JEE Capacitance Questions

AtPP, force on charge Q is zero if net electric field is zero.

Field due to dipole 1 (axial point):

$$E_1=\frac{1}{4\pi\epsilon_0}\frac{2pr}{(r^2-a^2)^2}$$

where

p=2aq

Direction is leftward (toward −q).

Field due to dipole 2 (equatorial point):

$$E_2=\frac{1}{4\pi\epsilon_0}\frac{p}{(r^2+a^2)^{3/2}}$$

Direction is rightward (opposite to dipole moment).

For zero net force,

$$E_1=E_2$$

So,

$$\frac{2r}{(r^2-a^2)^2}=\frac{1}{(r^2+a^2)^{3/2}}$$

Let

$$x=\frac{a}{r}$$

Then

$$r^2-a^2=r^2(1-x^2)$$

$$r^2+a^2=r^2(1+x^2)$$

Substitute:

$$\frac{2}{(1-x^2)^2}=\frac{1}{(1+x^2)^{3/2}}$$

Solving,

$$x=\sqrt{5}-2$$

moment of inertia about center:

$$I=2m\left(\frac{l}{2}\right)^2=\frac{ml^2}{2}$$

For small oscillations,

$$I\omega^2=−pE_0θ$$

with

p=ql

So

$$\omega=\sqrt{\frac{pE_0}{I}}=\sqrt{\frac{qlE_0}{ml^2/2}}=\sqrt{\frac{2qE_0}{ml}}$$

Hence

$$T=\frac{2\pi}{\omega}$$

$$T=2\pi\sqrt{\frac{ml}{2qE_0}}$$

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

Electron with $$\vec{v}=v_0\hat{i}$$ enters field $$\vec{E}=-E_0\hat{k}$$. Find de Broglie wavelength after time $$t$$.

The force on the electron is $$\vec{F}=-e\vec{E}=eE_0\hat{k}$$.

Integrating acceleration gives $$v_z=\frac{eE_0}{m}t$$ while $$v_x=v_0$$ remains constant.

Hence the speed is $$v=\sqrt{v_0^2+\frac{e^2E_0^2t^2}{m^2}}$$.

The de Broglie wavelength is $$\lambda=\frac{h}{mv}=\frac{h}{m\sqrt{v_0^2+e^2E_0^2t^2/m^2}}$$.

This can be written as $$\lambda=\frac{h}{mv_0\sqrt{1+e^2E_0^2t^2/(m^2v_0^2)}}$$.

Therefore $$\lambda=\frac{\lambda_0}{\sqrt{1+\frac{e^2E_0^2t^2}{m^2v_0^2}}}$$.

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

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

We need to evaluate the Assertion and Reason about the photoelectric effect.

Assertion (A): "Emission of electrons in photoelectric effect can be suppressed by applying a sufficiently negative electric potential to the photoemissive substance."

This is TRUE. When a negative potential (called the stopping potential) is applied to the photoemissive surface, it repels the emitted electrons back. If the potential is sufficiently negative (equal to or greater than the stopping potential $$V_0$$), all photoelectrons are pushed back and the photocurrent becomes zero.

Reason (R): "A negative electric potential, which stops the emission of electrons from the surface of a photoemissive substance, varies linearly with frequency of incident radiation."

This is TRUE. From Einstein's photoelectric equation:

$$eV_0 = h\nu - \phi$$

$$V_0 = \frac{h}{e}\nu - \frac{\phi}{e}$$

This shows that the stopping potential $$V_0$$ varies linearly with the frequency $$\nu$$ of incident radiation (with slope $$h/e$$ and y-intercept $$-\phi/e$$).

While both A and R are true, R explains the relationship between stopping potential and frequency. However, R does not directly explain why emission can be suppressed by a negative potential (which is explained by the work done by the electric field on the electrons). The reason behind A is that the electric field does negative work on electrons, not the linear relationship between $$V_0$$ and $$\nu$$.

Therefore, both A and R are true, but R is not the correct explanation of A.

The correct answer is Option (4): Both (A) and (R) are true but (R) is not the correct explanation of (A).

The electric flux is given as $$\phi = \alpha \sigma + \beta \lambda$$, where $$\sigma$$ is the surface charge density and $$\lambda$$ is the linear charge density. We need to find what $$\frac{\alpha}{\beta}$$ represents.

Electric flux has dimensions derived from Gauss's law. The dimensional formula for electric flux is $$[\phi] = [\text{kg} \cdot \text{m}^3 \cdot \text{s}^{-2} \cdot \text{C}^{-1}]$$.

The surface charge density $$\sigma = \frac{\text{charge}}{\text{area}}$$, so its dimensional formula is $$[\sigma] = [\text{C} \cdot \text{m}^{-2}]$$.

The linear charge density $$\lambda = \frac{\text{charge}}{\text{length}}$$, so its dimensional formula is $$[\lambda] = [\text{C} \cdot \text{m}^{-1}]$$.

In the expression $$\phi = \alpha \sigma + \beta \lambda$$, both terms must have the same dimensions as flux.

For the term $$\alpha \sigma$$:

$$[\alpha \sigma] = [\alpha] \cdot [\sigma] = [\alpha] \cdot [\text{C} \cdot \text{m}^{-2}] = [\text{kg} \cdot \text{m}^3 \cdot \text{s}^{-2} \cdot \text{C}^{-1}]$$

Solving for $$[\alpha]$$:

$$[\alpha] = \frac{[\text{kg} \cdot \text{m}^3 \cdot \text{s}^{-2} \cdot \text{C}^{-1}]}{[\text{C} \cdot \text{m}^{-2}]} = [\text{kg} \cdot \text{m}^5 \cdot \text{s}^{-2} \cdot \text{C}^{-2}]$$

For the term $$\beta \lambda$$:

$$[\beta \lambda] = [\beta] \cdot [\lambda] = [\beta] \cdot [\text{C} \cdot \text{m}^{-1}] = [\text{kg} \cdot \text{m}^3 \cdot \text{s}^{-2} \cdot \text{C}^{-1}]$$

Solving for $$[\beta]$$:

$$[\beta] = \frac{[\text{kg} \cdot \text{m}^3 \cdot \text{s}^{-2} \cdot \text{C}^{-1}]}{[\text{C} \cdot \text{m}^{-1}]} = [\text{kg} \cdot \text{m}^4 \cdot \text{s}^{-2} \cdot \text{C}^{-2}]$$

Now, the ratio $$\frac{\alpha}{\beta}$$ has dimensions:

$$\left[\frac{\alpha}{\beta}\right] = \frac{[\alpha]}{[\beta]} = \frac{[\text{kg} \cdot \text{m}^5 \cdot \text{s}^{-2} \cdot \text{C}^{-2}]}{[\text{kg} \cdot \text{m}^4 \cdot \text{s}^{-2} \cdot \text{C}^{-2}]} = [\text{m}]$$

The dimension $$[\text{m}]$$ corresponds to length. Comparing with the options:

• A. Electric field has dimensions $$[\text{kg} \cdot \text{m} \cdot \text{s}^{-2} \cdot \text{C}^{-1}]$$
• B. Area has dimensions $$[\text{m}^2]$$
• C. Charge has dimensions $$[\text{C}]$$
• D. Displacement has dimensions $$[\text{m}]$$

Displacement has the same dimension as length, so $$\frac{\alpha}{\beta}$$ represents displacement.

Therefore, the correct answer is displacement (option D).

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:

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

$$=4+5$$

$$=9\mu F$$

(A) Uniformly charged spherical shell ($$r < R$$):

$$q_{\text{enclosed}} = 0 \implies E_{\text{in}} = 0$$

Matches: (III)

(B) Infinite plane sheet:

$$E = \frac{\sigma}{2\varepsilon_0}$$

Matches: (II)

(C) Uniformly charged spherical shell ($$r > R$$):

$$q_{\text{total}} = \sigma (4\pi R^2)$$

$$E_{\text{out}} = \frac{q_{\text{total}}}{4\pi\varepsilon_0 r^2} \implies E_{\text{out}} = \frac{\sigma (4\pi R^2)}{4\pi\varepsilon_0 r^2} = \frac{\sigma R^2}{\varepsilon_0 r^2}$$

Matches: (IV)

(D) Between 2 oppositely charged infinite parallel sheets:

$$E_{\text{net}} = E_+ + E_- \implies E_{\text{net}} = \frac{\sigma}{2\varepsilon_0} + \frac{\sigma}{2\varepsilon_0} = \frac{\sigma}{\varepsilon_0}$$

Matches: (I)

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

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

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

Configuration (1): charges at four corners.

Side pairs (4):

$$4\cdot\frac{kq_0^2}{a}$$

Diagonal pairs (2):

$$2\cdot\frac{kq_0^2}{a\sqrt{2}}$$

So,

$$U_1=\frac{kq_0^2}{a}(4+\sqrt{2})$$

Configuration (2): charges at midpoints of four sides.

Distance between adjacent midpoints:

$$\frac{a}{\sqrt{2}}$$

4 such pairs:

$$4\cdot\frac{kq_0^2}{a/\sqrt{2}}=\frac{4\sqrt{2}kq_0^2}{a}$$

Distance between opposite midpoints:

a

2 such pairs:

$$2\cdot\frac{kq_0^2}{a}$$

Thus,

$$U_2=\frac{kq_0^2}{a}(4\sqrt{2}+2)$$

Work required:

$$W=U_2-U_1$$

$$=\frac{kq_0^2}{a}(3\sqrt{2}-2)$$

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.

By Gauss law,

$$\Phi=\frac{q}{\epsilon_0}$$

So

$$q=ϵ_0Φ$$

Given

$$\Phi=-2\times10^4$$

$$\epsilon_0=8.85\times10^{-12}$$

Thus

$$q=(8.85\times10^{-12})(-2\times10^4)$$

$$=-17.7\times10^{-8}$$

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 $$5$$.

Field due to infinite positive sheet is uniform and directed away from sheet.

Dipole moment is from −q to +q, which here is also away from sheet, so

$$⃗\vec{p}\parallel\vec{E}$$

Torque on dipole:

$$\tau=pE\sin\theta$$

Here

$$\theta=0$$

so

$$τ=0$$

Potential energy:

U=−p​⋅E=−pEcosθ

$$U=-pE$$

which is minimum.

Also forces on +q and −q are equal and opposite, so net force is zero.

Hence correct statement:

Potential energy is minimum and torque is zero.

Given:

• AB=30 cm
• AC=AD=40 cm (circular path centered at A, so $$q_1$$​ is at center)

Thus distance of $$q_3$$ from $$q_1$$​ stays constant during motion:

So potential energy due to $$q_1$$​ does not change.

Only contribution from $$q_2$$ changes.

Distances from $$q_2$$:

At C:

$$r_{2c}=\sqrt{\ 30^2+40^2}=50cm$$

At D:

Since AD=40 and AB=30,

BD=10 cm

so

$$r_{2D}=10cm$$

Change in potential energy:

$$\Delta U=\frac{1}{4\pi\varepsilon_0}q_3q_2\left(\frac{1}{r_{2D}}-\frac{1}{r_{2C}}\right)$$

Substitute:

$$=\frac{1}{4\pi\varepsilon_0}q_3q_2\left(\frac{1}{10}-\frac{1}{50}\right)=\frac{1}{4\pi\varepsilon_0}q_3q_2\times\ \left(\frac{4}{50}\right)$$
convert 50 cm to m
$$=\frac{1}{4\pi\varepsilon_0}q_3q_2\times\ \left(\frac{4}{0.5}\right)$$
final answer is 
$$=\frac{8}{4\pi\varepsilon_0}q_3q_2$$
now comparing it with the given equation 
$$K=8q_2$$

We need to find the ratio of electrostatic force to gravitational force between the two ions. The electrostatic force (Coulomb's law) is given by $$ F_e = \frac{1}{4\pi\epsilon_0}\frac{q_1 q_2}{r^2} $$. The gravitational force is $$ F_g = \frac{Gm_1 m_2}{r^2} $$. Dividing these gives $$ \frac{F_e}{F_g} = \frac{q_1 q_2}{4\pi\epsilon_0 G m_1 m_2} $$.

The charges are $$q_1 = 6.67 \times 10^{-19}$$ C and $$q_2 = 9.6 \times 10^{-10}$$ C, and the masses are $$m_1 = 19.2 \times 10^{-27}$$ kg and $$m_2 = 9 \times 10^{-27}$$ kg. We also use $$\frac{1}{4\pi\epsilon_0} = 9 \times 10^{9}$$ Nm$$^2$$C$$^{-2}$$ and $$G = 6.67 \times 10^{-11}$$ Nm$$^2$$kg$$^{-2}$$.

Substituting these values into the ratio yields: $$ \frac{F_e}{F_g} = \frac{9 \times 10^{9} \times 6.67 \times 10^{-19} \times 9.6 \times 10^{-10}}{6.67 \times 10^{-11} \times 19.2 \times 10^{-27} \times 9 \times 10^{-27}} $$.

The numerator simplifies as $$ 9 \times 6.67 \times 9.6 \times 10^{9-19-10} = 9 \times 6.67 \times 9.6 \times 10^{-20} = 576.288 \times 10^{-20} = 5.76288 \times 10^{-18} $$. The denominator becomes $$ 6.67 \times 19.2 \times 9 \times 10^{-11-27-27} = 6.67 \times 19.2 \times 9 \times 10^{-65} = 1152.576 \times 10^{-65} = 1.152576 \times 10^{-62} $$.

Therefore, $$ \frac{F_e}{F_g} = \frac{5.76288 \times 10^{-18}}{1.152576 \times 10^{-62}} = 5 \times 10^{44} = 0.5 \times 10^{45} $$. Hence, $$P = 0.5$$ and $$10P = 5$$, giving the final answer as 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.

Use solid angle idea.

Flux through any surface due to point charge is

$$\Phi=\frac{q}{4\pi\varepsilon_0}\Omega$$

where Ω\OmegaΩ is solid angle subtended by the surface at the charge.

The charge is located on the normal through the center of the square, and from the geometry shown, the whole square subtends a solid angle corresponding to a face of a cube with charge at cube center.

Flux through entire square is therefore

$$\Phi_{\text{square}}=\frac{q}{6\varepsilon_0}$$

since total flux

$$\frac{q}{\varepsilon_0}$$

gets equally shared among 6 cube faces.

Given

q=1C

so

$$\Phi_{\text{square}}=\frac{1}{6\varepsilon_0}$$

Now look at shaded region.

its area is

$$\frac{5a^2}{8}$$

So shaded fraction of the square is

$$\frac{5}{8}$$

Flux through the whole square (one face of the imaginary cube) is

$$\frac{q}{6\varepsilon_0}$$

Hence flux through shaded region is

$$\Phi=\frac{5}{8}\cdot\frac{q}{6\varepsilon_0}$$

With

q=1

$$\Phi=\frac{5}{48\varepsilon_0}$$

Comparing with

$$\frac{5}{p}\cdot\frac{1}{\varepsilon_0}$$

we get

$$p=48$$

$$C_0 = \frac{\varepsilon_0 A}{d} = \frac{8.85 \times 10^{-12} \times 4 \times 10^{-4}}{1.77 \times 10^{-3}}$$

$$C_0 = 5 \times 4 \times 10^{-13} \text{ F} = 20 \times 10^{-13} \text{ F} = 2 \text{ pF}$$

Top slab: $$C_1 = \frac{k_1 \varepsilon_0 A}{d/2} = 2k_1 C_0 = 2 \times 5 \times 2 \text{ pF} = 20 \text{ pF}$$

Bottom slab: $$C_2 = \frac{k_2 \varepsilon_0 A}{d/2} = 2k_2 C_0 = 2 \times 3 \times 2 \text{ pF} = 12 \text{ pF}$$

$$C_{\text{eq}} = \frac{C_1 C_2}{C_1 + C_2} = \frac{20 \times 12}{20 + 12} = \frac{240}{32} = 7.5 \text{ pF}$$

$$C_{\text{total}} = C_{\text{eq}} + 7.5 \text{ pF} = 7.5 + 7.5 = 15 \text{ pF}$$

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 find the force between charges in a dielectric medium at a reduced distance.

In vacuum at distance $$r$$ the force is given by $$F = \frac{kq_1q_2}{r^2}$$. Since the dielectric constant is $$K = 5$$ and the separation is reduced to $$r/5$$, Coulomb's law in the medium yields:

$$F' = \frac{kq_1q_2}{K \cdot (r/5)^2} = \frac{kq_1q_2}{5 \cdot r^2/25} = \frac{25kq_1q_2}{5r^2} = \frac{5kq_1q_2}{r^2} = 5F$$

This shows that although the dielectric reduces the force by a factor of $$K = 5$$, the shorter separation increases it by a factor of 25, resulting in a net factor of $$\frac{25}{5} = 5$$ times the original force. Therefore, the correct answer is Option 2: $$5F$$.

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).

The electrostatic potential due to an electric dipole at a point at distance $$r$$ from the center of the dipole is given by:

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

where $$p$$ is the dipole moment and $$\theta$$ is the angle between the dipole axis and the line joining the center of the dipole to the point.

From this expression, we can see that:

$$V \propto \frac{1}{r^2}$$

Note: This is different from a point charge, where $$V \propto \frac{1}{r}$$, and from the electric field of a dipole, where $$E \propto \frac{1}{r^3}$$.

The correct answer is $$\frac{1}{r^2}$$.

This question asks why vehicles carrying inflammable (flammable) fluids have metallic chains that touch the ground.

When a vehicle moves along a road, friction between the tyres and the road surface, as well as friction between the vehicle body and the air, causes a build-up of static electric charge on the vehicle's metallic body. This is similar to how rubbing a balloon on cloth creates static charge.

For ordinary vehicles, this small accumulation of charge is not a major concern. However, for vehicles carrying inflammable fluids (such as petroleum tankers), even a tiny spark can ignite the vapours and cause a catastrophic explosion.

A spark occurs when the accumulated charge reaches a high enough potential difference to ionise the surrounding air and discharge suddenly. To prevent this, metallic chains are attached to the vehicle so that they drag along the ground. Since metals are good conductors of electricity, the chain provides a continuous conducting path from the vehicle body to the ground (earth). This allows the excess charge to flow harmlessly to the earth as it builds up, keeping the vehicle at ground potential and preventing any dangerous accumulation of charge.

This is an application of the principle of earthing or grounding, which is widely used in electrical safety.

Therefore, the correct answer is Option (4): To conduct excess charge due to air friction to ground and prevent sparking.

Initial: P and S each have Q. Force = $$kQ^2/d^2 = 16$$ N.

R (uncharged) touches P: charge on each = Q/2. Then R (Q/2) touches S (Q): charge on each = 3Q/4.

New charges: P = Q/2, S = 3Q/4.

Force = $$k(Q/2)(3Q/4)/d^2 = \frac{3}{8}kQ^2/d^2 = \frac{3}{8}×16 = 6$$ N.

The correct answer is Option (2): 6 N.

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)

$$C=C_1+C_2+C_3$$

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

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

Charge q is at the center of one face of the cube.

Imagine attaching an identical cube to this face so the charge lies on the common face center, which becomes inside a rectangular box made of two cubes.

Now the charge is at the center of this combined box.

By Gauss law, total flux through the combined closed surface is

$$\Phi_{\text{total}}=\frac{q}{\varepsilon_0}$$

The two cubes are identical and symmetric, so flux splits equally between them.

Hence flux linked with one cube is

$$\Phi_{\text{total}}=\frac{q}{2\varepsilon_0}$$

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

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

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

Electric flux is

$$⃗\phi=\vec{E}\cdot\vec{A}$$

Given

$$E=\left(6i+5j^​+3k\right)$$

Surface lies in YZ-plane, so area vector is along x-axis:

$$A⃗=30i$$

Therefore,

Dot product would be $$=\ 6\times\ 30$$

=180

We need to find the electric field at any point on the surface of a thin spherical shell with uniform surface charge density $$\sigma$$ and radius $$R$$.

The total charge on the shell is $$Q = \sigma \cdot 4\pi R^2$$.

By Gauss's law, the electric field just outside the shell (at distance $$R$$ from center) is:

$$E_{\text{outside}} = \frac{Q}{4\pi\epsilon_0 R^2} = \frac{\sigma \cdot 4\pi R^2}{4\pi\epsilon_0 R^2} = \frac{\sigma}{\epsilon_0}$$

The electric field just inside the shell is $$E_{\text{inside}} = 0$$ (no enclosed charge for a Gaussian surface inside).

For a conducting shell or a charged surface, the electric field AT the surface is taken as the value just outside the surface, which is $$\frac{\sigma}{\epsilon_0}$$. This is the standard convention used in JEE for the field "on the surface" of a spherical shell.

The correct answer is Option B: $$\sigma/\epsilon_0$$.

Two conducting spheres of radii $$a$$ and $$b$$ are connected by a wire. We need the ratio of their charges. When two conductors are connected by a wire, charge flows until they reach the same electric potential.

The potential of a conducting sphere with charge $$Q$$ and radius $$r$$ is:

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

Equating the potentials of the two spheres gives $$V_1 = V_2 \implies \frac{Q_1}{4\pi\varepsilon_0 a} = \frac{Q_2}{4\pi\varepsilon_0 b},$$ which simplifies to $$\frac{Q_1}{a} = \frac{Q_2}{b} \implies \frac{Q_1}{Q_2} = \frac{a}{b}.$$

The correct answer is Option (1): $$\frac{a}{b}$$.

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.

Electric field due to an infinite line charge:

$$E=\frac{\lambda}{2\pi\epsilon_0r}$$

Force on electron provides centripetal force:

$$eE=\frac{mv^2}{r}$$

Substitute E:

$$e\cdot\frac{\lambda}{2\pi\epsilon_0r}=\frac{mv^2}{r}$$

r cancels:

$$\frac{e\lambda}{2\pi\epsilon_0}=mv^2$$

So

$$v^2=\frac{e\lambda}{2\pi\epsilon_0m}$$

which is constant.

Kinetic energy:

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

Also constant, independent of r.

so its constant

Two charges $$q$$ and $$3q$$ are separated by a distance $$r$$. We need to find the point between them where the resultant electric field is zero. Let the charge $$q$$ be at position O and the charge $$3q$$ at position $$r$$; let the point where the electric field vanishes be at a distance $$x$$ from $$q$$ (and hence at distance $$(r - x)$$ from $$3q$$). Since both charges are positive, the zero-field point lies between them where their fields are oppositely directed.

At distance $$x$$ from $$q$$, its field points away from $$q$$ (toward $$3q$$), while the field due to $$3q$$ at distance $$(r-x)$$ points away from $$3q$$ (toward $$q$$). Setting their magnitudes equal gives:

$$\frac{kq}{x^2} = \frac{k(3q)}{(r-x)^2}$$

Canceling $$kq$$ yields $$\frac{1}{x^2} = \frac{3}{(r-x)^2}$$. Cross-multiplying gives $$(r-x)^2 = 3x^2$$, and taking the positive root (since $$x>0$$ and $$r-x>0$$) leads to $$r - x = \sqrt{3}\,x$$, hence $$r = x(1 + \sqrt{3})$$ and

$$x = \frac{r}{1 + \sqrt{3}}$$

Numerically, with $$\sqrt{3}\approx1.732$$, this yields $$x\approx\frac{r}{2.732}\approx0.366r$$, confirming the point is between the charges and closer to the smaller charge $$q$$, as expected.

The correct answer is Option (3): $$\frac{r}{1 + \sqrt{3}}$$

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.

$$\text{Einstein's photoelectric equation: } eV_0 = \frac{hc}{\lambda} - \phi \implies V_0 = \frac{hc}{e\lambda} - \frac{\phi}{e}$$

Since the wavelength $$\lambda$$ and work function $$\phi$$ remain identical for both cases, their stopping potential $$-V_0$$ must be exactly the same

$$\text{The saturation photoelectric current is directly proportional to intensity: } i \propto I \implies \text{Since } I_1 < I_2\text{, the curve for } I_2 \text{ reaches a higher saturation current Level.}$$

Therefore, both curves must originate from the same stopping potential point $$-V_0$$ on the potential axis and separate as current rises

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 $$x = 86\,$$.

$$C_{\text{net}} = C_1 + C_2 = \frac{K_1 C_0}{2} + \frac{K_2 C_0}{2} = \frac{(K_1 + K_2)C_0}{2}$$

$$C_{\text{net}} = \frac{(2 + 3) \times 10 \times 10^{-6}}{2} = 25 \times 10^{-6}\ \text{F}$$

$$F = \frac{1}{2}\frac{C_{\text{net}} V^2}{d} \implies 8 = \frac{1}{2} \times \frac{25 \times 10^{-6} \times V^2}{10^{-2}}$$

$$8 = 1.25 \times 10^{-3} V^2$$

$$V^2 = \frac{8}{1.25 \times 10^{-3}} = 6400 \implies V = 80\ \text{V}$$

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 $$\boxed{2}$$.

Take away from sheet as positive.

Then electron starts moving away:

u=1

Displacement to reach sheet:

s=−1

Acceleration is toward sheet, hence negative:

$$a=-\frac{e\sigma}{2m\epsilon_0}$$

Using

$$s=ut+\frac{1}{2}at^2$$

with t=1,

$$-1=1+\frac{1}{2}a$$

$$\frac{a}{2}=-2$$

a=−4

Now

$$-\frac{e\sigma}{2m\epsilon_0}=-4$$

$$\frac{e\sigma}{2m\epsilon_0}=4$$

$$\sigma=8\frac{m\epsilon_0}{e}$$

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

Given

$$E=2\times10^4N/C$$

$$m=2g=2\times10^{-3}kg$$

l=20cm

Particle stays 10 cm from wall, so horizontal displacement is

x=10cm

From geometry,

$$\sin\theta=\frac{10}{20}=\frac{1}{2}$$

$$\theta=30^{\circ}$$

For equilibrium,

$$T\sin\theta=qE$$

$$T\cos\theta=mg$$

Dividing,

$$\tan\theta=\frac{qE}{mg}$$

So

$$q=\frac{mg\tan\theta}{E}$$

Substitute,

$$q=\frac{(2\times10^{-3})(10)\cdot\frac{1}{\sqrt{3}}}{2\times10^4}$$

$$=\frac{10^{-6}}{\sqrt{3}}C$$

$$=\frac{1}{\sqrt{3}}\mu C$$

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

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

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

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

Dipole moment magnitude:

$$p=q(2r)=2qr$$

At point P on axial line, distance from center is

$$OP=r$$

Axial field of dipole:

$$E_P=\frac{2kp}{r^3}$$

Given this is

E

so

$$E=\frac{2kp}{r^3}$$

Now point R lies on equatorial line at distance

OR=2r

Equatorial field of dipole:

Then

$$E_R=\frac{kp}{(2r)^3}$$

$$=\frac{kp}{8r^3}$$

Using

$$E=\frac{2kp}{r^3}$$

we get

$$E_R=\frac{1}{16}E$$

So

$$\frac{E}{x}=\frac{E}{16}$$

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.

The electric field ($$E$$) due to an infinitely long thin charged sheet with surface charge density $$\sigma$$ is  $$E = \frac{\sigma}{2\epsilon_0}$$.

Field points away from a positive charge ($$+\sigma$$)

Field points towards a negative charge ($$-\sigma$$)

Sheet 1 (at $$x = -a$$, charge $$-\sigma$$):

Point $$P$$ is to the right of this sheet. Since it is negatively charged, the field points left. $$\vec{E}_1 = -\frac{\sigma}{2\epsilon_0} \hat{i}$$

Sheet 2 (at $$x = a$$, charge $$-2\sigma$$):

Point $$P$$ is to the right of this sheet. Since it is negatively charged, the field points left. $$\vec{E}_2 = -\frac{2\sigma}{2\epsilon_0} \hat{i} = -\frac{\sigma}{\epsilon_0} \hat{i}$$

Sheet 3 (at $$x = 3a$$, charge $$\sigma$$):

Point $$P$$ is to the left of this sheet. Since it is positively charged, the field points left (away from the sheet). $$\vec{E}_3 = -\frac{\sigma}{2\epsilon_0} \hat{i}$$

$$\vec{E}_{net} = \vec{E}_1 + \vec{E}_2 + \vec{E}_3$$

$$\vec{E}_{net} = \left( -\frac{\sigma}{2\epsilon_0} - \frac{\sigma}{\epsilon_0} - \frac{\sigma}{2\epsilon_0} \right) \hat{i}$$

$$\vec{E}_{net} = \left( -\frac{2\sigma}{2\epsilon_0} - \frac{\sigma}{\epsilon_0} \right) \hat{i} = -\frac{2\sigma}{\epsilon_0} \hat{i}$$

$$|\vec{E}_{net}| = \frac{2\sigma}{\epsilon_0}$$

$$x = 2$$

Let the square be $$ABCD$$, taken in the clockwise sense.
Since the total length of the wire has resistance $$16\ \Omega$$, each side of the square has resistance

$$R = \frac{16\ \Omega}{4} = 4\ \Omega$$

The battery (emf $$E = 9\ \text{V}$$, internal resistance $$r = 1\ \Omega$$) is connected across side $$AB$$. Thus the terminals of the battery are joined to the same two nodes $$A$$ and $$B$$ that are already connected by the 4 $$\Omega$$ side.

For easier calculation, replace the battery by its Norton equivalent across the nodes $$A$$ and $$B$$.

Step 1: Norton conversion An emf $$E$$ in series with resistance $$r$$ is equivalent to a current source $$I_N$$ in parallel with the same resistance $$r$$, where

$$I_N = \frac{E}{r} = \frac{9}{1} = 9\ \text{A}$$

Therefore, across the nodes $$A$$ (positive) and $$B$$ (negative) we now have

• a current source of $$9\ \text{A}$$ from $$A \rightarrow B$$, and
• a resistor $$1\ \Omega$$ in parallel with that source.

Step 2: Find the equivalent resistance between $$A$$ and $$B$$ of the rest of the square.

Besides the internal $$1\ \Omega$$ resistor, two purely resistive paths join $$A$$ and $$B$$:

• Direct side $$AB$$ : $$4\ \Omega$$
• The remaining three sides $$A \rightarrow D \rightarrow C \rightarrow B$$ : $$4 + 4 + 4 = 12\ \Omega$$

Hence the three resistances $$1\ \Omega, 4\ \Omega, 12\ \Omega$$ are all in parallel between the same nodes. Their combined resistance is

$$\frac{1}{R_{\text{eq}}} \;=\; \frac{1}{1} + \frac{1}{4} + \frac{1}{12} \;=\; 1 + 0.25 + 0.0833 \;=\; 1.3333$$

$$R_{\text{eq}} = \frac{1}{1.3333} = 0.75\ \Omega$$

Step 3: Node voltage between $$A$$ and $$B$$

By KCL at the parallel combination, the current source of $$9\ \text{A}$$ drives the currents through the parallel resistors, producing a node voltage

$$V_{AB} = I_N \, R_{\text{eq}} = 9\ \text{A} \times 0.75\ \Omega = 6.75\ \text{V}$$

Take $$V_B = 0$$; hence

$$V_A = +6.75\ \text{V}$$

Step 4: Currents and potentials around the other three sides

The only path from $$A$$ to $$B$$ that does not go through the direct side is the series string $$A \rightarrow D \rightarrow C \rightarrow B$$ of total resistance $$12\ \Omega$$. The current through this path is

$$I = \frac{V_{AB}}{12\ \Omega} = \frac{6.75}{12} = 0.5625\ \text{A}$$

Voltage drop on each of the three equal resistors (4 $$\Omega$$) is

$$\Delta V = I \times 4 = 0.5625 \times 4 = 2.25\ \text{V}$$

Therefore, moving clockwise:

$$V_D = V_A - 2.25 = 6.75 - 2.25 = 4.50\ \text{V}$$
$$V_C = V_D - 2.25 = 4.50 - 2.25 = 2.25\ \text{V}$$

Step 5: Potential difference across a diagonal

Place the $$4\ \mu\text{F}$$ capacitor across any diagonal; choose diagonal $$AC$$. The voltage across this diagonal is

$$V_{AC} = V_A - V_C = 6.75 - 2.25 = 4.50\ \text{V}$$

(If the capacitor had been connected across the other diagonal $$BD$$, the magnitude is the same, $$4.50\ \text{V}$$.)

Step 6: Energy stored in the capacitor

The energy in a capacitor is $$U = \tfrac{1}{2} C V^2$$.

Here $$C = 4\ \mu\text{F} = 4 \times 10^{-6}\ \text{F}$$, $$V = 4.50\ \text{V}$$.

Hence

$$U = \tfrac{1}{2} \times 4 \times 10^{-6} \times (4.5)^2$$ $$\qquad= 2 \times 10^{-6} \times 20.25$$ $$\qquad= 40.5 \times 10^{-6}\ \text{J}$$

Expressing in microjoules: $$U = 40.5\ \mu \text{J}$$.

The question states this energy equals $$\dfrac{x}{2}\ \mu \text{J}$$, so

$$\frac{x}{2} = 40.5 \quad\Longrightarrow\quad x = 81$$

Therefore, $$x = 81$$.

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:

$$I_R=\frac{13}{3}-\frac{10}{3}=1\text{ A}$$

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

$$Q=CV=150\times\frac{8}{3}=400μC$$

In air: $$\tan(\theta/2) = \frac{F_e}{mg}$$ where $$F_e$$ is the electrostatic force.

In liquid: The angle remains the same, so $$\tan(\theta/2)$$ is unchanged. This means:

$$\frac{F_e'}{m'g} = \frac{F_e}{mg}$$

where $$F_e' = \frac{F_e}{K}$$ (dielectric constant $$K$$) and $$m'g = mg - \text{buoyancy} = mg(1 - \frac{\rho_l}{\rho_s})$$.

$$\frac{F_e/K}{mg(1 - \rho_l/\rho_s)} = \frac{F_e}{mg}$$

$$\frac{1}{K} = 1 - \frac{\rho_l}{\rho_s} = 1 - \frac{0.7}{1.4} = 1 - 0.5 = 0.5$$

$$K = 2$$.

Therefore, the answer is $$\boxed{2}$$.

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.

Coulomb force:

$$F_e=\frac{1}{4\pi\epsilon_0}\frac{e^2}{r^2}$$

Gravitational force:lkncoa

$$F_g=\frac{Gm_pm_e}{r^2}$$

Ratio,

$$\frac{F_e}{F_g}=\frac{1}{4\pi\epsilon_0}\frac{e^2}{Gm_pm_e}$$

Using approximate values,

$$\frac{1}{4\pi\epsilon_0}=9\times10^9$$

$$e=1.6\times10^{-19}$$

$$G=6.67\times10^{-11}$$

$$m_p=1.67\times10^{-27}$$

$$m_e=9.1\times10^{-31}$$

Substituting gives order

$$\frac{F_e}{F_g}\sim10^{39}$$

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.

The electrostatic force between two charges in a medium with dielectric constant $$k$$ is given by Coulomb's law:

$$F_{\text{medium}} = \frac{1}{4\pi\epsilon_0 k} \frac{q_1 q_2}{d^2}$$

In air (which is approximately vacuum), the force is:

$$F_{\text{air}} = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r^2}$$

where $$r$$ is the equivalent distance in air for the same force.

Since the force is the same in both cases, set $$F_{\text{air}} = F_{\text{medium}}$$:

$$\frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r^2} = \frac{1}{4\pi\epsilon_0 k} \frac{q_1 q_2}{d^2}$$

Cancel the common terms $$\frac{1}{4\pi\epsilon_0}$$ and $$q_1 q_2$$:

$$\frac{1}{r^2} = \frac{1}{k d^2}$$

Take the reciprocal of both sides:

$$r^2 = k d^2$$

Take the square root of both sides:

$$r = d \sqrt{k}$$

Thus, the equivalent distance in air is $$d\sqrt{k}$$.

Comparing with the options, the correct answer is A. $$d\sqrt{k}$$.

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.

By Gauss's law, the electric flux through any closed surface equals the total charge enclosed divided by $$\varepsilon_0$$:

$$\Phi = \frac{Q_{enclosed}}{\varepsilon_0}$$

The positive plate of the capacitor carries charge $$Q = CV$$.

Since the closed surface encloses only the positive plate:

$$\Phi = \frac{CV}{\varepsilon_0}$$

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.

At steady state, capacitor branch carries no current.

So middle branch is open, and only top and bottom branches determine potential across capacitor.

Top branch: battery  with resistor $$4Ω$$

Bottom branch: resistor $$6Ω$$

They form one closed loop.

Current:

$$I=\frac{3}{4+6}=0.3A$$

Potential drop across bottom $$6Ω$$ resistor:

V=IR

$$=0.3(6)=1.8V$$

This is the potential difference between left and right nodes.

Since in steady state no current flows through middle $$6Ω$$, no voltage drop occurs across that resistor, so capacitor gets full node voltage:

$$V_C=1.8V$$

Charge stored:

$$Q=CV$$

$$=(4\mu F)(1.8)$$

$$=7.2μC$$

Kirchhoff's laws and Thevenin's theorem determine circuit behavior during transient and steady-state conditions of a charging $$R-C$$ branch.

Given: $$I_{1\Omega} = 1\text{ A}$$ (when $$K$$ is open), $$C = 2\ \mu\text{F}$$, $$e^{-1} = 0.36$$

At $$t < t_0$$ (Key $$K$$ open, bottom branch disconnected), using nodal analysis at the common right node relative to the left node ($$0\text{ V}$$):

$$I_{1\Omega} = \frac{V - 5}{1} = 1\text{ A} \implies V = 6\text{ V}$$

$$\frac{15 - V}{R} = \frac{V - 5}{1} + \frac{V - 0}{3} \implies \frac{15 - 6}{R} = 1 + \frac{6}{3} = 3 \implies R = 3\ \Omega$$

$$I_1 = \frac{V}{3} = \frac{6}{3} = 2\text{ A}$$

when $$K$$ is closed: $$R_{\text{total}} = 0.6 + 3 = 3.6\ \Omega$$

$$\tau = R_{\text{total}}C = 3.6 \times 2\ \mu\text{F} = 7.2\ \mu\text{s}$$

Evaluating capacitor branch current at $$t = t_0 + 7.2\ \mu\text{s}$$ ($$t - t_0 = \tau$$):

$$I_{\text{max}} = \frac{V_{\text{th}}}{R_{\text{total}}} = \frac{6}{3.6} = \frac{5}{3}\text{ A}$$

$$I(\tau) = I_{\text{max}}e^{-1} = \frac{5}{3} \times 0.36 = 0.6\text{ A}$$

Evaluating steady-state charge ($$t \rightarrow \infty$$): $$q_{\infty} = C V_{\text{th}} = 2\ \mu\text{F} \times 6\text{ V} = 12\ \mu\text{C}$$

Answer: Option (A), Option (B), Option (C), Option (D)

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

$$V_{\text{small}} = \frac{kq}{r} = 10\text{ mV}$$

Equating total volume to find the radius $$R$$ of the combined large drop:

$$\frac{4}{3}\pi R^3 = 64 \times \frac{4}{3}\pi r^3 \implies R = 4r$$

$$Q = 64q$$

$$V_{\text{large}} = \frac{kQ}{R} = \frac{k(64q)}{4r} = 16 \left(\frac{kq}{r}\right)$$

$$V_{\text{large}} = 16 \times 10 = 160\text{ mV}$$

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.

$$C_1 = \frac{\varepsilon_0 A}{d} = 15 \text{ pF}$$

$$C_2 = \frac{K_2\varepsilon_0 A}{d_2} = \frac{3.5 \cdot \varepsilon_0 A}{2d}$$

$$C_2 = \frac{3.5}{2} \times 15 = \frac{52.5}{2} = 26.25 \text{ pF}$$

$$x = 26.25 \times 4 = 105$$

Solution :

Given electrostatic potential inside the spherical ball :

$$V = 2ar^2 + b$$

Electric field is related to potential by :

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

Therefore,

$$E = -\frac{d}{dr}(2ar^2+b)$$

$$E = -4ar$$

Using Gauss law in differential form :

$$\nabla \cdot \vec{E} = \frac{\rho}{\epsilon}$$

For spherical symmetry :

$$\nabla \cdot \vec{E} = \frac{1}{r^2}\frac{d}{dr}(r^2E_r)$$

Substituting,

$$E_r = -4ar$$

$$\nabla \cdot \vec{E} = \frac{1}{r^2}\frac{d}{dr}(r^2(-4ar))$$

$$= \frac{1}{r^2}\frac{d}{dr}(-4ar^3)$$

$$= \frac{1}{r^2}(-12ar^2)$$

$$= -12a$$

Hence,

$$\frac{\rho}{\epsilon} = -12a$$

Therefore,

$$\rho = -12a\epsilon$$

Comparing with :

$$-\lambda a\epsilon$$

we get,

$$\lambda = 12$$

Final Answer :

$$12$$

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

$$C_{eq} = (x + 5) \mu\text{F}$$

$$Q = C_{eq}V$$: $$100 = (x + 5) \times 10$$

$$x = 5$$

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

At steady state capacitor carries no current, so ignore capacitor branch while finding voltages.

The diamond becomes two parallel branches across 12 V source.

Left branch:

$$3\Omega+9\Omega=12\Omega$$

Current through left branch:

$$I_1=\frac{12}{12}=1A$$

Potential drop across $$3Ω$$:

$$V=1(3)=3V$$

So potential at left capacitor terminal (junction of $$3Ω\ and\ 9Ω$$) is

$$12-3=9V$$

Right branch:

$$4Ω+2Ω=6Ω$$

Current:

$$I_2=\frac{12}{6}=2$$

$$Dropnacrossn4Ω:V=2(4)=8V$$

Potential at right capacitor terminal:

$$12−8=4V$$

Voltage across capacitor:

$$V_C=9-4=5V$$

Capacitance:

$$C=6μF$$

Energy stored:

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

$$=\frac{1}{2}(6)(5^2)\mu J$$

$$=75μJ$$

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 $$C_4$$:

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

$$=4\times1=4$$

Two parallel plate capacitors $$C_1$$ and $$C_2$$, each with capacitance 10 $$\mu$$F, are charged to 100 V. A dielectric slab of constant $$K = 10$$ is inserted in each under different conditions.

When $$C_1$$ remains connected to the 100 V source and a dielectric is inserted:

New capacitance: $$C_1' = KC_1 = 10 \times 10 = 100$$ $$\mu$$F

Voltage stays at 100 V (source connected).

Charge: $$Q_1 = C_1' \times V = 100 \times 100 = 10{,}000$$ $$\mu$$C

Initial charge: $$Q_2 = C_2 \times V = 10 \times 100 = 1{,}000$$ $$\mu$$C

After disconnection, charge remains $$Q_2 = 1{,}000$$ $$\mu$$C.

After dielectric insertion: $$C_2' = KC_2 = 100$$ $$\mu$$F

Total charge: $$Q_{total} = Q_1 + Q_2 = 10{,}000 + 1{,}000 = 11{,}000$$ $$\mu$$C

Total capacitance: $$C_{total} = C_1' + C_2' = 100 + 100 = 200$$ $$\mu$$F

$$ V_{common} = \frac{Q_{total}}{C_{total}} = \frac{11{,}000}{200} = 55 \text{ V} $$

The common potential of the combination is 55 V.

Steady State: In a DC circuit, once a capacitor is fully charged, no current flows through its branch ($$I_C = 0$$). This effectively removes the $$1\text{ }\mu\text{F}$$ capacitor and the $$1\text{ }\Omega$$ resistor in series with it from the active current path.

The circuit simplifies into two parallel branches ($$ABC$$ and $$ADC$$). We can find the potential at nodes $$B$$ and $$D$$ using the voltage divider rule.

In the upper branch $$ABC$$, the capacitor branch is open. Current only flows through the $$1\text{ }\Omega$$ (top-left) and the main $$2\text{ }\Omega$$ (top-right) resistors. Total resistance of branch $$ABC = 1\text{ }\Omega + 2\text{ }\Omega = 3\text{ }\Omega$$.

Using the voltage divider rule relative to $$C$$: $$V_B = V_A \times \frac{R_{BC}}{R_{AB} + R_{BC}} = 6\text{ V} \times \frac{2\text{ }\Omega}{1\text{ }\Omega + 2\text{ }\Omega}$$

$$V_B = 6 \times \frac{2}{3} = \mathbf{4\text{ V}}$$

In the lower branch $$ADC$$, the $$10\text{ }\Omega$$ and $$2\text{ }\Omega$$ resistors are in series. Total resistance of branch $$ADC = 10\text{ }\Omega + 2\text{ }\Omega = 12\text{ }\Omega$$. Using the voltage divider rule relative to $$C$$: $$V_D = V_A \times \frac{R_{DC}}{R_{AD} + R_{DC}} = 6\text{ V} \times \frac{2\text{ }\Omega}{10\text{ }\Omega + 2\text{ }\Omega}$$

$$V_D = 6 \times \frac{2}{12} = \mathbf{1\text{ V}}$$

$$|V_B - V_D| = |4\text{ V} - 1\text{ V}| = \mathbf{3\text{ V}}$$

We need to determine the correct graph representing the variation of potential difference ($$V$$) across a condenser's plates with respect to the charge ($$Q$$) on it.

1. Relate Potential Difference ($$V$$) and Charge ($$Q$$)

The fundamental relationship between the charge stored in a capacitor, its capacitance ($$C$$), and the voltage across its plates is given by:

$$Q = C \times V$$

Rearranging the equation to solve for $$V$$ (since $$V$$ is plotted on the vertical $$y$$-axis and $$Q$$ is plotted on the horizontal $$x$$-axis):

$$V = \left(\frac{1}{C}\right) \cdot Q$$

This matches the equation of a straight line passing through the origin ($$y = mx$$), where:

• The slope of the line is $$m = \frac{1}{C}$$.
• Since capacitance $$C$$ is a positive constant, the slope is a constant positive value, making the graph a straight, linearly rising line starting from $$(0,0)$$.

2. Calculate the Coordinates at Maximum Charge

From the problem 

• Capacitance ($$C$$) = $$2\ \mu\text{F} = 2 \times 10^{-6}\text{ F}$$
• Maximum charge ($$Q$$) = $$5\text{ C}$$

Substitute these values into our equation to find the corresponding maximum voltage:

$$V = \frac{5}{2 \times 10^{-6}} = 2.5 \times 10^6\text{ V}$$

The vertical axis on the graph is scaled by a factor of $$10^6\text{ V}$$. Therefore, the value shown on the $$y$$-axis at $$Q = 5\text{ C}$$ must be $$2.5$$.

Conclusion

The correct graph is a straight line starting from zero and reaching a peak coordinate of $$(5, 2.5)$$. This matches Option A perfectly.

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.

Two identical positive charges $$Q$$ are fixed at a distance $$2a$$ apart, and a negative charge $$q_0$$ (mass $$m$$) is placed at the midpoint. We wish to find the time period of small oscillations of $$q_0$$ along the line joining the charges.

We place the two fixed charges at $$x=-a$$ and $$x=+a$$, with the negative charge $$q_0$$ initially at the origin. Suppose it is displaced by a small distance $$x$$ (where $$x\ll a$$) toward the charge at $$+a$$.

Since $$q_0$$ is negative and both fixed charges are positive, it experiences attractive forces from each. The force due to the charge at $$+a$$ (pulling toward $$+a$$) is $$F_{\text{right}}=\frac{1}{4\pi\epsilon_0}\frac{q_0Q}{(a-x)^2},$$ while the force due to the charge at $$-a$$ (pulling toward $$-a$$) is $$F_{\text{left}}=\frac{1}{4\pi\epsilon_0}\frac{q_0Q}{(a+x)^2}.$$

Taking the $$+x$$ direction as positive, the net force on $$q_0$$ is $$F_{\text{net}}=\frac{q_0Q}{4\pi\epsilon_0}\Bigl[\frac{1}{(a-x)^2}-\frac{1}{(a+x)^2}\Bigr].$$ Because the charge is closer to $$+a$$ when displaced, the pull from that side is stronger, while the farther charge at $$-a$$ pulls back toward the origin. For negative $$q_0$$, this difference produces a restoring force toward $$x=0$$.

For small $$x$$ ($$x\ll a$$), we expand by the binomial theorem: $$\frac{1}{(a-x)^2}=\frac{1}{a^2}\Bigl(1-\frac{x}{a}\Bigr)^{-2}\approx\frac{1}{a^2}\Bigl(1+\frac{2x}{a}\Bigr),$$ $$\frac{1}{(a+x)^2}=\frac{1}{a^2}\Bigl(1+\frac{x}{a}\Bigr)^{-2}\approx\frac{1}{a^2}\Bigl(1-\frac{2x}{a}\Bigr).$$

Subtracting these approximations gives $$\frac{1}{(a-x)^2}-\frac{1}{(a+x)^2}\approx\frac{1}{a^2}\Bigl(\frac{4x}{a}\Bigr)=\frac{4x}{a^3}.$$

Substituting back into the expression for the force yields the magnitude of the restoring force $$|F|=\frac{q_0Q}{4\pi\epsilon_0}\cdot\frac{4x}{a^3}=\frac{q_0Q}{\pi\epsilon_0a^3}\,x.$$ This is of the form $$F=-kx$$, where $$k=\frac{q_0Q}{\pi\epsilon_0a^3}.$$

From the above, the angular frequency satisfies $$\omega^2=\frac{k}{m}=\frac{q_0Q}{\pi\epsilon_0ma^3},$$ and the time period is $$T=\frac{2\pi}{\omega}=2\pi\sqrt{\frac{\pi\epsilon_0ma^3}{q_0Q}}=\sqrt{\frac{4\pi^3\epsilon_0ma^3}{q_0Q}}\,. $$

The correct answer is Option A: $$\sqrt{\dfrac{4\pi^3 \epsilon_0 m a^3}{q_0 Q}}$$.