides by 5, we obtain

$$C_1 = 2\;\mu\text{F}$$

To find $$C_2$$ we return to equation (2):

$$C_2 = 4C_1 = 4 \times 2\;\mu\text{F} = 8\;\mu\text{F}$$

Having identified the individual capacitances $$C_1 = 2\;\mu\text{F}$$ and $$C_2 = 8\;\mu\text{F}$$, we now connect them in series and determine their new equivalent capacitance. For capacitors in series, the standard formula is

$$\dfrac{1}{C_{\text{series}}} = \dfrac{1}{C_1} + \dfrac{1}{C_2}$$

Substituting the numerical values, we write

$$\dfrac{1}{C_{\text{series}}} = \dfrac{1}{2\;\mu\text{F}} + \dfrac{1}{8\;\mu\text{F}}$$

We express both fractions with a common denominator of $$8\;\mu\text{F}$$ to combine them easily:

$$\dfrac{1}{2\;\mu\text{F}} = \dfrac{4}{8\;\mu\text{F}}, \quad\text{and}\quad \dfrac{1}{8\;\mu\text{F}} = \dfrac{1}{8\;\mu\text{F}}$$

Adding these two gives

$$\dfrac{1}{C_{\text{series}}} = \dfrac{4}{8\;\mu\text{F}} + \dfrac{1}{8\;\mu\text{F}} = \dfrac{5}{8\;\mu\text{F}}$$

To solve for $$C_{\text{series}}$$ we now invert the fraction:

$$C_{\text{series}} = \dfrac{8\;\mu\text{F}}{5}$$

Carrying out the division, we find

$$C_{\text{series}} = 1.6\;\mu\text{F}$$

Hence, the correct answer is Option C.

We have two capacitors. The first has capacitance $$C$$ and was originally charged to the potential difference $$V$$, while the second has capacitance $$2C$$ and was originally charged to the potential difference $$2V$$.

For every capacitor the charge on its positive plate is related to the potential difference by the basic relation

$$Q = C \, \Delta V,$$

where $$\Delta V$$ is the potential of the positive plate minus that of the negative plate.

Hence the initial charges are

$$Q_1 = C \times V = CV,$$

$$Q_2 = (2C) \times (2V) = 4CV.$$

The plates therefore carry

positive plate of $$C$$ : $$+CV,$$    negative plate of $$C$$ : $$-CV,$$

positive plate of $$2C$$ : $$+4CV,$$    negative plate of $$2C$$ : $$-4CV.$$

Now the capacitors are connected in parallel but with opposite polarity: the positive plate of the first is joined to the negative plate of the second, and simultaneously the remaining two plates are joined together. Let us call the common point of the joined positive-negative plates node X, and the common point of the other two plates node Y.

The total charge that already resides on each node before any redistribution is obtained by simple addition:

$$Q_X^{\text{(initial)}} = (+CV) + (-4CV) = -3CV,$$

$$Q_Y^{\text{(initial)}} = (-CV) + (+4CV) = +3CV.$$

After the connection the charges are free to move inside each node until both plates in the same node come to a common potential. However, no charge can leave a node, so these total charges stay the same in the final state:

$$Q_X^{\text{(final)}} = -3CV, \qquad Q_Y^{\text{(final)}} = +3CV.$$

Let the final potentials of the two nodes be $$V_X$$ and $$V_Y$$, so that the final potential difference across each capacitor is

$$V_f = V_X - V_Y.$$

Using the basic formula again, the charges on the plates after equilibrium are

on node X from capacitor $$C$$ : $$Q_{1X}=C(V_X - V_Y),$$

on node X from capacitor $$2C$$ : the plate here is negative for this capacitor, so its charge is $$Q_{2X}=+2C(V_X - V_Y).$$

Therefore the total charge on node X in the final state is

$$Q_X^{\text{(final)}} = Q_{1X} + Q_{2X} = C(V_X - V_Y) + 2C(V_X - V_Y)= 3C(V_X - V_Y).$$

But this must equal the value already fixed for the node, namely $$-3CV$$. Equating,

$$3C(V_X - V_Y) = -3CV \quad\Longrightarrow\quad V_X - V_Y = -V.$$

The magnitude of the final potential difference is therefore

$$|V_f| = |V_X - V_Y| = V.$$

Thus each capacitor is finally at a potential difference of just $$V$$ in magnitude.

The energy stored in a capacitor is given by the well-known expression

$$U = \frac12 C (\Delta V)^2.$$

Applying this to the two capacitors after equilibrium:

$$U_1 = \frac12\,C\,(V)^2 = \frac12\,CV^2,$$

$$U_2 = \frac12\,(2C)\,(V)^2 = CV^2.$$

The total final energy is therefore

$$U_{\text{final}} = U_1 + U_2 = \frac12\,CV^2 + CV^2 = \frac32\,CV^2.$$

Hence, the correct answer is Option B.

We have a capacitor of capacitance $$C_1 = 60\ \text{pF} = 60 \times 10^{-12}\ \text{F}$$ that is first charged to a potential difference $$V_1 = 20\ \text{V}$$ by a battery.

For a capacitor charged to a voltage $$V$$, the energy stored is given by the formula

$$U = \dfrac{1}{2} C V^{2}.$$

Substituting the given values, the initial electrostatic energy is

$$ U_{\text{initial}} = \dfrac{1}{2}\,(60 \times 10^{-12})\,(20)^{2} = \dfrac{1}{2}\,(60 \times 10^{-12})\,(400) = 30 \times 10^{-12} \times 400 \times \dfrac{1}{2}. $$

Simplifying step by step,

$$ \dfrac{1}{2}\,(60 \times 10^{-12})\,(400) = 30 \times 10^{-12} \times 400 = 12 \times 10^{-9}\ \text{J}. $$

Thus $$U_{\text{initial}} = 12\ \text{nJ}.$$

Now the battery is disconnected, so the total charge $$Q$$ on the first capacitor remains fixed. Using $$Q = C V$$ we get

$$ Q = C_1 V_1 = (60 \times 10^{-12})(20) = 1200 \times 10^{-12}\ \text{C} = 1.2 \times 10^{-9}\ \text{C}. $$

This charged capacitor is then connected in parallel to a second identical but uncharged capacitor $$C_2 = 60\ \text{pF}$$. In a parallel combination the equivalent capacitance is the sum, so

$$ C_{\text{eq}} = C_1 + C_2 = 60\ \text{pF} + 60\ \text{pF} = 120\ \text{pF} = 120 \times 10^{-12}\ \text{F}. $$

Since no charge can escape the system, the total charge $$Q$$ now spreads over both capacitors. The final common voltage $$V_f$$ is obtained from

$$ V_f = \dfrac{Q}{C_{\text{eq}}} = \dfrac{1.2 \times 10^{-9}}{120 \times 10^{-12}} = \dfrac{1.2}{120} \times 10^{3} = 0.01 \times 10^{3} = 10\ \text{V}. $$

The final energy stored in both capacitors together is again given by $$U = \tfrac{1}{2}CV^{2}$$, now with the equivalent capacitance and the common voltage, so

$$ U_{\text{final}} = \dfrac{1}{2}\,(120 \times 10^{-12})\,(10)^{2} = \dfrac{1}{2}\,(120 \times 10^{-12})\,(100) = 60 \times 10^{-12} \times 100 = 6 \times 10^{-9}\ \text{J}. $$

Thus $$U_{\text{final}} = 6\ \text{nJ}.$$

The energy lost in the process is the difference between the initial and final energies:

$$ \Delta U = U_{\text{initial}} - U_{\text{final}} = 12\ \text{nJ} - 6\ \text{nJ} = 6\ \text{nJ}. $$

So, the answer is $$6\ \text{nJ}$$.

We are dealing with two identical positive charges. One charge $$q$$ is smeared uniformly on a ring of radius $$R = 3a$$ that lies in the $$x\text{-}y$$ plane and is centred at the origin. The second charge, also of magnitude $$q$$ and mass $$m$$, starts on the axis of the ring (the $$z$$-axis) at the point $$z_0 = 4a$$ with speed $$v$$ directed towards the origin. Only electrostatic forces act, so the mechanical energy of the moving charge is conserved.

First we need the electric potential due to the ring on its axis. For a ring of total charge $$q$$ and radius $$R$$, the standard formula for the potential at an axial distance $$z$$ from the centre is

$$ V(z) \;=\; \frac{1}{4\pi\varepsilon_0}\; \frac{q}{\sqrt{R^{2}+z^{2}}}. $$

The electrostatic potential energy of our point charge $$q$$ at a position $$z$$ is therefore

$$ U(z) \;=\; q\,V(z) \;=\; \frac{1}{4\pi\varepsilon_0}\; \frac{q^{2}}{\sqrt{R^{2}+z^{2}}}. $$

We now evaluate this energy at the two positions of interest.

Initial position $$z_0 = 4a$$:

$$ U_{\text{initial}} = \frac{1}{4\pi\varepsilon_0}\; \frac{q^{2}}{\sqrt{R^{2}+z_0^{2}}} = \frac{1}{4\pi\varepsilon_0}\; \frac{q^{2}}{\sqrt{(3a)^{2} + (4a)^{2}}} = \frac{1}{4\pi\varepsilon_0}\; \frac{q^{2}}{\sqrt{9a^{2} + 16a^{2}}} = \frac{1}{4\pi\varepsilon_0}\; \frac{q^{2}}{5a}. $$

Final position (the origin) $$z = 0$$:

$$ U_{\text{final}} = \frac{1}{4\pi\varepsilon_0}\; \frac{q^{2}}{\sqrt{R^{2}+0}} = \frac{1}{4\pi\varepsilon_0}\; \frac{q^{2}}{R} = \frac{1}{4\pi\varepsilon_0}\; \frac{q^{2}}{3a}. $$

Because the ring and the charge have the same sign, the potential energy nearer the ring is higher: $$U_{\text{final}} > U_{\text{initial}}.$$ To reach the origin, the moving charge must supply exactly the additional energy difference out of its initial kinetic energy. We therefore employ the principle of conservation of mechanical energy:

$$ \text{Initial kinetic} + \text{Initial potential} \;=\; \text{Final kinetic} + \text{Final potential}. $$

For the minimum speed required to just cross the origin, the charge will momentarily come to rest there, so the final kinetic energy is zero. Writing this out gives

$$ \frac{1}{2}\,m\,v^{2} + U_{\text{initial}} \;=\; U_{\text{final}}. $$

Substituting the two potential energies derived above, we have

$$ \frac{1}{2}\,m\,v^{2} \;=\; U_{\text{final}} - U_{\text{initial}} = \frac{1}{4\pi\varepsilon_0}\,q^{2} \left( \frac{1}{3a} - \frac{1}{5a} \right). $$

Now we simplify the bracket:

$$ \frac{1}{3a} - \frac{1}{5a} = \frac{5 - 3}{15a} = \frac{2}{15a}. $$

Substituting this back,

$$ \frac{1}{2}\,m\,v^{2} = \frac{1}{4\pi\varepsilon_0}\,q^{2}\; \frac{2}{15a}. $$

Multiplying both sides by 2 and dividing by $$m$$ to isolate $$v^{2}$$ gives

$$ v^{2} = \frac{2}{m}\; \frac{1}{4\pi\varepsilon_0}\,q^{2}\; \frac{2}{15a} = \frac{4}{15m}\; \frac{q^{2}}{4\pi\varepsilon_0\,a}. $$

Taking the square root to obtain the minimum speed, we find

$$ v = \sqrt{\frac{4}{15m}\; \frac{q^{2}}{4\pi\varepsilon_0\,a}} = \sqrt{\frac{2}{m}}\; \sqrt{\frac{2}{15}\; \frac{q^{2}}{4\pi\varepsilon_0\,a}}. $$

This matches option D exactly.

Hence, the correct answer is Option D.

We have two large, parallel conducting plates of a capacitor. Because the plates are very close and very wide, the electric field between them is practically uniform. For such a configuration, electrostatics gives the direct relation between the electric field $$E$$ and the surface charge density $$\sigma$$ on either plate:

$$E \;=\; \dfrac{\sigma}{\epsilon_0}.$$

Here $$\epsilon_0$$ is the permittivity of free space. We are given

$$E = 100 \; \text{N/C}, \qquad \epsilon_0 = 8.85 \times 10^{-12} \; \dfrac{\text{C}^2}{\text{N}\cdot\text{m}^2}.$$

Re-arranging the formula to solve for $$\sigma$$, we write

$$\sigma = \epsilon_0 \, E.$$

Substituting the numerical values,

$$\sigma = \left( 8.85 \times 10^{-12} \; \dfrac{\text{C}^2}{\text{N}\cdot\text{m}^2} \right) \!\times\! \left( 100 \; \dfrac{\text{N}}{\text{C}} \right).$$

Now we multiply the powers of ten and the coefficients step by step:

$$8.85 \times 10^{-12} \times 100 \;=\; 8.85 \times 10^{-12} \times 10^{2} \;=\; 8.85 \times 10^{-10}.$$

Hence,

$$\sigma = 8.85 \times 10^{-10} \; \dfrac{\text{C}}{\text{m}^2}.$$

The plates each have an area

$$A = 1 \; \text{m}^2.$$

The total charge $$Q$$ residing on one plate is the surface charge density times the area:

$$Q = \sigma \, A.$$

Substituting $$\sigma$$ and $$A$$,

$$Q = \left( 8.85 \times 10^{-10} \; \dfrac{\text{C}}{\text{m}^2} \right) \!\times\! \left( 1 \; \text{m}^2 \right) = 8.85 \times 10^{-10} \; \text{C}.$$

This value represents the magnitude of charge on each plate of the capacitor. The separation between the plates (0.1 m) is immaterial in this calculation because the relation $$E = \sigma / \epsilon_0$$ already embodies the correct dependence for an ideal parallel-plate capacitor.

Hence, the correct answer is Option A.

 Step 1: Use relation

$$q=CV$$

 Slope of q-V graph = Capacitance

 Step 2: Identify graphs

• Steeper line (A) ⇒ parallel combination (higher capacitance)
• Less steep line (B) ⇒ series combination

 Step 3: Read values from graph

$$At\ V=10V:$$

• For A: q=500 μCq C

$$Cp​=\ \frac{\ 500}{10}=50μF$$

• For B: q=80 μCq 

$$C_s=\ \frac{\ 80}{10}​=8μF$$

 Step 4: Match with given capacitors

Let capacitors be C1=40 μF,  C2=10 μFC

• Parallel:

Cp=40+10=50 μFC

• Series:

Cs=40×1040+10=8 μFC

Let the common surface charge density on every shell be $$\sigma$$ (in $$\text{C m}^{-2}$$). For a spherical conducting shell of radius $$R$$, the charge $$q$$ on it is related to $$\sigma$$ by the elementary relation

$$q \;=\; \sigma \,\times\, 4\pi R^{2}.$$

Therefore, if $$q_{a},\,q_{b},\,q_{c}$$ are the charges on the shells of radii $$a,\,b,\,c$$ respectively, we have

$$\begin{aligned} q_{a} &= \sigma \, 4\pi a^{2},\\ q_{b} &= \sigma \, 4\pi b^{2},\\ q_{c} &= \sigma \, 4\pi c^{2}. \end{aligned}$$

The total given charge is $$Q$$, so

$$Q \;=\; q_{a}+q_{b}+q_{c} \;=\; \sigma \, 4\pi \left(a^{2}+b^{2}+c^{2}\right).$$

Solving this expression for $$\sigma$$ gives

$$\sigma \;=\; \dfrac{Q}{4\pi\left(a^{2}+b^{2}+c^{2}\right)}.$$

Next we calculate the electric potential at a point whose distance from the common centre is $$r$$, with $$r<a$$. For a charged conducting spherical shell, the standard electrostatic result states:

$$\text{Potential at any interior point} \;=\; \dfrac{1}{4\pi\varepsilon_{0}}\dfrac{q}{R},$$

because the entire shell’s charge may be treated as if it were concentrated at its centre for points lying inside the shell.

Hence the contributions of the three shells to the required potential $$V(r)$$ are

$$\begin{aligned} V_{a} &= \dfrac{1}{4\pi\varepsilon_{0}}\,\dfrac{q_{a}}{a},\\[4pt] V_{b} &= \dfrac{1}{4\pi\varepsilon_{0}}\,\dfrac{q_{b}}{b},\\[4pt] V_{c} &= \dfrac{1}{4\pi\varepsilon_{0}}\,\dfrac{q_{c}}{c}. \end{aligned}$$

Adding these, the total potential is

$$\begin{aligned} V(r) &= \dfrac{1}{4\pi\varepsilon_{0}}\left(\dfrac{q_{a}}{a}+\dfrac{q_{b}}{b}+\dfrac{q_{c}}{c}\right)\\[6pt] &= \dfrac{1}{4\pi\varepsilon_{0}}\left[\dfrac{\sigma\,4\pi a^{2}}{a}+\dfrac{\sigma\,4\pi b^{2}}{b}+\dfrac{\sigma\,4\pi c^{2}}{c}\right]\\[6pt] &= \dfrac{1}{4\pi\varepsilon_{0}}\;4\pi\sigma\left(a+b+c\right)\\[6pt] &= \dfrac{\sigma\left(a+b+c\right)}{\varepsilon_{0}}. \end{aligned}$$

Now we substitute the value of $$\sigma$$ obtained earlier:

$$\begin{aligned} V(r) &= \dfrac{1}{\varepsilon_{0}}\left[\dfrac{Q}{4\pi\left(a^{2}+b^{2}+c^{2}\right)}\right]\left(a+b+c\right)\\[8pt] &= \dfrac{Q\left(a+b+c\right)}{4\pi\varepsilon_{0}\left(a^{2}+b^{2}+c^{2}\right)}. \end{aligned}$$

This expression matches Option D.

Hence, the correct answer is Option D.

We have a solid conducting sphere of radius $$R_1$$ carrying a charge $$+Q$$. It is concentrically surrounded by a conducting hollow spherical shell whose inner radius is $$R_2$$ and outer radius is $$R_3$$. Initially the shell is uncharged.

Statement of the electrostatic condition inside a conductor: the electric field inside the material of a conductor must be zero. Therefore any free charge present resides only on its surface(s), and the potential is the same at every point of the metallic body.

Because the inner solid sphere has charge $$+Q$$, a charge $$-Q$$ is induced on the inner surface of the hollow shell (radius $$R_2$$), leaving a compensating charge $$+Q$$ on the outer surface (radius $$R_3$$) so that the shell as a whole remains neutral. Thus the initial charge distribution is

$$\begin{aligned} \text{Sphere (}\,r=R_1\,\text{)} &: +Q,\\ \text{Inner surface of shell (}\,r=R_2\,\text{)} &: -Q,\\ \text{Outer surface of shell (}\,r=R_3\,\text{)} &: +Q. \end{aligned}$$

We now evaluate the initial potential difference between the surface of the sphere ($$r=R_1$$) and the outer surface of the shell ($$r=R_3$$).

Formula stated: the potential at a point due to a point charge is $$V = k\displaystyle\frac{q}{r}$$ with $$k=\dfrac{1}{4\pi\varepsilon_0}$$. A charged spherical shell behaves as though its entire charge were concentrated at the centre for an external point, while the potential is uniform inside the shell.

Initial potential on the sphere’s surface:

$$\begin{aligned} V_{R_1} &= k\frac{(+Q)}{R_1} \;+\; k\frac{(-Q)}{R_2} \;+\; k\frac{(+Q)}{R_3}\\ &= kQ\left(\frac{1}{R_1}-\frac{1}{R_2}+\frac{1}{R_3}\right). \end{aligned}$$

Initial potential on the outer surface of the shell (all three charges are now “inside” the Gaussian surface of radius $$R_3$$):

$$\begin{aligned} V_{R_3} &= k\frac{(+Q)}{R_3} \;+\; k\frac{(-Q)}{R_3} \;+\; k\frac{(+Q)}{R_3}\\ &= kQ\left(\frac{1}{R_3}-\frac{1}{R_3}+\frac{1}{R_3}\right)\\ &= kQ\left(\frac{1}{R_3}\right). \end{aligned}$$

Hence the initial potential difference is

$$\begin{aligned} V &= V_{R_1}-V_{R_3}\\ &= kQ\left(\frac{1}{R_1}-\frac{1}{R_2}+\frac{1}{R_3}-\frac{1}{R_3}\right)\\ &= kQ\left(\frac{1}{R_1}-\frac{1}{R_2}\right). \end{aligned}$$

Now the shell is given an additional charge $$-\,4Q$$. The shell already possessed $$0$$ net charge; after accepting $$-4Q$$ its net charge becomes $$-4Q$$. However, the electrostatic shielding requirement inside the conductor remains: the electric field in the metal of the shell must still be zero. Consequently the charge on the inner surface must stay at exactly $$-Q$$ to cancel the field arising from the central $$+Q$$. Whatever is left, namely $$-4Q-(-Q) = -3Q$$, must appear on the outer surface.

The new charge distribution therefore is

$$\begin{aligned} \text{Sphere (}\,r=R_1\,\text{)} &: +Q,\\ \text{Inner surface of shell (}\,r=R_2\,\text{)} &: -Q,\\ \text{Outer surface of shell (}\,r=R_3\,\text{)} &: -3Q. \end{aligned}$$

We now compute the new potentials with exactly the same method.

New potential on the sphere’s surface:

$$\begin{aligned} V'_{R_1} &= k\frac{(+Q)}{R_1} \;+\; k\frac{(-Q)}{R_2} \;+\; k\frac{(-3Q)}{R_3}\\ &= kQ\left(\frac{1}{R_1}-\frac{1}{R_2}-\frac{3}{R_3}\right). \end{aligned}$$

New potential on the outer surface of the shell:

$$\begin{aligned} V'_{R_3} &= k\frac{(+Q)}{R_3} \;+\; k\frac{(-Q)}{R_3} \;+\; k\frac{(-3Q)}{R_3}\\ &= kQ\left(\frac{1}{R_3}-\frac{1}{R_3}-\frac{3}{R_3}\right)\\ &= kQ\left(-\frac{3}{R_3}\right). \end{aligned}$$

Finally, the new potential difference is

$$\begin{aligned} \Delta V' &= V'_{R_1}-V'_{R_3}\\ &= \left[kQ\left(\frac{1}{R_1}-\frac{1}{R_2}-\frac{3}{R_3}\right)\right] - \left[kQ\left(-\frac{3}{R_3}\right)\right]\\ &= kQ\left(\frac{1}{R_1}-\frac{1}{R_2}-\frac{3}{R_3}+\frac{3}{R_3}\right)\\ &= kQ\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\\ &= V. \end{aligned}$$

The potential difference between the surface of the solid sphere and the outer surface of the hollow shell remains exactly the same as before, namely $$V$$.

Hence, the correct answer is Option C.

We have two large square plates of side $$a$$ separated by a small distance $$d$$, with $$d \ll a$$ so that edge effects can be ignored and the electric field between the plates is practically uniform in the vertical direction. The space between the plates is partly air and partly filled with a dielectric of constant $$K$$. The interface between the dielectric and the air is a straight line joining the bottom left corner of the plates to the top right corner, so that at the left edge there is no dielectric while at the right edge the entire gap is filled with dielectric. In the cross-section (an $$x\!-\!y$$ plane) the dielectric therefore occupies the lower triangular region, and the air occupies the upper triangular region.

To find the total capacitance we slice the plates into thin vertical strips, each of width $$dx$$, parallel to the interface. Every strip behaves as an independent capacitor because all strips have the same potential difference $$V$$ across the plates. Hence the capacitances of the strips are in parallel and will add algebraically.

For a strip situated at a horizontal coordinate $$x$$ (measured from the left edge, $$0 \le x \le a$$), the diagonal interface meets the strip at a height that is proportional to $$x$$. Since the interface goes from the bottom left corner $$(x=0,y=0)$$ to the top right corner $$(x=a,y=d)$$, the thickness of the dielectric layer inside this strip is

$$t_2=\frac{x}{a}\,d,$$

and the remaining thickness is air, namely

$$t_1=d-t_2=d-\frac{x}{a}d=d\Bigl(1-\frac{x}{a}\Bigr).$$

Thus, along the field direction, the air layer and the dielectric layer are connected in series. For a series combination of two capacitors of plate area $$A$$ and thicknesses $$t_1$$ and $$t_2$$ with dielectric constants $$\varepsilon_1$$ and $$\varepsilon_2$$, the formula is first stated:

$$\frac{1}{C_{\text{series}}}=\frac{t_1}{\varepsilon_1 A}+\frac{t_2}{\varepsilon_2 A}.$$

Here $$\varepsilon_1=\varepsilon_0$$ (air), $$\varepsilon_2=K\varepsilon_0$$ (dielectric), and the area of the strip is

$$A= (\text{depth})\times (\text{width}) = a\;dx.$$

Substituting these into the formula we get

$$\frac{1}{dC(x)}=\frac{t_1}{\varepsilon_0 A}+\frac{t_2}{K\varepsilon_0 A}

=\frac{d\bigl(1-\dfrac{x}{a}\bigr)}{\varepsilon_0 a\,dx}

+\frac{d\dfrac{x}{a}}{K\varepsilon_0 a\,dx}.$$

Combining the two fractions in the numerator:

$$\frac{1}{dC(x)}=\frac{d}{\varepsilon_0 a\,dx}

\Bigl[\,1-\frac{x}{a}+\frac{x}{aK}\Bigr]

=\frac{d}{\varepsilon_0 a\,dx}

\Bigl[1-\frac{x}{a}\Bigl(1-\frac{1}{K}\Bigr)\Bigr].$$

Hence the capacitance of the strip is

$$dC(x)=\frac{\varepsilon_0 a\,dx}

{d\Bigl[1-\dfrac{x}{a}\Bigl(1-\dfrac{1}{K}\Bigr)\Bigr]}.$$

For convenience define

$$\alpha=1-\frac{1}{K},$$

so that

$$dC(x)=\frac{\varepsilon_0 a\,dx}{d\bigl[1-\dfrac{\alpha x}{a}\bigr]}.$$

All these strip capacitances are in parallel, so the total capacitance is obtained by integrating $$dC(x)$$ from $$x=0$$ to $$x=a$$:

$$C=\int_{0}^{a} dC(x)=\frac{\varepsilon_0 a}{d}\int_{0}^{a}

\frac{dx}{1-\dfrac{\alpha x}{a}}.$$

We now carry out the integral. First rewrite the denominator:

$$1-\frac{\alpha x}{a}=1-\alpha\left(\frac{x}{a}\right).$$

Let us substitute

$$u=1-\alpha\left(\frac{x}{a}\right)\quad\Longrightarrow\quad

du=-\frac{\alpha}{a}\,dx\quad\Longrightarrow\quad

dx=-\frac{a}{\alpha}\,du.$$

When $$x=0$$, $$u=1$$; when $$x=a$$, $$u=1-\alpha.$$ Therefore

$$

\begin{aligned}

C & =\frac{\varepsilon_0 a}{d}\int_{x=0}^{x=a}

\frac{dx}{1-\dfrac{\alpha x}{a}}

=\frac{\varepsilon_0 a}{d}\int_{u=1}^{u=1-\alpha}

\frac{-\dfrac{a}{\alpha}\,du}{u} \\

& =\frac{\varepsilon_0 a}{d}\cdot\frac{a}{\alpha}

\int_{u=1-\alpha}^{u=1}\frac{du}{u} \\

& =\frac{\varepsilon_0 a^2}{d\alpha}

\bigl[\ln u\bigr]_{u=1-\alpha}^{u=1}.

\end{aligned}

$$

Evaluating the limits:

$$

\begin{aligned}

\bigl[\ln u\bigr]_{1-\alpha}^{1}

&=\ln(1)-\ln(1-\alpha)\\

&=0-\ln(1-\alpha)\\

&=-\ln(1-\alpha).

\end{aligned}

$$

So

$$C=\frac{\varepsilon_0 a^2}{d\alpha}\,\bigl[-\ln(1-\alpha)\bigr]

=\frac{\varepsilon_0 a^2}{d\alpha}\,\ln\!\Bigl(\frac{1}{1-\alpha}\Bigr).$$

Recall that $$\alpha=1-\dfrac{1}{K} \; \Rightarrow \; 1-\alpha=\dfrac{1}{K}.$$ Substituting this back in gives

$$

C=\frac{\varepsilon_0 a^2}{d\alpha}\,\ln(K)

=\frac{\varepsilon_0 a^2}{d}\;

\frac{1}{1-\dfrac{1}{K}}\,\ln K.

$$

Simplifying the fraction:

$$

\frac{1}{1-\dfrac{1}{K}}

=\frac{1}{\dfrac{K-1}{K}}

=\frac{K}{K-1}.

$$

Therefore

$$

C=\frac{\varepsilon_0 a^2}{d}\,

\frac{K}{K-1}\,\ln K

=\frac{K\varepsilon_0 a^2}{d\,(K-1)}\,\ln K.

$$

This matches Option 2 in the given list.

Hence, the correct answer is Option 2.

We have the electric field in the region given as $$\vec E = (Ax + B)\,\hat i$$, where the constants are $$A = 20\text{ (SI units)}$$ and $$B = 10\text{ (SI units)}$$.

For one-dimensional situations, the relation between electric field and electric potential is stated first:

$$E_x = -\frac{dV}{dx}.$$

This can be rewritten as

$$dV = -E_x\,dx.$$

The potential difference between two points $$x = x_1$$ and $$x = x_2$$ is therefore obtained by integrating:

$$V(x_2) - V(x_1) = -\int_{x_1}^{x_2} E_x \, dx.$$

In our problem the two positions are $$x_1 = -5\text{ m}$$ and $$x_2 = 1\text{ m}$$, with corresponding potentials $$V_2$$ and $$V_1$$. Substituting these limits,

$$V_1 - V_2 = -\int_{-5}^{1} (Ax + B)\,dx.$$

Now we substitute the numerical values of $$A$$ and $$B$$:

$$V_1 - V_2 = -\int_{-5}^{1} \bigl(20x + 10\bigr)\,dx.$$

We carry out the algebraic integration term by term. The integral of $$20x$$ is $$20 \cdot \dfrac{x^2}{2} = 10x^2$$, and the integral of $$10$$ is $$10x$$. So,

$$\int (20x + 10)\,dx \;=\; 10x^2 + 10x.$$

Placing the limits gives

$$V_1 - V_2 = -\Bigl[\,10x^2 + 10x\,\Bigr]_{x=-5}^{x=1}.$$

First we evaluate the bracket at the upper limit $$x = 1$$:

$$10(1)^2 + 10(1) = 10 + 10 = 20.$$

Next we evaluate at the lower limit $$x = -5$$:

$$10(-5)^2 + 10(-5) = 10(25) - 50 = 250 - 50 = 200.$$

Now we find the difference of the two bracketed values:

$$\Bigl(10x^2 + 10x\Bigr)_{1} - \Bigl(10x^2 + 10x\Bigr)_{-5} = 20 - 200 = -180.$$

Finally, we remember the overall minus sign outside the bracket:

$$V_1 - V_2 = -(-180) = 180\ \text{V}.$$

Hence, the correct answer is Option C.

First we note the basic relation for any capacitor, namely the charge-voltage relation $$Q = C\,V$$ where $$C$$ is the capacitance, $$V$$ the potential difference across its plates and $$Q$$ the magnitude of charge stored on either plate.

The two given capacitors are air-filled parallel-plate capacitors having capacitances

$$C_1 = C \qquad\text{and}\qquad C_2 = nC.$$

Because they are connected in parallel to the same battery, the potential difference across each capacitor is initially the battery voltage $$V$$. Using $$Q = C\,V$$ we obtain their individual charges just after connection to the battery:

$$Q_1^{\text{(initial)}} = C_1 V = C\,V,$$

$$Q_2^{\text{(initial)}} = C_2 V = nC\,V.$$

The total charge residing on the combination is therefore

$$Q_{\text{total}}^{\text{(initial)}} \;=\; Q_1^{\text{(initial)}} + Q_2^{\text{(initial)}} \;=\; C\,V + nC\,V \;=\; (n + 1)\,C\,V.$$

Now the battery is carefully removed. When the external circuit is opened in this way, no further charge can flow into or out of the parallel combination, so the total charge remains fixed at the value just calculated:

$$Q_{\text{total}} = (n + 1)\,C\,V \quad\text{(conserved)}.$$

After the battery has been detached, a dielectric slab of dielectric constant $$K$$ is inserted fully between the plates of the first capacitor only. Inserting a dielectric multiplies the capacitance by $$K$$, so the new capacitances are

$$C_1' = K\,C \qquad\text{and}\qquad C_2' = nC \;(\text{unchanged}).$$

Because the connecting wires joining the two capacitors are still intact, the two capacitors remain in parallel and must therefore share a common new potential difference. Let us denote this unknown final potential difference by $$V'$$.

Using $$Q = C\,V$$ once more, the individual charges after insertion of the dielectric become

$$Q_1' = C_1' V' = K\,C\,V',$$

$$Q_2' = C_2' V' = nC\,V'.$$

The total charge after the insertion is thus

$$Q_{\text{total}}^{\text{(final)}} = Q_1' + Q_2' = K\,C\,V' + nC\,V' = (K + n)\,C\,V'.$$

But charge is conserved, so this final total charge must equal the initial total charge:

$$(K + n)\,C\,V' = (n + 1)\,C\,V.$$

We divide both sides by $$C$$ to eliminate the common factor and then solve for $$V'$$:

$$(K + n)\,V' = (n + 1)\,V,$$

$$V' = \frac{(n + 1)\,V}{K + n}.$$

This is the new potential difference across each capacitor in the parallel combination after the dielectric has been introduced.

Hence, the correct answer is Option A.

We are dealing with a parallel-plate capacitor completely filled with a dielectric. For such a capacitor the capacitance is given by the well-known formula

$$C \;=\; \kappa \,\varepsilon_0 \,\dfrac{A}{d},$$

where $$C$$ is the capacitance, $$\kappa$$ is the dielectric constant (relative permittivity) we have to find, $$\varepsilon_0$$ is the permittivity of free space, $$A$$ is the plate area and $$d$$ is the separation between the plates.

The separation $$d$$ is not given directly, but we do know the largest voltage the capacitor can sustain (its voltage rating) and the largest electric field the dielectric can withstand. The electric field in a parallel-plate capacitor is related to the voltage by the basic relation

$$E \;=\; \dfrac{V}{d}.$$

Re-arranging this relation to express $$d$$ in terms of the voltage $$V$$ and the maximum permissible field $$E_{\text{max}}$$, we get

$$d \;=\; \dfrac{V_{\text{max}}}{E_{\text{max}}}.$$

Now we substitute the given numerical values. The maximum voltage is $$V_{\text{max}} = 500 \text{ V}$$ and the maximum field is $$E_{\text{max}} = 10^{6} \text{ V m}^{-1}$$. Hence

$$d \;=\; \dfrac{500}{10^{6}} \text{ m} \;=\; 5 \times 10^{-4} \text{ m}.$$

Next, we substitute this $$d$$ into the capacitance formula to solve for the dielectric constant $$\kappa$$. First rewrite the capacitance formula explicitly for $$\kappa$$:

$$\kappa \;=\; \dfrac{C\,d}{\varepsilon_0\,A}.$$

All quantities in this expression are now known:

Capacitance: $$C = 15 \text{ pF} = 15 \times 10^{-12} \text{ F}$$.

Plate separation: $$d = 5 \times 10^{-4} \text{ m}$$ (found above).

Permittivity of free space: $$\varepsilon_0 = 8.86 \times 10^{-12} \text{ C}^2\text{ N}^{-1}\text{ m}^{-2}$$.

Plate area: $$A = 10^{-4} \text{ m}^2$$.

Substituting each value carefully we obtain

$$\kappa \;=\; \dfrac{(15 \times 10^{-12})\,(5 \times 10^{-4})}{(8.86 \times 10^{-12})\,(10^{-4})}.$$

First, multiply the numbers in the numerator:

$$15 \times 5 = 75,$$

and combine the powers of ten:

$$10^{-12} \times 10^{-4} = 10^{-16}.$$

Thus the numerator becomes

$$75 \times 10^{-16} = 7.5 \times 10^{-15}.$$

For the denominator, multiply the constants:

$$8.86 \times 1 = 8.86,$$

and combine the powers of ten:

$$10^{-12} \times 10^{-4} = 10^{-16}.$$

Hence, the denominator is

$$8.86 \times 10^{-16}.$$

Now we divide numerator by denominator:

$$\kappa \;=\; \dfrac{7.5 \times 10^{-15}}{8.86 \times 10^{-16}}.$$

Because both numerator and denominator contain a factor of $$10^{-16}$$, canceling that common factor leaves a single factor of $$10^{+1}$$ (since $$10^{-15}/10^{-16} = 10^{1}$$):

$$\kappa \;=\; \dfrac{7.5}{8.86}\,\times\,10^{1}.$$

Evaluating the numerical fraction,

$$\dfrac{7.5}{8.86} \approx 0.846.$$

Multiplying by $$10^{1}$$ (that is, by 10) gives

$$\kappa \approx 0.846 \times 10 = 8.46.$$

The value rounds to $$\kappa \approx 8.5$$, matching one of the listed choices.

Hence, the correct answer is Option B.

We begin by noting that the charge on the capacitor is kept constant, because the plates are merely pulled apart without any external circuit connected. So throughout the process we have $$Q = 5\;\mu\text{C} = 5 \times 10^{-6}\;\text{C}.$$

The electrostatic energy stored in a charged capacitor in terms of its charge and capacitance is given by the formula

$$U = \frac{Q^{2}}{2C}.$$

Initially, the capacitance is $$C_i = 5\;\mu\text{F} = 5 \times 10^{-6}\;\text{F}.$$ Substituting the known values into the energy formula, we find

$$U_i \;=\; \frac{Q^{2}}{2C_i} \;=\; \frac{(5 \times 10^{-6})^{2}}{2 \times 5 \times 10^{-6}}\; \text{J}.$$

Carrying out the algebra step by step, first square the charge:

$$Q^{2} = (5 \times 10^{-6})^{2} = 25 \times 10^{-12}.$$

Now form the denominator:

$$2 \times C_i = 2 \times 5 \times 10^{-6} = 10 \times 10^{-6} = 1 \times 10^{-5}.$$

So

$$U_i = \frac{25 \times 10^{-12}}{1 \times 10^{-5}} = 25 \times 10^{-12}\times 10^{5} = 25 \times 10^{-7}\;\text{J} = 2.5 \times 10^{-6}\;\text{J}.$$

Next, the plates are separated so that the capacitance becomes $$C_f = 2\;\mu\text{F} = 2 \times 10^{-6}\;\text{F}.$$ The charge is still $$Q = 5 \times 10^{-6}\;\text{C},$$ so the final energy is

$$U_f \;=\; \frac{Q^{2}}{2C_f} \;=\; \frac{25 \times 10^{-12}}{2 \times 2 \times 10^{-6}}\; \text{J}.$$

Simplifying the denominator gives

$$2 \times 2 \times 10^{-6} = 4 \times 10^{-6},$$

and therefore

$$U_f = \frac{25 \times 10^{-12}}{4 \times 10^{-6}} = 25 \times 10^{-12} \times \frac{1}{4} \times 10^{6} = \frac{25}{4} \times 10^{-6}\;\text{J} = 6.25 \times 10^{-6}\;\text{J}.$$

The work done by the external agent in pulling the plates apart equals the increase in the electrostatic energy, because the system gains this additional energy:

$$W = U_f - U_i.$$

Substituting the computed energies, we have

$$W = 6.25 \times 10^{-6}\;\text{J} \;-\; 2.5 \times 10^{-6}\;\text{J} = 3.75 \times 10^{-6}\;\text{J}.$$

Hence, the correct answer is Option D.

We have an isolated parallel-plate capacitor whose capacitance is given as $$C = 1\;\mu\text{F} = 1 \times 10^{-6}\,\text{F}.$$

The problem states that one conducting plate is provided with a charge of $$+2\;\mu\text{C}$$ while the other plate is provided with a charge of $$+4\;\mu\text{C}.$$ Let us denote these charges as

$$Q_1 = +2\;\mu\text{C}, \qquad Q_2 = +4\;\mu\text{C}.$$

For any capacitor, the electric field (and therefore the potential difference) is produced only by that part of the charge which is opposite and equal on the two plates. Any charge that is common to both plates does not contribute to the field between them because its electric effects cancel out internally.

Mathematically we decompose the charges on the plates into two parts:

1. A common charge that is present, with the same sign and magnitude, on both plates.
2. An opposite charge that is equal in magnitude and opposite in sign on the two plates; this part alone decides the potential difference.

The common charge is obtained by averaging the two given charges:

$$Q_{\text{common}} = \frac{Q_1 + Q_2}{2} = \frac{(+2\;\mu\text{C}) + (+4\;\mu\text{C})}{2} = \frac{6\;\mu\text{C}}{2} = 3\;\mu\text{C}.$$

Subtracting this common charge from each plate’s total charge leaves us with the opposite (effective) charge:

For plate 1: $$Q_{1,\text{eff}} = Q_1 - Q_{\text{common}} = 2\;\mu\text{C} - 3\;\mu\text{C} = -1\;\mu\text{C}.$$

For plate 2: $$Q_{2,\text{eff}} = Q_2 - Q_{\text{common}} = 4\;\mu\text{C} - 3\;\mu\text{C} = +1\;\mu\text{C}.$$

Clearly, the two effective charges are equal in magnitude (1 μC) and opposite in sign, which is exactly the situation for an ordinarily charged capacitor.

Hence the magnitude of the charge that actually participates in producing the potential difference is

$$Q_{\text{eff}} = 1\;\mu\text{C} = 1 \times 10^{-6}\,\text{C}.$$

Now we use the fundamental capacitor formula relating charge, capacitance and potential difference:

$$V = \frac{Q_{\text{eff}}}{C}.$$

Substituting the known values,