Capacitor

Given,

Capacitance of $A=C$

Capacitance of $B=2 C$

$A$ and $B$ are in series, series capacitance is given by

$ \frac{1}{C_{\mathrm{eq}}}=\frac{1}{C}+\frac{1}{2 C} \Rightarrow C_{\mathrm{eq}}=\frac{2}{3} C $

Charge on each cell before removing battery

$ Q=C_{\mathrm{eq}} E=\frac{2 C E}{3} $

Energy before disconnecting battery.

$ E=\frac{1}{2} C_{\mathrm{eq}} V^2=\frac{1}{2} \times \frac{2}{3} \times C E^2=\frac{1}{3} C E^2 $

When battery is disconnected and plates are joined in opposite polarity

$ C=\frac{Q-Q}{C+2 C}=0 $

So, all energy is lost $=\frac{1}{3} C E^2$

Hence, statement (ii) and (iii) are correct and correct option is (ii).

$ \text { Given circuit } $

Each capacitor has $C$ capacitance. Draw the circuit in simplified manner.

Now, From (1)

Capacitor are in parallel connect

$ C_{\mathrm{eq}_1}=C_1+C_2=2 C $

From (3) [both capacitor are parallely connected]

$ C_{\mathrm{eq}_2}=C_1+C_2=2 C $

The circuit will resolves as

$C_{\mathrm{eq}}$ and $C_{\mathrm{eq}}$ are in series connection

$ \begin{array}{ll} \therefore & C_{\mathrm{eq}_3}=\frac{1}{C_{\mathrm{eq}_1}}+\frac{1}{C_{\mathrm{eq}_2}} \\ \Rightarrow & C_{\mathrm{eq}_3}=\frac{1}{2 C}+\frac{1}{2 C}=C \end{array} $

Now, $C_{\mathrm{eq}_3}$ and $C$ are parallely connected. So, net capacitance is $=C+C=2 C$

The given circuit diagram is given as,

$ \therefore \quad C_{A B}=\frac{C_1 C_2}{C_1+C_2}+\frac{C_3 C_4}{C_3+C_4} $

$ \begin{aligned} & =\frac{20 \times 20}{20+20}+\frac{10 \times 10}{10+10} \\ & =10+5=15 \mu \mathrm{~F} \end{aligned} $

We know that, displacement current

$ \begin{array}{rlr} I_D & =\varepsilon_0 \frac{d \phi}{d t} & \\ & =\varepsilon_0 \frac{d}{d t}(E A) & (\because \phi=E A) \\ & =\varepsilon_0 A \frac{d E}{d t} & \\ & =\varepsilon_0 A \frac{d}{d t}\left(\frac{V}{d}\right) & \left(\because E=\frac{V}{d}\right) \\ & =\frac{\varepsilon_0 A}{d} \frac{d V}{d t} & \\ I_D & =C \frac{d V}{d t} & \left(\because C=\frac{\varepsilon_0 A}{d}\right) \\ \frac{d V}{d t} & =\frac{I_D}{C} & \end{array} $

Here, $I_D=150 \mu \mathrm{~A}=1.5 \times 10^{-4} \mathrm{~A}$

$ \begin{aligned} & C =30 \mu \mathrm{~F}=3 \times 10^{-5} \mathrm{~F} \\ \therefore & \frac{d V}{d t} =\frac{1.5 \times 10^{-4}}{3 \times 10^{-4}}=5 \mathrm{Vs}^{-1} \end{aligned} $

Given:

Capacitance $ C = 100 \, \mu\text{F} $

Charging rate $ q = 100 \, \mu\text{C/s} $

Desired potential difference $ V = 100 \, \text{V} $

To find the time $ t $ required to achieve a potential difference of 100 V, use the relationship between charge, capacitance, and potential difference:

$ Q = C \cdot V $

Since charge $ Q $ is also given by:

$ Q = q \cdot t $

Equate the two expressions:

$ C \cdot V = q \cdot t $

Rearrange to solve for time $ t $:

$ t = \frac{C \cdot V}{q} $

Insert the given values:

$ t = \frac{100 \times 10^{-6} \, \text{F} \times 100 \, \text{V}}{100 \times 10^{-6} \, \text{C/s}} $

$ t = 100 \, \text{s} $

Therefore, the time required to produce a potential difference of 100 V between the capacitor plates is 100 seconds.

Charge on the left surface of plate $\mathrm{A} = {{\mathrm{Total\,charge}} \over 2}$

$ = {{{q_1} + {q_2}} \over 2}$

Let right surface of plate A has charge $= x$

And total charge on plate $A = {q_1}$

$\therefore$ $q_1$ = Charge on left surface of plate A + Charge on right surface of plate A

$ = {{{q_1} + {q_2}} \over 2} + x$

$ \Rightarrow x = {q_1} - {{{q_1} + {q_2}} \over 2}$

$ = {{2{q_1} - {q_1} - {q_2}} \over 2}$

$ = {{{q_1} - {q_2}} \over 2}$

Let potential difference between two plates $= V$

For capacitor we know,

$q = CV$

$\therefore$ ${{{q_1} - {q_2}} \over 2} = CV$

$ \Rightarrow V = {{{q_1} - {q_2}} \over {2C}}$

${C_0} = {{{\varepsilon _0}A} \over d}$

$C = {{{\varepsilon _0}A} \over {d - {{3d} \over 4} + {{3d} \over {4K}}}} = {{4{\varepsilon _0}AK} \over {3d + Kd}}$

$ = {{4K{C_0}} \over {3 + K}}$

${{40K \times 40} \over {40K + 40}} = 24$

$40K = 24(K + 1)$

$40K = 24K + 24$

$16K = 24$

$K = {{24} \over {16}} = {3 \over 2} = 1.5$

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

$ \Rightarrow {E_1} = C{V^2}$ ...... (i)

${E_2} = {1 \over 2}(5C){V^2} + {1 \over 2}{{{{(CV)}^2}} \over {5C}}$

$ = {{13} \over 5}C{V^2}$

$ \Rightarrow {{{E_1}} \over {{E_2}}} = {5 \over {13}}$

Equivalent $C = \sum {{C_i}} $

$ = 10\,\mu F$

$\Rightarrow$ Charge $Q = CV$

$ = 200\,\mu C$

Initially $ = {C_0} = 4\pi {\varepsilon _0}{R_1}$

Finally ${{4\pi {\varepsilon _0}{R_1}{R_2}} \over {{R_2} - {R_1}}} = n{C_0} = 4\pi {\varepsilon _0}n{R_1}$

$ \Rightarrow $ ${{{R_2}} \over {{R_2} - {R_1}}} = n$

$ \Rightarrow $$1 - {{{R_1}} \over {{R_2}}} = {1 \over n}$

$ \Rightarrow $ ${{{R_1}} \over {{R_2}}} = {{n - 1} \over n}$

$ \Rightarrow $ ${{{R_2}} \over {{R_1}}} = {n \over {n - 1}}$

Q = CV

As capacitance is constant so, Q $\propto$ V

So graph between V and Q will be a straight line.

Initially voltage, ${V_i} = {{{Q_i}} \over C} = {0 \over {2 \times {{10}^{ - 6}}}} = 0 V$

and ${V_f} = {{{Q_f}} \over C} = {5 \over {2 \times {{10}^{ - 6}}}} = 2.5 \times {10^6}\,V$

So correct graph will be A.

Capacitance of Fig. 1, ${C_0} = {{{\varepsilon _0}A} \over d}$

In Fig. 2, C₁ and C₂ are parallel with each other,

$\therefore$ Equivalent capacitance,

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

$ = {{{\varepsilon _0}A/2} \over d} + {{{\varepsilon _0}{\varepsilon _r}A/2} \over d}$

$ = {{{\varepsilon _0}A} \over {2d}} + {{{\varepsilon _0}{\varepsilon _r}A} \over {2d}}$

$ = {{A{\varepsilon _0}} \over {2d}}\left( {1 + {\varepsilon _r}} \right) = {{{C_0}} \over 2}\left( {1 + {\varepsilon _r}} \right)$

For a discharging capacitor when energy reduces to half the charge would become ${1 \over {\sqrt 2 }}$ times the initial value.

$ \Rightarrow {\left( {{1 \over 2}} \right)^{1/2}} = {e^{ - {t_1}/\tau }}$

Similarly, ${\left( {{1 \over 2}} \right)^3} = {e^{ - {t_2}/\tau }}$

$ \Rightarrow {{{t_1}} \over {{t_2}}} = {1 \over 6}$

$U = {1 \over 2}(k{C_0}){V^2}$

$ \Rightarrow {{U'} \over U} = 1.5$

$\Rightarrow$ Energy increases by 50%

E between two plates is ${\sigma \over {{\varepsilon _0}}}$ and due to one plate is ${\sigma \over {2{\varepsilon _0}}}$ so the force will be halved

So new force F = 5 N

${C_{eq}} = {{120} \over {26}}\mu F$

$ \Rightarrow {Q_{flown}}$ or $Q = {{13 \times 120} \over {26}}\mu C = 60\mu C$

$\Rightarrow$ Charge on 15 $\mu$F capacitor = 60 $\mu$C

As all the capacitors are in series.

Equivalent circuit is

Now, ${{{\varepsilon _0}A} \over d} = 4\,\mu F$

$ \Rightarrow {{12} \over 8}{{{\varepsilon _0}A} \over d} = 6\,\mu F$

${C_{eq}} = {{{\varepsilon _0}A} \over {d - {d \over 2} + {d \over {2k}}}} = {{{\varepsilon _0}A} \over {{d \over 2}}} = {{2{\varepsilon _0}A} \over d}$

If $C = {{{\varepsilon _0}A} \over d}$

$ \Rightarrow {C_{eq}} = 2C$ or ${{{C_{new}}} \over {{C_{old}}}} = {2 \over 1}$

Let initially the charge is q so

${1 \over 2}{{{q^2}} \over C} = {U_i}$

And ${1 \over 2}{{{{(q + 2)}^2}} \over C} = {U_f}$

Given ${{{U_f} - {U_i}} \over {{U_i}}} \times 100 = 44$

${{{{(q + 2)}^2} - {q^2}} \over q} = .44$

$ \Rightarrow q = 10C$

Field inside the dielectric $ = {\sigma \over {k{\varepsilon _0}}}$

According to the given information,

${\sigma \over {k{\varepsilon _0}}} = 3.6 \times {10^7}$

$ \Rightarrow {{{Q \over A}} \over {k{\varepsilon _0}}} = 3.6 \times {10^7}$

$ \Rightarrow k = 2.33$

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

$ \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)}$

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

Given, plate area, $A=10 \mathrm{~cm}^2$

$ =10^{-3} \mathrm{~m}^2 $

Plate separation, $d=3 \mathrm{~mm}=3 \times 10^{-3} \mathrm{~m}$

Potential difference, $V=12 \mathrm{~V}$

Capacitance of parallel plate capacitor,

$ \begin{aligned} & C=\frac{\varepsilon_0 A}{d}=\frac{\varepsilon_0 \times 10^{-3}}{3 \times 10^{-3}}=\frac{\varepsilon_0}{3} \\ \Rightarrow \quad & C=\frac{\varepsilon_0}{3} \end{aligned} $

Charge stored in capacitor,

$ q=C V=\frac{\varepsilon_0}{3} \times 12=4 \varepsilon_0 C $

∵ Work done, $W=q V$

$ =4 \varepsilon_0 12=48 \varepsilon_0 $

When dielectric material (with $K=3$ ) is introduced between the parallel plate capacitor, then new value of capacitance,

$ C^{\prime}=\frac{K \varepsilon_0 A}{d}=K C=3 \times \frac{\varepsilon_0}{3}=\varepsilon_0 $

Charge stored $q^{\prime}=C^{\prime} V^{\prime}$

$ \Rightarrow \quad q^{\prime}=\varepsilon_0 \cdot \frac{V}{K}=\varepsilon_0 \frac{12}{3}=4 \varepsilon_0 $

Work done $W=q^{\prime} V^{\prime}$

$ \begin{aligned} & =4 \varepsilon_0 \cdot \frac{V}{K}=4 \varepsilon_0 \frac{12}{3}=16 \varepsilon_0 \\ & =\alpha \varepsilon_0\,\,\,\,\text { (Given) } \\ \alpha & =16 \end{aligned} $

The given circuit diagram is redrawn as,

Equivalent capacitance between point $A$ And $B$ is given as,

$ \begin{aligned} C_{A B} & =\frac{\left(C_2+C_3\right) \times C_1}{\left(C_2+C_3\right)+C_1} \\ & =\frac{(3 C+2 C) \times C}{3 C+2 C+C} \\ & =\frac{5 C \times C}{6 C}=\frac{5 C}{6} \end{aligned} $

Given, $C_1=C_2=C_3=1 \mu \mathrm{~F}=10^{-6} \mathrm{~F}$

$ V=9 \mathrm{~V} $

When switch is thrown to the right side, then left side of switch is open. Hence, capacitors $C_2$ and $C_3$ are uneffective. Therefore, only capacitor $C_2$ is in valid connection.

∴ Charge on capacitor $C_1$,

$ Q_1=C_1 V=1 \times 10^{-6} \times 9=9 \times 10^{-6} \mathrm{C}=9 \mu \mathrm{C} $

Now, when switch is thrown to left, then capacitors $C_1, C_2$ and $C_3$ are connected in series and charge of charged capacitor $C_1$ is flowing towards uncharged capacitors $C_2$ and $C_3$.

Since, $C_1=C_2=C_3$, hence $V_{C_1}=V_{C_2}=V_{C_3}$

Therefore, $Q_2=Q_3=\frac{Q_1}{3}=\frac{9}{3}=3 \mu \mathrm{C}$

The given circuit diagram can be represented as

Since, $\frac{C_{\mathrm{r}}}{C_2}=\frac{C_3}{C_4}=\frac{1}{3}$

Hence, given circuit follows the condition of balance wheatstone bridge.

Hence, capacitor $C_5$ is ineffective.

Now, the circuit diagram is redrawn as

$\begin{aligned} C_{A B} & =\frac{C_1 C_2}{C_1+C_2}+\frac{C_3 C_4}{C_3+C_4}=\frac{1 \times 3}{1+3}+\frac{2 \times 6}{2+6} \\ & =\frac{3}{4}+\frac{12}{8}=\frac{9}{4} \mu \mathrm{F} \end{aligned}$

Given, $C_1=1 \mu \mathrm{F}, V=9 \mathrm{~V}$,

$C_2=2 \mu \mathrm{F}, C_3=3 \mu \mathrm{F}$

Equivalent capacitance of $C_2$ and $C_3$

$C^{\prime}=C_2+C_3=2+3=5 \mu \mathrm{F}$

Common potential difference when $C_1$ and $C^{\prime}$ are combined, is given as

$\begin{aligned} & V^{\prime}=\frac{\text { total charge }(Q)}{\text { total copacitance }\left(C_1+C^{\prime}\right)} \\ & V^{\prime}=\frac{Q}{C_1+C^{\prime}} \end{aligned}$

$\begin{aligned} & \text { But } Q=C_1 V \\ & \therefore V^{\prime}=\frac{C_1 V}{C_1+C^{\prime}}=\frac{1 \times 9}{1+5}=\frac{9}{6}=1.5 \mathrm{~V} \end{aligned}$

Charge on capacitor $C_3$ is given as

$\begin{aligned} Q_3 & =C_3 V^{\prime} \\ & =3 \times 10^{-6} \times 1.5=4.5 \times 10^{-6} \mathrm{C} \end{aligned}$