Capacitor

$C_a=\frac{K \varepsilon_0 A}{d}$

$ \therefore $ $q_a=\frac{K \varepsilon_0 A}{d} V$

$ E_a =\frac{q_a}{K A \varepsilon_0}$

$=\frac{K \varepsilon_0 A V}{d K \varepsilon_0 A} $

$ =\frac{V}{d}$

$\begin{aligned} C_b & =\frac{\varepsilon_0 A}{d+\left(\frac{d}{K}\right)} \\\\ & =\frac{\varepsilon_0 A K}{d(K+1)}\end{aligned}$

$q_b=\frac{\varepsilon_0 A K V}{d(K+1)}$

$\left(E_b\right)_{\text {dielectric }} $

$=\frac{E_{\text {air }}}{K} $

$=\frac{q_b}{K A \varepsilon_0} $

$=\frac{\varepsilon_0 A K V}{d(K+1)\left(K A \varepsilon_0\right)}$

$ =\frac{V}{d(K+1)}$

$\begin{aligned} &\therefore\text { Capacitance decrease by a factor of } \frac{1}{K+1} \\\\ & \text { Work done in the process }=U_f-U_i \\\\ &=\frac{1}{2}\left(C_f-C_i\right) V^2 \\\\ &=\frac{1}{2}\left(\frac{\varepsilon_0 A K}{d(K+1)}-\frac{K \varepsilon_0 A}{d}\right) V^2 \\\\ &=\frac{1}{2} V^2 \frac{\varepsilon_0 A K}{d}\left(\frac{1}{K+1}-1\right) \\\\ &=\frac{1}{2} \frac{\varepsilon_0 A K V^2}{d} \frac{1-K-1}{K+1} \\\\ &=\frac{1}{2} \frac{\varepsilon_0 A V^2}{d}\left(\frac{-K^2}{K+1}\right)\end{aligned}$

We can think of this configuration to be made up of three parts:

(a) Capacitor of plate area s/2, plate separation d, and filled with a dielectric of relative permittivity $\in$₁ = 2 (lower half). The capacitance of this part is ${C_a} = { \in _1}{ \in _0}{{s/2} \over d} = { \in _0}{s \over d} = {C_1}$.

(b) Capacitor of plate area s/2, plate separation d/2, and filled with a dielectric of relative permittivity $\in$₁ = 2 (left upper half). The capacitance of this part is ${C_b} = { \in _1}{ \in _0}{{s/2} \over {d/2}} = 2{ \in _0}{s \over d} = 2{C_1}$.

(c) Capacitor of plate area s/2, plate separation d/2, and filled with a dielectric of relative permittivity $\in$₂ = 4 (right upper half). The capacitance of this part is ${C_c} = { \in _2}{ \in _0}{{s/2} \over {d/2}} = 4{ \in _0}{s \over d} = 4{C_1}$.

The capacitors C_b and C_c are connected in series with equivalent capacitance

${C_{bc}} = {{{C_b}{C_c}} \over {{C_b} + {C_c}}} = {{(2{C_1})(4{C_1})} \over {(2{C_1}) + (4{C_1})}} = {4 \over 3}{C_1}$.

The capacitor C_bc is connected in parallel with C_a. The equivalent capacitance is

$\left. {{C_2} = {C_{abc}} = {C_a}} \right\|{C_{bc}} = {C_1} + {4 \over 3}{C_1} = {7 \over 3}{C_1}$.

Let A be area of each plate and d is the distance between the plates.

The given capacitor is equivalent to two capacitors in parallel with capacitances

${C_1} = {{K{\varepsilon _0}(A/3)} \over d} = {{K{\varepsilon _0}A} \over {3d}}$

${C_2} = {{{\varepsilon _0}(2A/3)} \over d} = {{2{\varepsilon _0}A} \over {3d}}$

$\therefore$ $C = {C_1} + {C_2}$

$ = {{K{\varepsilon _0}A} \over {3d}} + {{2{\varepsilon _0}A} \over {3d}} = {{{\varepsilon _0}A} \over {3d}}(K + 2)$

$\therefore$ ${C \over {{C_1}}} = {{K + 2} \over K}$

Hence, option (d) is correct.

Let V be potential difference between the plates. Then

${E_1} = {V \over d}$ and ${E_2} = {V \over d}$

$\therefore$ ${{{E_1}} \over {{E_2}}} = 1$

Hence, option (a) is correct and option (b) is incorrect.

${Q_1} = {C_1}V = {{K{\varepsilon _0}A} \over {3d}}V$ and ${Q_2} = {C_2}V = {{2{\varepsilon _0}A} \over {3d}}V$

$\therefore$ ${{{Q_1}} \over {{Q_2}}} = {K \over 2}$

Hence, option (c) is incorrect.

Let the charges on the $ 3 \mu \mathrm{F} $ and $ 2 \mu \mathrm{F} $ capacitors be $ q $ and $ q' $ respectively. The charge on the lower plate of the $ 4 \mu \mathrm{F} $ capacitor is $ -80 \mu \mathrm{C} $. The lower plate of the $ 4 \mu \mathrm{F} $ capacitor and the upper plates of the $ 2 \mu \mathrm{F} $ and $ 3 \mu \mathrm{F} $ capacitors form an isolated system. Therefore, the net charge on this system must be zero, which can be represented as :

$ q + q' - 80 \mu \mathrm{C} = 0 \quad \text{..........(i)} $

The potentials ($ V = q / C $) across these two capacitors are equal, which gives :

$ \frac{q}{3} = \frac{q'}{2} \quad \text{..........(ii)} $

By substituting $ q' $ from equation (ii) into equation (i), we can solve for $ q $ and find :

$ q = 48 \mu \mathrm{C} $

When switch S is connected to terminal 1, the potential difference across the 2 $\mu$F capacitor is V volt. Therefore, energy stored in the system is

${U_1} = {1 \over 2}{C_1}{V^2} = {1 \over 2} \times 2 \times {V^2}$

$ = {V^2}\,\mu J$

When switch S is turned to terminal 2, the charge will flow from 2 $\mu$F capacitor to 8 $\mu$F capacitor until their potentials are equalized. The common potential is

${V^2} = {q \over {{C_1} + {C_2}}} = {{{C_1}V} \over {{C_1} + {C_2}}}$

$ = {{2V} \over {(2 + 8)}} = {V \over 5}$ volt

$\therefore$ Energy stored in the system now will be

${U_2} = {1 \over 2}({C_1} + {C_2})V_2^2$

$ = {1 \over 2}(2 + 8) \times {\left( {{V \over 5}} \right)^2} = {{{V^2}} \over 5}\mu J$

$\therefore$ Percentage loss of energy is

${{{U_1} - {U_2}} \over {{U_1}}} \times 100 = {{\left( {{V^2} - {{{V^2}} \over 5}} \right)} \over {{V^2}}} \times 100 = 80\% $

Step 1: Initial State (Switch Open)

Let $q_1$ and $V_1$ be the charge and voltage on capacitor $C_1 = 3\,\mu\mathrm{F}$. Let $q_2$ and $V_2$ be the charge and voltage on capacitor $C_2 = 6\,\mu\mathrm{F}$.

When the switch $S$ is open, the two capacitors are connected in series. This means that the charge stored on each capacitor is the same. So, $q_1 = q_2$.

The voltages on the capacitors add up to the battery voltage: $V_1 + V_2 = 9\,\mathrm{V}$

The charges and voltages on capacitors in series also follow this equation: $C_1 V_1 = C_2 V_2$ This means the charge on each capacitor is the same since $q_1 = C_1 V_1$ and $q_2 = C_2 V_2$.

Solving these two equations gives:

• $V_1 = 6\,\mathrm{V}$, $q_1 = 3\,\mu\mathrm{F} \times 6\,\mathrm{V} = 18\,\mu\mathrm{C}$
• $V_2 = 3\,\mathrm{V}$, $q_2 = 6\,\mu\mathrm{F} \times 3\,\mathrm{V} = 18\,\mu\mathrm{C}$

Total charge on the side connected to $X$ is $-q_1 + q_2 = 0$. No net charge has moved yet.

Step 2: After Closing the Switch

When the switch is closed, the resistors $R_1 = 3\,\Omega$ and $R_2 = 6\,\Omega$ are connected in series with the capacitors and battery.

Total resistance is $R_1 + R_2 = 9\,\Omega$.

The current from the battery is: $I = \\frac{9\,\mathrm{V}}{9\,\Omega} = 1\,\mathrm{A}$

The voltage drop across each resistor is:

• $V_1' = I \times R_1 = 1\,\mathrm{A} \times 3\,\Omega = 3\,\mathrm{V}$
• $V_2' = I \times R_2 = 1\,\mathrm{A} \times 6\,\Omega = 6\,\mathrm{V}$

The new charges on the capacitors are:

• $q_1' = C_1 \times V_1' = 3\,\mu\mathrm{F} \times 3\,\mathrm{V} = 9\,\mu\mathrm{C}$
• $q_2' = C_2 \times V_2' = 6\,\mu\mathrm{F} \times 6\,\mathrm{V} = 36\,\mu\mathrm{C}$

Final Calculation

The total charge on the plates connected to $X$ is now $-q_1' + q_2' = -9 + 36 = 27\,\mu\mathrm{C}$.

This means that $27\,\mu\mathrm{C}$ of charge flows from $Y$ to $X$ when the switch is closed.

(A) When capacitor is shorted, current will begin to flow and capacitor will start discharging. Because of discharging, magnitude of current in circuit and potential across the two ends will reduce continually until becomes zero. Thermal energy will be generated due to flow of charges.

(B) When wire is moved perpendicularly in magnetic field, due to change in flux associated with wire, a potential difference will be developed across its ends, but not current will flow as circuit is not complete and no heat will be generated.

(C) Since wire is placed in electric field, electrons inside the wire will align in such a way that net electric field inside the metal is zero. Net charges will be developed across the two ends of wire, creating a potential difference. Here, circuit is incomplete, so no current will flow and no heat will be generated.

(D) When a battery is connected, across two ends of wire, there will be constant potential difference. In this case circuit is complete, so current will start flowing and heat will be generated.

For (i) : The net capacitance

$ \begin{aligned} \mathrm{C} & =\frac{C_1 C_2}{C_1+C_2} \\ & =\frac{2 \times 4}{2+4}=\frac{8}{6}=\frac{4}{3} \mu \mathrm{~F} \end{aligned} $

And net resistance, $R=\frac{R_1 R_2}{R_1+R_2}$

$ =\frac{1 \times 2}{1+2}=\frac{2}{3} \Omega $

Thus, time constant, $\mathrm{T}=\mathrm{C} . \mathrm{R}=\frac{4}{3} \times \frac{2}{3}=\frac{8}{9} \mu \mathrm{~s}$

For (ii) :

$ \begin{aligned} \mathrm{C} & =\mathrm{C}_1+\mathrm{C}_2 \\ & =2+4=6 \mu \mathrm{~F} \text { and } \\ \mathrm{R} & =\mathrm{R}_1+\mathrm{R}_2 \\ & =1+2=3 \Omega \end{aligned} $

Time constant T = C.R

$ =6 \times 3=18 \mu \mathrm{~s} $

$ \begin{aligned} &\text { For (iii) : }\\ &\begin{aligned} \mathrm{C} & =\mathrm{C}_1+\mathrm{C}_2 \\ & =2+4=6 \mu \mathrm{~F} \text { and } \\ \mathrm{R} & =\frac{\mathrm{R}_1 \mathrm{R}_2}{\mathrm{R}_1+\mathrm{R}_2} \\ & =\frac{1 \times 2}{1+2}=\frac{2}{3} \Omega \\ \mathrm{~T} & =\mathrm{C} . \mathrm{R}=6 \times \frac{2}{3} \\ & =4 \mu \mathrm{~s} \end{aligned} \end{aligned} $

Given, $Q=Q_{0}\left(1-e^{-\alpha t}\right)$

Where, Q$_0$ is the steady state charge stored in the capacitor.

Q$_0$ = C[PD across capacitor in steady state]

$\mathrm{Q}_{0}=C\left(\frac{\mathrm{V}}{\mathrm{R}_{1}+\mathrm{R}_{2}}\right) \mathrm{R}_{2}$

$\therefore \quad \mathrm{Q}_{0}=\frac{\mathrm{CVR}_{2}}{\mathrm{R}_{1}+\mathrm{R}_{2}}$

From, $Q = {Q_0}(1 - {e^{ - \alpha t}})$

$\alpha$ is $\frac{1}{\tau}$ or $\frac{1}{\mathrm{CR}_{e q}}$

Here, $\mathrm{R}_{e q}$ is equivalent resistance across capacitor after short circuiting the battery.

$\begin{aligned} \therefore \quad R_{e q} & =\frac{\mathrm{R}_{1} R_{2}}{\mathrm{R}_{1}+\mathrm{R}_{2}} \\ \alpha & =\frac{1}{C\left(\frac{R_{1} R_{2}}{R_{1}+R_{2}}\right)}=\frac{R_{1}+R_{2}}{C_{1} R_{2}} \end{aligned}$