JEE Capacitance Questions

We need to find the potentials of three concentric conducting spherical shells A, B, and C with radii $$a$$, $$b$$, and $$c$$ (where $$c > b > a$$) carrying charges $$q_1$$, $$q_2$$, and $$q_3$$ respectively.

Since the key principle is that the potential at any point due to a conducting spherical shell of radius R carrying charge q is $$\frac{q}{4\pi\epsilon_0 R}$$ when the point is on the shell or inside it ($$r \le R$$), and $$\frac{q}{4\pi\epsilon_0 r}$$ when the point is outside ($$r > R$$).

For sphere A at radius $$a$$, the contribution from $$q_1$$ on A itself is $$V_1 = \frac{q_1}{4\pi\epsilon_0 a}$$ since the point lies on A, from $$q_2$$ on B is $$V_2 = \frac{q_2}{4\pi\epsilon_0 b}$$ as the point is inside B, and from $$q_3$$ on C is $$V_3 = \frac{q_3}{4\pi\epsilon_0 c}$$ because the point is inside C.

Therefore, the total potential at A is $$V_A = \frac{1}{4\pi\epsilon_0}\left(\frac{q_1}{a} + \frac{q_2}{b} + \frac{q_3}{c}\right)$$.

Turning to sphere B at radius $$b$$, the potential due to $$q_1$$ on A is $$\frac{q_1}{4\pi\epsilon_0 b}$$ since B lies outside A, the potential from $$q_2$$ on B itself is $$\frac{q_2}{4\pi\epsilon_0 b}$$, and the potential from $$q_3$$ on C is $$\frac{q_3}{4\pi\epsilon_0 c}$$ because B is inside C.

Hence, the total potential at B is $$V_B = \frac{1}{4\pi\epsilon_0}\left(\frac{q_1 + q_2}{b} + \frac{q_3}{c}\right)$$.

Finally, for sphere C at radius $$c$$, both A and B lie inside C, so the contributions from $$q_1$$ and $$q_2$$ are $$\frac{q_1}{4\pi\epsilon_0 c}$$ and $$\frac{q_2}{4\pi\epsilon_0 c}$$ respectively, and the self-potential from $$q_3$$ is $$\frac{q_3}{4\pi\epsilon_0 c}$$.

Thus, the total potential at C is $$V_C = \frac{1}{4\pi\epsilon_0}\left(\frac{q_1 + q_2 + q_3}{c}\right)$$.

Comparing these results with the given options shows that they correspond to Option 2.

We need to find the total electrostatic energy of a system of two charges in an external electric field.

First, the total electrostatic energy of a system of charges in an external field has three contributions: $$U_{\text{total}} = U_1^{\text{ext}} + U_2^{\text{ext}} + U_{12}^{\text{mutual}}$$ where $$U_i^{\text{ext}} = q_i V_{\text{ext}}(\vec{r}_i)$$ is the energy of charge $$q_i$$ in the external potential, and $$U_{12}^{\text{mutual}} = \frac{kq_1 q_2}{r_{12}}$$ is the mutual interaction energy.

Next, we find the external potential associated with the radial field $$\vec{E} = \frac{A}{r^2}\hat{r}$$ where $$A = 9 \times 10^5 \, \text{N/C·m}^2$$. By integrating from infinity to $$r$$, we obtain $$V(r) = -\int_\infty^r \vec{E} \cdot d\vec{r} = -\int_\infty^r \frac{A}{r'^2}\,dr' = \frac{A}{r}$$ with the reference $$V(\infty) = 0$$.

Then we calculate the positions and distances of the charges. The first charge is $$q_1 = 7 \, \mu\text{C} = 7 \times 10^{-6} \, \text{C}$$ located at $$(-9,0,0)\text{ cm}$$, so $$r_1 = 9 \, \text{cm} = 0.09 \, \text{m}$$. The second charge is $$q_2 = -2 \, \mu\text{C} = -2 \times 10^{-6} \, \text{C}$$ at $$(9,0,0)\text{ cm}$$, giving $$r_2 = 9 \, \text{cm} = 0.09 \, \text{m}$$. The distance between the charges is $$r_{12} = 18 \, \text{cm} = 0.18 \, \text{m}$$.

Substituting these into the expression for the potential, the external potential at each charge location is $$V(r_1) = \frac{A}{r_1} = \frac{9 \times 10^5}{0.09} = 10^7 \, \text{V}$$ and similarly $$V(r_2) = \frac{A}{r_2} = \frac{9 \times 10^5}{0.09} = 10^7 \, \text{V}$$.

Consequently, the energy of each charge in the external field becomes $$U_1^{\text{ext}} = q_1 V(r_1) = 7 \times 10^{-6} \times 10^7 = 70 \, \text{J}$$ and $$U_2^{\text{ext}} = q_2 V(r_2) = -2 \times 10^{-6} \times 10^7 = -20 \, \text{J}$$.

Meanwhile, the mutual interaction energy between the two charges is $$U_{12} = \frac{kq_1q_2}{r_{12}} = \frac{9 \times 10^9 \times 7 \times 10^{-6} \times (-2 \times 10^{-6})}{0.18} = \frac{9 \times 10^9 \times (-14 \times 10^{-12})}{0.18} = \frac{-126 \times 10^{-3}}{0.18} = -0.7 \, \text{J}$$.

Finally, summing all contributions yields $$U_{\text{total}} = 70 + (-20) + (-0.7) = 49.3 \, \text{J}$$.

The correct answer is Option (3): 49.3 J.

Let the node between $$C_1$$ and $$C_2$$ be P. 
The node between $$C_2$$ and $$C_3$$ is A (since that is shorted)
The two capacitors $$C_1 , C_2 $$ are connected end-to-end between the terminals A and P, so they are in parallel.
The simplified circuit looks like 

$$(C_1 \parallel C_2) \text{ in series with } C_4$$
$$\text{and the above eq. cap in parallel with } C_3$$

$$C_{eq} = \frac{(C_1 + C_2)C_4}{(C_1 + C_2) + C_4} + C_3$$

$$= \frac{(1 + 1)(2)}{1 + 1 + 2} + 1$$

$$= \frac{4}{4} + 1$$

$$= 2\mu F$$

Option C which is: $$2,\mu F$$

Density of oil $$\rho = 1.5 \,\text{g cm}^{-3} = 1.5 \times 10^{3}\,\text{kg m}^{-3}$$

Radius of droplet $$r = 1\,\text{mm} = 1 \times 10^{-3}\,\text{m}$$

Volume of a sphere $$V = \frac{4}{3}\pi r^{3} = \frac{4}{3}\pi(1 \times 10^{-3})^{3} = \frac{4}{3}\pi \times 10^{-9}\,\text{m}^{3}$$

Mass of one droplet
$$m = \rho V = 1.5 \times 10^{3} \times \frac{4}{3}\pi \times 10^{-9} = 2\pi \times 10^{-6}\,\text{kg}$$

Weight acting downward
$$W = mg = 2\pi \times 10^{-6} \times 10 = 2\pi \times 10^{-5}\,\text{N} \approx 6.28 \times 10^{-5}\,\text{N}$$

Charge on each droplet $$q = 5\,\text{nC} = 5 \times 10^{-9}\,\text{C}$$

For the droplet to float, the upward electric force must equal its weight:
$$qE = mg \quad\Longrightarrow\quad E = \frac{mg}{q} = \frac{6.28 \times 10^{-5}}{5 \times 10^{-9}} = 1.26 \times 10^{4}\,\text{V m}^{-1}$$

Separation between the plates
$$d = \frac{12}{\pi}\,\text{cm} = \frac{12}{\pi} \times 10^{-2}\,\text{m} = \frac{0.12}{\pi}\,\text{m} \approx 3.82 \times 10^{-2}\,\text{m}$$

Required potential difference
$$\Delta V = E d = (1.26 \times 10^{4})(3.82 \times 10^{-2}) \approx 4.8 \times 10^{2}\,\text{V} \approx 480\,\text{V}$$

Direction of field: the electric force must act upward (toward plate A, which is on top). For a positive charge, $$\mathbf{E}$$ points from higher to lower potential. Hence the bottom plate B must be at the higher potential.

Choosing convenient values that differ by $$480\,\text{V}$$ with $$V_B \gt V_A$$,
$$V_A = 100\,\text{V}, \quad V_B = 580\,\text{V}$$

These match Option A.

Option A which is: 100 V and 580 V

For a parallel-plate air capacitor of plate area $$A$$ and separation $$x$$, the capacitance is given by
$$C = \frac{\varepsilon_0 A}{x}\quad -(1)$$

The capacitor remains connected to the battery, so the potential difference across the plates is a constant $$V$$.

Electrostatic energy stored in a capacitor at constant voltage is
$$U = \tfrac12 C V^{2}\quad -(2)$$

Substituting $$C$$ from $$(1)$$ into $$(2)$$:
$$U = \tfrac12 \varepsilon_{0} A \frac{V^{2}}{x}\quad -(3)$$

Differentiate $$(3)$$ with respect to time to find the rate of change of energy:
$$\frac{dU}{dt} = \tfrac12 \varepsilon_{0} A V^{2}\,\frac{d}{dt}\!\left(\frac{1}{x}\right)$$
Using $$\dfrac{d}{dt}\!\left(\frac{1}{x}\right)= -\dfrac1{x^{2}}\dfrac{dx}{dt}$$, we get
$$\frac{dU}{dt}= -\tfrac12 \varepsilon_{0} A V^{2}\,\frac{1}{x^{2}}\frac{dx}{dt}\quad -(4)$$

The plates are being pulled apart at a uniform speed $$v$$, so $$\dfrac{dx}{dt}=v$$ (a positive constant). Substituting this into $$(4)$$:
$$\frac{dU}{dt}= -\tfrac12 \varepsilon_{0} A V^{2}\,v\,\frac{1}{x^{2}}$$

The magnitude of the time rate of change of electrostatic energy is therefore proportional to $$x^{-2}$$. Hence
$$\left|\frac{dU}{dt}\right|\propto x^{-2}$$ and the required exponent is $$\alpha=-2$$.

Option A which is: $$-2$$

$$U_i = \frac{1}{2} C V^2$$

$$U_i = \frac{1}{2} \times (100 \times 10^{-12}) \times (100)^2 = \frac{1}{2} \times 10^{-10} \times 10^4 = 5 \times 10^{-7}\text{ J}$$

$$V_c = \frac{\text{Total Charge}}{\text{Total Capacitance}} = \frac{C \cdot V + 0}{C + C} = \frac{V}{2}$$

$$V_c = \frac{100}{2} = 50\text{ V}$$

$$U_f = \frac{1}{2} C_{\text{combined}} V_c^2$$

$$U_f = \frac{1}{2} \times (200 \times 10^{-12}) \times (50)^2 = 100 \times 10^{-12} \times 2500 = 2.5 \times 10^{-7}\text{ J}$$

$$\Delta U = U_i - U_f = (5 \times 10^{-7}) - (2.5 \times 10^{-7}) = 2.5 \times 10^{-7}\text{ J}$$

$$\Delta U = \frac{5}{2} \times 10^{-7}$$

We need to find the work done in bringing a 3 nC charge from infinity to the third corner of an equilateral triangle.

q₁ = 1 nC, q₂ = 2 nC at two corners, side = 3 cm = 0.03 m, q₃ = 3 nC.

Work done = potential energy of q₃ due to q₁ and q₂:

$$W = q_3(V_1 + V_2) = q_3\left(\frac{kq_1}{r} + \frac{kq_2}{r}\right) = \frac{kq_3(q_1 + q_2)}{r}$$

$$W = \frac{9 \times 10^9 \times 3 \times 10^{-9} \times (1 + 2) \times 10^{-9}}{0.03}$$

$$= \frac{9 \times 10^9 \times 3 \times 10^{-9} \times 3 \times 10^{-9}}{0.03}$$

$$= \frac{81 \times 10^{-9}}{0.03} = 2700 \times 10^{-9} = 2.7 \times 10^{-6} \text{ J} = 2.7 \, \mu J$$

Therefore, the work done is Option 2: 2.7 μJ.

Let initial capacitance be:

$$C_0=\frac{\varepsilon_0A}{5}$$

(since separation = 5 mm)

When mica of thickness 2 mm is inserted, remaining air gap = 3 mm

So capacitor becomes series combination:

$$\frac{1}{C}=\frac{3}{\varepsilon_0A}+\frac{2}{k\varepsilon_0A}$$

Given: charge increases by 25%

$$Q'=1.25Q$$

Since battery connected ⇒ voltage constant ⇒

$$C'=1.25C_0$$

step 1: equate

$$\frac{\varepsilon_0A}{3+\frac{2}{k}}=1.25\cdot\frac{\varepsilon_0A}{5}$$

Cancel $$\varepsilon_0A$$:

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

step 2: solve

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

Three drops coalesce: volume conserved: $$3 \times \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3$$, so $$R = 3^{1/3}r$$. Charge: $$3q$$ total.

$$V_r/V_R = \frac{kq/r}{k(3q)/R} = \frac{R}{3r} = \frac{3^{1/3}}{3} = 3^{-2/3}$$. Ratio = $$1:3^{2/3}$$.

The answer is Option 2.

We are given capacitor P with capacitance $$C_P = 10 \times 10^{-6}$$ F charged to a potential difference of $$V_P = 6.0$$ V, and capacitor Q with capacitance $$C_Q = 20 \times 10^{-6}$$ F (initially uncharged). When P is disconnected from the battery and connected across Q, the charge on Q at equilibrium can be determined as follows.

Since the initial charge on P is given by $$Q = CV$$, substituting yields $$Q_{initial} = C_P \times V_P = 10 \times 10^{-6} \times 6.0 = 60 \times 10^{-6}\text{ C} = 60 \mu\text{C}$$.

After connection, the total charge is conserved because no external path allows charge to escape, and Q starts uncharged. Therefore $$Q_{total} = Q_P + Q_Q = 60 \mu\text{C}$$.

At equilibrium both capacitors share the same potential difference $$V_{common}$$, which follows from $$V_{common} = \frac{Q_{total}}{C_P + C_Q} = \frac{60 \times 10^{-6}}{(10 + 20) \times 10^{-6}} = \frac{60}{30} = 2\text{ V}$$.

Finally, the charge on Q is $$Q_Q = C_Q \times V_{common} = 20 \times 10^{-6} \times 2 = 40 \times 10^{-6}\text{ C} = 4 \times 10^{-5}\text{ C}$$. This shows that $$\alpha = 4$$.

The answer is 4.

$$W = q \int_{A}^{B} \vec{E} \cdot d\vec{r}$$

$$\vec{E} \cdot d\vec{r} = (2x\hat{i} + 3y^2\hat{j} + 4\hat{k}) \cdot (dx\hat{i} + dy\hat{j} + dz\hat{k}) = 2x\,dx + 3y^2\,dy + 4\,dz$$

$$W = 3 \int_{A}^{B} (2x\,dx + 3y^2\,dy + 4\,dz)$$

$$W = 3 \left[ \int_{0}^{5} 2x\,dx + \int_{-2}^{1} 3y^2\,dy + \int_{-5}^{2} 4\,dz \right]$$

$$W = 3 [25 + 9 + 28]$$

$$W = 186 \text{ J}$$

Let original capacitance (no dielectric):

$$C=\frac{\varepsilon_0A}{d}$$

step 1: understand geometry

• Top half (thickness d/2): dielectric $$K_1=2$$, area A
• Bottom half (thickness d/2): split into two equal areas:

• left: $$K_2=3$$, area A/2
• right: $$K_3=5$$, area A/2

step 2: bottom part (parallel combination)

Each has thickness d/2:

$$C_2=\frac{3\varepsilon_0(A/2)}{d/2}=\frac{3\varepsilon_0A}{d}$$

$$C_3=\frac{5\varepsilon_0(A/2)}{d/2}=\frac{5\varepsilon_0A}{d}$$

Parallel:

$$C_{bottom}=C_2+C_3=\frac{8\varepsilon_0A}{d}$$

step 3: top part

$$C_{top}=\frac{2\varepsilon_0A}{d/2}=\frac{4\varepsilon_0A}{d}$$

step 4: series combination

Top and bottom are in series:

$$\frac{1}{C'}=\frac{1}{C_{top}}+\frac{1}{C_{bottom}}=\frac{d}{4\varepsilon_0A}+\frac{d}{8\varepsilon_0A}$$

$$=\frac{3d}{8\varepsilon_0A}$$

$$C'=\frac{8\varepsilon_0A}{3d}$$

Circuit simplification at steady state

In steady state, the capacitor branch is an open circuit. We simplify the resistance from right to left:

• Right loop series: $$4\ \Omega+4\ \Omega+2\ \Omega=10\ \Omega$$
• Parallel with middle $$10\ \Omega$$ resistor: $$\frac{10 \times 10}{10 + 10} = 5\ \Omega$$
• Middle loop series: $$5\ \Omega + 5\ \Omega + 2\ \Omega = 12\ \Omega$$
• Parallel with first $$12\ \Omega$$ resistor: $$\frac{12 \times 12}{12 + 12} = 6\ \Omega$$

Total current from the battery:

$$I_{total} = \frac{12\ V}{6\ \Omega} = 2\ A$$

• The $$2\ A$$ current splits equally at the first branch ($$12\ \Omega$$ each), so $$1\ A$$ enters the middle section.
• That $$1\ A$$ splits equally at the second branch ($$10\ \Omega$$ each), so $$0.5\ A$$ flows through the rightmost $$4\ \Omega$$ vertical resistor.

The voltage across the capacitor ($$V_{C}$$) is the same as the voltage across the $$4\ \Omega$$ vertical resistor:

$$V_{C} = 0.5\ A \times 4\ \Omega = 2\ V$$

Final charge ($$Q$$):

$$Q = C \times V_{C}$$

$$Q = 100\ \mu F \times 2\ V = 200\ \mu C$$

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

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

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

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

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

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

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

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 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 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.

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: $$\sqrt{\dfrac{4\pi^3 \epsilon_0 m a^3}{q_0 Q}}$$.

Two conducting spheres $$A$$ (radius $$r_A = 5 \text{ mm}$$) and $$B$$ (radius $$r_B = 10 \text{ mm}$$) are connected by a conducting wire, so they reach the same potential in equilibrium. Thus $$V_A = V_B$$ gives $$\dfrac{kQ_A}{r_A} = \dfrac{kQ_B}{r_B}$$. From this, $$\dfrac{Q_A}{Q_B} = \dfrac{r_A}{r_B} = \dfrac{5}{10} = \dfrac{1}{2}$$.

The electric field at the surface of a conducting sphere is $$E = \dfrac{kQ}{r^2} = \dfrac{V}{r}$$. Since both spheres are at the same potential $$V$$, $$\dfrac{E_A}{E_B} = \dfrac{V/r_A}{V/r_B} = \dfrac{r_B}{r_A} = \dfrac{10}{5} = \dfrac{2}{1}$$. Therefore, $$E_A : E_B = 2 : 1$$.

The correct answer is Option B: $$2:1$$.

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

Equivalent capacitance:

$$C_{eq}=2C$$

Initial energy:

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

$$E_1=CV^2$$

Now switch is opened and dielectric constant

k=5

is inserted.

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

New capacitance:

$$C_1'=5C$$

Energy in left capacitor:

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

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

Right capacitor becomes isolated, so charge remains constant.

Initial charge on it was

Q=CV

After dielectric insertion,

$$C_2'=5C$$

Energy:

$$E_R=\frac{Q^2}{2C_2'}=\frac{(CV)^2}{2(5C)}=2(5C)(CV)2​$$

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

Total final energy:

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

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

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

Therefore

$$\frac{E_1}{E_2}=\frac{CV^2}{(13/5)CV^2}$$

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

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

Write the capacitance of an isolated sphere.

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

Write the capacitance of the spherical capacitor.

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

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

Apply the condition $$C_2 = nC_1$$.

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

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

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

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

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

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

The mass of each particle is $$m = 10$$ g $$= 0.01$$ kg, the charge is $$q = 2.0 \times 10^{-7}$$ C, the coefficient of friction is $$\mu = 0.25$$, and $$g = 10$$ m/s$$^2$$.

The two identical charged particles repel each other with the Coulomb force, and they remain in equilibrium because friction balances this repulsive force. At limiting equilibrium the Coulomb repulsive force equals the maximum static friction force:

$$F_{\text{Coulomb}} = F_{\text{friction}}$$ $$\frac{kq^2}{L^2} = \mu mg$$

Solving for $$L^2$$ gives $$L^2 = \frac{kq^2}{\mu mg}$$. Substituting numerical values: $$L^2 = \frac{9 \times 10^9 \times (2.0 \times 10^{-7})^2}{0.25 \times 0.01 \times 10}$$. Calculating the numerator: $$(2.0 \times 10^{-7})^2 = 4.0 \times 10^{-14}$$ and $$9 \times 10^9 \times 4.0 \times 10^{-14} = 36 \times 10^{-5} = 3.6 \times 10^{-4}$$. The denominator is $$0.25 \times 0.01 \times 10 = 0.025$$. Therefore, $$L^2 = \frac{3.6 \times 10^{-4}}{0.025} = \frac{3.6 \times 10^{-4}}{2.5 \times 10^{-2}} = 1.44 \times 10^{-2} \text{ m}^2$$.

Thus, $$L = \sqrt{1.44 \times 10^{-2}} = 0.12 \text{ m} = 12 \text{ cm}$$, so the separation $$L = 12$$ cm.

The correct answer is Option A.

We need to find the potential difference $$V$$ between two parallel metal plates that carry charges $$q_1$$ and $$q_2$$ and form a capacitor of capacitance $$C$$.

When two large conducting plates carrying charges $$q_1$$ and $$q_2$$ are brought close together, the total charge is distributed across their outer and inner surfaces due to electrostatic induction.

The charge on the outermost surfaces of both plates must be equal and is given by the average of the total charge: $$q_{outer} = \frac{q_1 + q_2}{2}$$.

The charge on the inner surface of the first plate ($$q_{inner1}$$) is found by subtracting the outer surface charge from its total charge: $$q_{inner1} = q_1 - \frac{q_1 + q_2}{2} = \frac{q_1 - q_2}{2}$$.

By conservation of charge and induction, the inner surface of the second plate carries an equal and opposite charge: $$q_{inner2} = -\left(\frac{q_1 - q_2}{2}\right)$$.

The potential difference between the plates of a capacitor depends solely on the magnitude of the charge residing on their inner facing surfaces, which is $$q = \frac{q_1 - q_2}{2}$$.

Using the standard definition of capacitance $$C = \frac{q}{V}$$, we can rearrange the equation to find the potential difference: $$V = \frac{q}{C}$$.

Substituting the inner surface charge expression into the potential formula yields: $$V = \frac{q_1 - q_2}{2C}$$.

Therefore, the correct answer is Option C: (q_1 - q_2) / 2C.

Two point charges $$A = +8 \times 10^{-6}$$ C and $$B = -8 \times 10^{-6}$$ C are placed at a distance $$d$$ apart. The electric field at the midpoint $$O$$ is $$6.4 \times 10^4$$ N C$$^{-1}$$. We need to find the distance $$d$$.

Understand the configuration and direction of electric fields at the midpoint.

The midpoint $$O$$ is at a distance of $$d/2$$ from each charge.

The electric field due to the positive charge $$A$$ at point $$O$$ points away from $$A$$ (i.e., from $$A$$ toward $$B$$).

The electric field due to the negative charge $$B$$ at point $$O$$ points toward $$B$$ (i.e., also from $$A$$ toward $$B$$).

Since both fields point in the same direction, they add up.

Calculate the electric field due to each charge at the midpoint.

The magnitude of the electric field due to charge $$A$$ at the midpoint:

$$E_A = \frac{kq}{(d/2)^2} = \frac{4kq}{d^2}$$

Similarly, the magnitude of the electric field due to charge $$B$$ at the midpoint:

$$E_B = \frac{kq}{(d/2)^2} = \frac{4kq}{d^2}$$

where $$q = 8 \times 10^{-6}$$ C and $$k = 9 \times 10^9$$ N m$$^2$$ C$$^{-2}$$.

Write the total electric field at the midpoint.

Since both fields are in the same direction:

$$E_{total} = E_A + E_B = \frac{4kq}{d^2} + \frac{4kq}{d^2} = \frac{8kq}{d^2}$$

Substitute the known values and solve for $$d$$.

$$6.4 \times 10^4 = \frac{8 \times 9 \times 10^9 \times 8 \times 10^{-6}}{d^2}$$

Calculating the numerator:

$$8 \times 9 \times 10^9 \times 8 \times 10^{-6} = 8 \times 72 \times 10^{3} = 576 \times 10^{3} = 5.76 \times 10^5$$

So:

$$6.4 \times 10^4 = \frac{5.76 \times 10^5}{d^2}$$

Solve for $$d$$.

$$d^2 = \frac{5.76 \times 10^5}{6.4 \times 10^4} = \frac{576}{64} = 9$$

$$d = 3 \text{ m}$$

The correct answer is Option B.

Between the two plates of a parallel plate capacitor, the electric field is due to both plates. Each plate produces a uniform electric field of $$\frac{\sigma}{2\epsilon_0}$$.

Between the plates, the fields from both plates add up:

$$E_{between} = \frac{\sigma}{2\epsilon_0} + \frac{\sigma}{2\epsilon_0} = \frac{\sigma}{\epsilon_0}$$

The force on the charged particle between the plates is:

$$F = qE_{between} = \frac{q\sigma}{\epsilon_0} = 10 \text{ N}$$

When one plate is removed, only one plate remains. The electric field due to a single infinite plate is:

$$E_{single} = \frac{\sigma}{2\epsilon_0}$$

The force on the charged particle now becomes:

$$F' = qE_{single} = \frac{q\sigma}{2\epsilon_0} = \frac{F}{2} = \frac{10}{2} = 5 \text{ N}$$

The correct answer is Option A.

A long cylindrical volume of radius $$R$$ contains a uniformly distributed charge of density $$\rho$$. A charge particle $$q$$ revolves around the cylinder in a circular path.

Find the electric field using Gauss's Law: Consider a cylindrical Gaussian surface of radius $$r$$ (where $$r > R$$) and length $$L$$, coaxial with the charged cylinder.

The charge enclosed by the Gaussian surface:

$$Q_{\text{enc}} = \rho \cdot \pi R^2 \cdot L$$

By Gauss's Law:

$$\oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enc}}}{\varepsilon_0}$$

$$E \cdot (2\pi r L) = \frac{\rho \pi R^2 L}{\varepsilon_0}$$

Solving for $$E$$:

$$E = \frac{\rho R^2}{2\varepsilon_0 r}$$

Apply the condition for circular motion: For the charged particle $$q$$ revolving in a circular orbit of radius $$r$$, the electrostatic force provides the centripetal force:

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

Find the kinetic energy: The kinetic energy is:

$$KE = \frac{1}{2}mv^2 = \frac{1}{2}qEr$$

Substituting the expression for $$E$$:

$$KE = \frac{1}{2} \cdot q \cdot \frac{\rho R^2}{2\varepsilon_0 r} \cdot r$$

$$KE = \frac{1}{2} \cdot \frac{q\rho R^2}{2\varepsilon_0}$$

$$KE = \frac{q\rho R^2}{4\varepsilon_0}$$

The kinetic energy of the particle is $$\dfrac{q\rho R^2}{4\varepsilon_0}$$.

The correct answer is Option A.

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

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

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

The capacitance with dielectric is given by

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

The maximum voltage before dielectric breakdown is

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

Using $$Q = CV$$ yields

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

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

Substituting the given values:

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

Computing the denominator step by step:

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

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

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

Therefore, the correct answer is Option D.

We consider each statement about conductors in electrostatic equilibrium.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

We have a parallel plate capacitor maintained at a fixed voltage $$V$$ by a battery, and its dielectric constant is changed from $$K_1 = 10$$ to $$K_2 = 15$$.

Because the battery keeps the voltage constant, the energy stored in the capacitor is given by the well-known formula

$$U = \frac{1}{2} C V^2$$

Inserting a dielectric multiplies the vacuum capacitance $$C_0$$ by the dielectric constant $$K$$ so that $$C = K C_0$$. Hence the stored energy becomes

$$U = \frac{1}{2} K C_0 V^2$$

Comparing the energies before and after introducing the dielectric gives

$$\frac{U_2}{U_1} = \frac{K_2}{K_1} = \frac{15}{10} = 1.5$$

Therefore, the percentage increase is

$$\text{Percentage increase} = \left(\frac{U_2 - U_1}{U_1}\right)\times 100 = (1.5 - 1)\times 100 = 50\%$$

Thus the energy stored in the capacitor increases by 50%, corresponding to Option A.

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

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

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

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

The new energy is:

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

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

$$E' = 1.44 \times E$$

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

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

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

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

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

Multiply through by 100 to clear decimals:

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

Divide by 4:

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

Solve the quadratic equation: Using the quadratic formula:

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

Taking the positive root (charge must be positive):

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

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

The correct answer is Option A.

Here $$B_1,\ B_2,and\ B_3$$​ form an equilateral triangle.

Let the force between any two balls be F.

Consider ball $$B_1$$. It experiences two repulsive forces:

• Force F due to $$B_2$$​
• Force F due to $$B_3$$

These two forces act along $$B_1B_2\ and\ B_1B_3$$

Since $$B_1B_2B_3$$​ is an equilateral triangle, the angle between these two forces is

$$60^{\circ}$$

Hence the resultant force on $$B_1$$ is

$$R=\sqrt{F^2+F^2+2F^2\cos60^{\circ}}$$

$$R=\sqrt{2F^2+F^2}$$

$$R=\sqrt{3}F$$

Therefore,

Net force on one ballForce between any two balls=

$$\frac{\text{Net force on one ball}}{\text{Force between any two balls}}=\frac{\sqrt{\ 3}F}{F}$$

$$=\sqrt{3}:1$$

Hence the required ratio is

Let the battery of emf $$8 \text{ V}$$ be connected between the upper junction $$X$$ (positive) and the lower junction $$Y$$ (zero potential). The network between $$X$$ and $$Y$$ consists of two series branches kept in parallel and a bridging capacitor $$C_3$$ joining the mid-points of the two branches.

Branch-1 (left): $$C_1 = 12 \,\mu\text{F}$$ in series with $$C_2 = 4 \,\mu\text{F}$$. Branch-2 (right): $$C_4 = 2 \,\mu\text{F}$$ in series with $$C_5 = 2 \,\mu\text{F}$$. The mid-points of the two branches are the plates joined by $$C_3 = 4 \,\mu\text{F}$$.

Step 1 : Equivalent capacitance of each series branch
For two capacitors in series, $$C_{\text{eq}} = \frac{C_a\,C_b}{C_a + C_b}$$.

Left branch: $$C_{L} = \frac{12 \times 4}{12 + 4} = \frac{48}{16} = 3 \,\mu\text{F}$$.

Right branch: $$C_{R} = \frac{2 \times 2}{2 + 2} = \frac{4}{4} = 1 \,\mu\text{F}$$.

Step 2 : Charge flowing through each branch
Because the two branches are in parallel across the same 8 V, $$Q_{L} = C_{L}\,V = 3 \times 8 = 24 \,\mu\text{C}$$ $$Q_{R} = C_{R}\,V = 1 \times 8 = 8 \,\mu\text{C}$$.

Step 3 : Potentials of the mid-points
In a series combination the charge on each capacitor is the same, so

Left branch: Voltage across $$C_2$$ (lower capacitor) $$V_{C_2} = \frac{Q_L}{C_2} = \frac{24}{4} = 6 \text{ V}$$. Therefore the potential of its upper plate (mid-point $$A$$) with respect to $$Y$$ is $$6 \text{ V}$$.

Right branch: Voltage across $$C_5$$ (lower capacitor) $$V_{C_5} = \frac{Q_R}{C_5} = \frac{8}{2} = 4 \text{ V}$$. Therefore the potential of its upper plate (mid-point $$B$$) with respect to $$Y$$ is $$4 \text{ V}$$.

Step 4 : Potential difference across $$C_3$$
$$\Delta V_{3} = V_A - V_B = 6 \text{ V} - 4 \text{ V} = 2 \text{ V}$$.

Step 5 : Charge on $$C_3$$
Using $$Q = C\,\Delta V$$, $$Q_3 = 4 \,\mu\text{F} \times 2 \text{ V} = 8 \,\mu\text{C}$$.

Hence, the charge stored in $$C_3$$ is $$\mathbf{8 \,\mu C}$$.

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

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

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

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

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

The electrostatic energy loss is 125 nJ.

Initial charge on $$C_1$$:   $$Q_i = C_1 V_1 = 5\ \mu\text{F} \times 30\ \text{V} = 150\ \mu\text{C}$$

Common potential after closing the switch:

$$V_c = \frac{Q_i}{C_1 + C_2} = \frac{150}{5 + 10} = \frac{150}{15} = 10\ \text{V}$$

Final charge on $$C_2$$:    $$Q_2 = C_2 V_c = 10\ \mu\text{F} \times 10\ \text{V} = 100\ \mu\text{C}$$

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

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

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

V =100V

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

so,

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

Substitute:

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

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

Multiply by 12:

$$2:3$$

Total voltage is

$$V_1+V_2=100$$

So

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

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

V=60V

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

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

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

Step 1: Individual capacitances

$$C=\frac{\varepsilon_0A}{d}​$$

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

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

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

Step 2: Equivalent capacitance (parallel)

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

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

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

Step 3: Compare with given form

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

$$\Rightarrow x=23$$

Final Answer:

23

In a potentiometer, the balancing length is proportional to the emf of the cell: $$E \propto l$$. Let the emf of the first cell be $$E_1$$ and of the second be $$E_2$$ with $$\frac{E_1}{E_2} = \frac{3}{2}$$. The first cell gives a balancing length of $$l_1 = 75$$ cm, and since emf is proportional to balancing length, we have $$\frac{E_1}{E_2} = \frac{l_1}{l_2}$$, which leads to $$\frac{3}{2} = \frac{75}{l_2}$$. Solving for $$l_2$$ yields $$l_2 = \frac{75 \times 2}{3} = 50 \text{ cm}$$. The difference in balancing lengths is $$\Delta l = l_1 - l_2 = 75 - 50 = 25 \text{ cm}$$. The answer is 25 cm.

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

Total initial charge:

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

The battery is disconnected, so charge is conserved.

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

$$C' = KC = 9C$$

The new total capacitance of the parallel combination is:

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

Since charge is conserved, the new potential difference is:

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

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

We have a long cylindrical volume with uniform charge density $$\rho \text{ C m}^{-3}$$. We need to find the electric field at a distance $$x = \dfrac{2\epsilon_0}{\rho} \text{ m}$$ from the axis.

Consider a cylindrical Gaussian surface of radius $$x$$ and length $$l$$, coaxial with the charged cylinder.

By Gauss's law:

$$\oint \vec{E} \cdot d\vec{A} = \frac{q_{enc}}{\epsilon_0}$$

The enclosed charge is:

$$q_{enc} = \rho \times \pi x^2 l$$

The flux through the curved surface of the Gaussian cylinder is:

$$E \times 2\pi x l = \frac{\rho \pi x^2 l}{\epsilon_0}$$

$$E = \frac{\rho x}{2\epsilon_0}$$

$$E = \frac{\rho}{2\epsilon_0} \times \frac{2\epsilon_0}{\rho} = 1 \text{ V m}^{-1}$$

The electric field at the given distance is $$1 \text{ V m}^{-1}$$.

We need to find the charge stored in the capacitor under steady-state conditions for the given circuit.

1. Understand the Steady-State Condition

In a DC circuit, once a capacitor is fully charged and reaches a steady state, it acts as an open circuit (infinite resistance). This means:

• No current flows through the branch containing the capacitor ($$C$$) and the resistor ($$R'$$).
• Current only flows through the closed loop containing the battery ($$E$$), its internal resistance ($$r$$), and the parallel resistor ($$R$$).

2. Calculate the Circuit Current ($$I$$)

Using the values given in the circuit diagram 

• EMF of the battery ($$E$$) = $$10\text{ V}$$
• Internal resistance ($$r$$) = $$10\ \Omega$$
• External resistance ($$R$$) = $$100\ \Omega$$

The total current flowing through the main loop is:

$$I = \frac{E}{R + r} = \frac{10}{100 + 10} = \frac{10}{110} = \frac{1}{11}\text{ A}$$

3. Determine the Potential Difference Across the Capacitor ($$V_C$$)

Since no current flows through the branch with $$R' = 200\ \Omega$$, there is no potential drop across it ($$V_{R'} = 0$$). Therefore, the potential difference ($$V_C$$) across the capacitor plates is exactly equal to the potential difference across the parallel resistor ($$R$$):

$$V_C = I \times R$$

$$V_C = \frac{1}{11}\text{ A} \times 100\ \Omega = \frac{100}{11}\text{ V}$$

4. Calculate the Stored Charge ($$Q$$)

The capacitance value given is $$C = 1.1\ \mu\text{F} = 1.1 \times 10^{-6}\text{ F}$$. Using the formula for charge stored in a capacitor ($$Q = C \times V_C$$):

$$Q = 1.1 \times 10^{-6}\text{ F} \times \frac{100}{11}\text{ V}$$

We can rewrite $$1.1$$ as $$\frac{11}{10}$$ to simplify the expression easily:

$$Q = \left(\frac{11}{10} \times 10^{-6}\right) \times \frac{100}{11}$$

$$Q = \frac{100}{10} \times 10^{-6}\text{ C} = 10 \times 10^{-6}\text{ C}$$

Conclusion

The charge stored in the capacitor in steady state is 10 $$\times 10^{-6}\text{ C}$$.

In a potentiometer, the balancing length is proportional to the EMF of the cell:

$$\dfrac{E_1}{E_2} = \dfrac{l_1}{l_2}$$

Given: $$E_1 = 1.20 \text{ V}$$, $$l_1 = 36 \text{ cm}$$, and $$E_2 = 1.80 \text{ V}$$.

Finding $$l_2$$:

$$l_2 = l_1 \times \dfrac{E_2}{E_1} = 36 \times \dfrac{1.80}{1.20} = 36 \times 1.5 = 54 \text{ cm}$$

The difference in balancing lengths:

$$\Delta l = l_2 - l_1 = 54 - 36 = 18 \text{ cm}$$

Therefore, the difference is $$\boxed{18}$$ cm.

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

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

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

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

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

$$R = 3r$$

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

Total charge on the bigger drop:

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

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

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

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

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

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

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

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

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

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

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

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

Hence

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

and

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

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

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

Thus, the answer is 240.

$$\text{The first, second, and third capacitors are connected in parallel between terminal } A \text{ and node } P.$$

$$\text{The fourth and fifth capacitors are connected across the same node } P $$ on both sides, which short-circuits them

$$\text{The sixth capacitor is connected in series between node } P \text{ and terminal } B.$$

Parallel combination of the first 3 active capacitors: $$C_p = 8 + 8 + 8 = 24\ \mu\text{F}$$

Equivalent capacitance of the entire network with the final $$8\ \mu\text{F}$$ capacitor in series:

$$C_{AB} = \frac{C_p \times 8}{C_p + 8} = \frac{24 \times 8}{24 + 8} = \frac{24 \times 8}{32} = 6\ \mu\text{F}$$