Capacitor

The capacitance of a parallel plate capacitor without any dielectric is given by:

$ C=\frac{\epsilon_0 A}{d} $

where A is the area of the plates and d is the separation between them.

Capacitor $\mathrm{C}_1$

Dielectric constant $\mathrm{K}_1=2$.

Thickness $\mathrm{d}_1=\frac{\mathrm{d}}{2}$.

Area $\mathrm{A}_1=\mathrm{A}$

So, the capacitance is;

$ C_1=\frac{K_1 \epsilon_0 A_1}{d_1}=\frac{2 \epsilon_0 A}{d / 2}=4\left(\frac{\epsilon_0 A}{d}\right)=4 C $

Capacitor $\mathrm{C}_2$

Dielectric constant $\mathrm{K}_2=3$.

Thickness $\mathrm{d}_2=\frac{\mathrm{d}}{2}$.

Area $\mathrm{A}_1=\mathrm{A} / 2$

So, the capacitance is;

$\mathrm{C}_2=\frac{\mathrm{K}_2 \epsilon_0 \mathrm{~A}_2}{\mathrm{~d}_2}=\frac{3 \epsilon_0(\mathrm{~A} / 2)}{\mathrm{d} / 2}=3\left(\frac{\epsilon_0 \mathrm{~A}}{\mathrm{~d}}\right)=3 \mathrm{C}$

Capacitor $\mathrm{C}_3$

Dielectric constant $\mathrm{K}_3=5$.

Thickness $\mathrm{d}_3=\frac{\mathrm{d}}{2}$.

Area $\mathrm{A}_2=\mathrm{A} / 2$

So, the capacitance is;

$ \mathrm{C}_3=\frac{\mathrm{K}_3 \epsilon_0 \mathrm{~A}_3}{\mathrm{~d}_3}=\frac{5 \epsilon_0(\mathrm{~A} / 2)}{\mathrm{d} / 2}=5\left(\frac{\epsilon_0 \mathrm{~A}}{\mathrm{~d}}\right)=5 \mathrm{C} $

Capacitors $C_2$ and $C_3$ are in parallel.

$ C_{23}=C_2+C_3 $

$\Rightarrow $ $ C_{23}=3 C+5 C=8 C $

This combined section $\left(\mathrm{C}_{23}\right)$ is in series with the first capacitor $\left(\mathrm{C}_1\right)$.

$\frac{1}{\mathrm{C}_{\mathrm{eq}}}=\frac{1}{\mathrm{C}_1}+\frac{1}{\mathrm{C}_{23}}$

$\Rightarrow $ $\frac{1}{\mathrm{C}_{\mathrm{eq}}}=\frac{1}{4 \mathrm{C}}+\frac{1}{8 \mathrm{C}}$

$\Rightarrow $ $\frac{1}{C_{\mathrm{eq}}}=\frac{2+1}{8 C}=\frac{3}{8 C}$

$\Rightarrow $ $ C_{\mathrm{eq}}=\frac{8 C}{3}=\frac{n}{3} C \Rightarrow n=8 $

Therefore, the value of n is 8 . Hence, the correct answer is 8 .

When capacitor P is fully charged by the 6.0 V battery, it stores an initial amount of charge ( $\mathrm{Q}_{\text {initial }}$ ).

$ Q=C \cdot V $

Given the values for capacitor $P$ the capacitance is $C_P=10 \times 10^{-6} \mathrm{~F}$

$ Q_{\text {initial }}=\left(10 \times 10^{-6}\right) \times 6.0 \mathrm{C} $

$\Rightarrow $ $ Q_{\text {initial }}=60 \times 10^{-6} \mathrm{C} $

After being disconnected from the battery, capacitor $P$ is connected across an initially uncharged capacitor $Q$.

The charges flow from P to Q until both capacitors reach the same electrical potential.

According to the law of conservation of charge, the total charge in the system remains constant :

$ Q_{\text {total }}=Q_{\text {initial }}=60 \times 10^{-6} \mathrm{C} $

In a parallel connection, the equivalent capacitance ( $\mathrm{C}_{\mathrm{eq}}$ ) is the sum of the individual capacitances :

$ C_{\mathrm{eq}}=C_{\mathrm{P}}+C_{\mathrm{Q}} $

$\Rightarrow $ $C_{\mathrm{eq}}=\left(10 \times 10^{-6}\right)+\left(20 \times 10^{-6}\right)$

$\Rightarrow $ $C_{\mathrm{eq}}=30 \times 10^{-6} \mathrm{~F}$

$\Rightarrow $ $\mathrm{V}_{\text {common }}=\frac{\mathrm{Q}_{\text {total }}}{\mathrm{C}_{\mathrm{eq}}}$

$\Rightarrow $ $ \mathrm{V}_{\text {common }}=\frac{60 \times 10^{-6}}{30 \times 10^{-6}} \mathrm{~V}=2.0 \mathrm{~V} $

The final charge $\left(\mathrm{Q}_{\mathrm{Q}}\right)$ using the standard capacitor formula :

$ \mathrm{Q}_{\mathrm{Q}}=\mathrm{C}_{\mathrm{Q}} \cdot \mathrm{~V}_{\text {common }} $

$\Rightarrow $ $\mathrm{Q}_{\mathrm{Q}}=\left(20 \times 10^{-6}\right) \times 2.0 \mathrm{C}$

$\Rightarrow $ $ \mathrm{Q}_{\mathrm{Q}}=40 \times 10^{-6} \mathrm{C} $

So, the charge on capacitor Q will be $4 \times 10^{-5} \mathrm{C}$.

$ \alpha \times 10^{-5} \mathrm{C}=4.0 \times 10^{-5} \mathrm{C} \Rightarrow \alpha=4 $

$\begin{aligned} & \mathrm{C}_1=\frac{5 \times 4 \times 10^{-4} \times 8.85 \times 10^{-12}}{\frac{1.77}{2} \times 10^{-3}}=20 \mathrm{pF} \\ & \mathrm{C}_2=\frac{3 \times 4 \times 10^{-4} \times 8.85 \times 10^{-12}}{\frac{1.77}{2} \times 10^{-3}}=12 \mathrm{pF} \\ & \mathrm{C}_{\mathrm{eq}}=\frac{\mathrm{C}_1 \mathrm{C}_2}{\mathrm{C}_1+\mathrm{C}_2}=\frac{12 \times 20}{12+20}=7.5 \mathrm{pF} \end{aligned}$

Finally equivalent capacitance

$\left(\mathrm{C}_{\text {eq }}\right)_{\text {final }}=7.5+7.5=15 \mathrm{pF}$

To find the dielectric constant $ K $ of the slab, we use the formula for the induced charge $ Q_{\text{in}} $ in a parallel plate capacitor:

$ Q_{\text{in}} = Q \left(1 - \frac{1}{K}\right) $

Given:

Total charge $ Q = 5 \times 10^{-6} \, \text{C} $

Induced charge $ Q_{\text{in}} = 4 \times 10^{-6} \, \text{C} $

We substitute the values into the formula:

$ 4 \times 10^{-6} = 5 \times 10^{-6} \left(1 - \frac{1}{K}\right) $

To solve for the dielectric constant $ K $, simplify the equation:

Divide both sides by $ 5 \times 10^{-6} $:

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

Rearrange to solve for $ \frac{1}{K} $:

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

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

Invert both sides to find $ K $:

$ K = 5 $

Therefore, the dielectric constant of the slab is $ K = 5 $.

Given, $r = 10\,cm = {1 \over {10}}\,m$

$I = 0.15\,A$, and ${{dv} \over {dt}} = 7 \times {10^8}v/s$

We know, for a parallel plate capacitor,

$c = {{{\varepsilon _0}A} \over d} \Rightarrow c = {{{\varepsilon _0}\pi {r^2}} \over d}$ .... (1)

In a capacitor,

$Q = CV$

$ \Rightarrow V = {Q \over C}$

by differentiating w.r.t. t

$ \Rightarrow {{dv} \over {dt}} = {1 \over C}{{dQ} \over {dt}}$

$ \Rightarrow {{dv} \over {dt}} = {d \over {{\varepsilon _0}\pi {r^2}}}I$ (As $I = {{dQ} \over {dt}}$)

$ \Rightarrow d = {{{\varepsilon _0}\pi {r^2}} \over I}{{dv} \over {dt}}$ .... (From (1))

$ \Rightarrow d = {{9 \times {{10}^{ - 12}}} \over {0.15}} \times {{22} \over 7} \times {1 \over {100}} \times 7 \times 10$

$ = {{198} \over {0.15}} \times {10^{ - 6}}$

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

$ \Rightarrow d = 1320\,\mu m$

$\begin{aligned} & \mathrm{i}=\left(\frac{5}{25}\right) \\\\ & \mathrm{Q}=\mathrm{CV} \\\\ & \mathrm{Q}=\left(8 \times 10^{-6}\right)\left(\frac{5}{25} \times 10\right) \\\\ & \mathrm{Q}=\left(\frac{8 \times 5 \times 10^{-2}}{25}\right)=16 \mu \mathrm{C}\end{aligned}$

$\begin{array}{ll} C_1=10 \mu \mathrm{F} & C_2=15 \mu \mathrm{F} \\ F=F_1+F_2 & \end{array}$

$\begin{aligned} & F=\frac{\left(10 \times 10^{-6} V\right)^2}{2 \cdot \frac{A}{2} \cdot \epsilon_0}+\frac{(15~~10)}{2 \cdot \frac{A}{2} \cdot \epsilon_0}=8 \\ & V^2=0.9 \times 10^{-4} \\ & V=0.95 \times 10^{-2} \mathrm{~V} \\ \end{aligned}$

Data in consistent. Answer not matching.

To find the resistance of the wire connecting the two plates of the capacitor, we need to apply the formula that relates the time constant $\tau$ (in seconds) of a capacitor-resistor (CR) circuit to the capacitance C (in Farads) and resistance R (in Ohms). The time constant $\tau$ is given by:

$\tau = R \times C$

The time constant also defines the time it takes for the voltage across the capacitor (and therefore the electric field between the plates, since they are directly related) to drop to approximately $\frac{1}{e}$ (where $e$ is the base of natural logarithms, approximately equal to 2.718) of its initial value. However, the question states that the electric field drops to one-third of its initial value. Using the natural logarithm properties, we can relate this decay process to the concept of the time constant.

The formula for the voltage (or electric field) across a discharging capacitor as a function of time $t$ is:

$V(t) = V_0 \times e^{-\frac{t}{RC}}$

where:

• $V(t)$ is the voltage across the capacitor at time $t$,
• $V_0$ is the initial voltage,
• $R$ is the resistance,
• $C$ is the capacitance, and
• $t$ is the time.

Given that the electric field drops to one-third of its initial value in $6.6 \mu \mathrm{s}$, we can set $V(t)$ to $\frac{1}{3}V_0$ and solve for $R$:

$\frac{1}{3}V_0 = V_0 \times e^{-\frac{6.6 \mu s}{R \times 1.5 \mu F}}$

By dividing both sides by $V_0$, we simplify to:

$\frac{1}{3} = e^{-\frac{6.6}{R \times 1.5}}$

Taking the natural logarithm of both sides to solve for $R$:

$\ln\left(\frac{1}{3}\right) = -\frac{6.6}{R \times 1.5}$

Given that $\ln\left(\frac{1}{3}\right) = \ln(3^{-1}) = -\ln(3) = -1.1$ (since $\log 3 = 1.1$ and using the natural log instead of common log), we have:

$-1.1 = -\frac{6.6}{R \times 1.5}$

Solving for $R$:

$R = \frac{6.6}{1.5 \times 1.1}$

Now, calculate the value of $R$:

$R = \frac{6.6}{1.65} = 4\, \Omega$

Therefore, the resistance of the wire connecting the two plates of the capacitor is $4 \, \Omega$.

$E=\frac{1}{2}(25+30+45)(100)^2 \quad \text{.... (i)}$

$\text { Also, } \frac{9}{x} E=\frac{1}{2} \frac{1}{\left(\frac{1}{25}+\frac{1}{30}+\frac{1}{45}\right)}(100)^2 \quad \text{.... (ii)}$

From (i) and (ii)

$x=86$

$\begin{aligned} E_1 & =\frac{1}{2}\left(\frac{25}{2}\right) \times 10^{-12} \times 144 \\ & =900 \times 10^{-12} \mathrm{~J} \\ E_2 & =\frac{1}{2}\left(6 \times \frac{25}{2} \times 10^{-12}\right)\left(\frac{12}{6}\right)^2=150 \times 10^{-12} \mathrm{~J} \\ \Delta E & =750 \times 10^{-12} \mathrm{~J} \end{aligned}$

Without dielectric

$\mathrm{Q}=\frac{\mathrm{A} \in_0}{\mathrm{~d}} \mathrm{~V}$

with dielectric

$Q=\frac{A \in_0 V}{d-t+\frac{t}{K}}$

given

$\begin{aligned} & \frac{\mathrm{A} \in_0 \mathrm{~V}}{\mathrm{~d}-\mathrm{t}+\frac{\mathrm{t}}{\mathrm{K}}}=(1.25) \frac{\mathrm{A} \in_0 \mathrm{~V}}{\mathrm{~d}} \\ & \Rightarrow 1.25\left(3+\frac{2}{\mathrm{~K}}\right)=5 \\ & \Rightarrow \mathrm{K}=2 \end{aligned}$

$\begin{aligned} & \text { Energy loss }=\frac{1}{2} \frac{C_1 C_2}{C_1+C_2}\left(V_1-V_2\right)^2 \\ & =\frac{2}{3} \cdot E \\ & \therefore x=2 \end{aligned}$

At steady state, capacitor behaves as an open circuit and current flows in circuit as shown in the diagram.

$\begin{aligned} & \mathrm{R}_{\mathrm{eq}}=9 \Omega \\ & \mathrm{i}=\frac{9 \mathrm{~V}}{9 \Omega}=1 \mathrm{~A} \\ & \Delta \mathrm{V}_{6 \Omega}=1 \times 6=6 \mathrm{~V} \\ & \mathrm{~V}_{\mathrm{A}}=3 \mathrm{~V} \end{aligned}$

So, potential difference across 6$\mu$F is 6 V.

$\begin{aligned} \text { Hence } \mathrm{Q} & =\mathrm{C} \Delta \mathrm{V} \\ & =6 \times 6 \times 10^{-6} \mathrm{C} \\ & =36 \mu \mathrm{C} \end{aligned}$

$\begin{aligned} & I=\frac{V}{R_{\text {eq }}} I=\frac{V}{R_{\text {eq }}}=\frac{9}{1+\frac{12 \times 4}{12+4}}=\frac{9}{4} \\ & I_1=\frac{9}{4} \times \frac{4}{16}=\frac{9}{16} \\ & V_A-V_B=I_1 \times 8=\frac{9}{16} \times 8=\frac{9}{2} V \\ & \therefore U=\frac{1}{2} \times 4 \times \frac{81}{4} \mu J \\ & \therefore U=\frac{81}{2} \mu J \\ & \therefore x=81 \end{aligned}$

$\begin{aligned} & \mathrm{V}_{\mathrm{A}}+\frac{10}{3}(1)-6(1)=\mathrm{V}_{\mathrm{B}} \\ & \mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{B}}=6-\frac{10}{3}=\frac{8}{3} \mathrm{volt} \\ & \mathrm{Q}=\mathrm{C}\left(\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{B}}\right) \\ & =150 \times \frac{8}{3}=400 \mu \mathrm{C} \end{aligned}$

The energy stored in a capacitor can be calculated using the formula:

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

where:

• (U) is the energy,
• (C) is the capacitance,
• (V) is the voltage.

Initially, the energy stored in the first capacitor is:

$ U_{\text{initial}} = \frac{1}{2} C V^2 = \frac{1}{2} \times 600 \times 10^{-12} \, \text{F} \times (200 \, \text{V})^2 = 0.012 \, \text{J} = 12 \, \mu\text{J}. $

When the charged capacitor is connected to the uncharged capacitor, the charge will distribute equally between them because they have the same capacitance. Therefore, the final voltage across each capacitor is half of the initial voltage, i.e., 100 V.

The energy in each capacitor after the redistribution is:

$ U_{\text{final each}} = \frac{1}{2} C \left(\frac{V}{2}\right)^2 = \frac{1}{2} \times 600 \times 10^{-12} \, \text{F} \times (100 \, \text{V})^2 = 0.003 \, \text{J} = 3 \, \mu\text{J}. $

As there are two capacitors, the total final energy is:

$ U_{\text{final total}} = 2 \times U_{\text{final each}} = 2 \times 3 \, \mu\text{J} = 6 \, \mu\text{J}. $

The energy loss is the difference between the initial energy and the final energy:

$ \Delta U = U_{\text{initial}} - U_{\text{final total}} = 12 \, \mu\text{J} - 6 \, \mu\text{J} = 6 \, \mu\text{J}. $

${U_i} = {1 \over 2}C{V^2} = {1 \over 2} \times 900 \times {10^{ - 6}} \times {100^2} = 4.5$ J

As the other capacitor is identical therefore charge is equally divided and potential difference across the capacitors becomes half. So

${U_f} = {1 \over 2}2C{\left( {{V \over 2}} \right)^2} = {1 \over 2} \times 2 \times 900 \times {10^{ - 6}}{\left( {{{100} \over 2}} \right)^2}$

$ = {9 \over 4}$ J = 2.25 J

So, loss in energy $\Delta {U_{loss}} = {U_i} - {U_f}$

= 2.25 J

= 225 $\times$ 10$^{-2}$ J

Initially

$ \frac{\varepsilon_{0} A}{d}=15 \times 10^{-12} \mathrm{~F} $

Finally

At steady state potential difference across capacitor

${V_c} = {{10 \times 100} \over {110}}\,V$

$Q = C{V_c}$

$ = {{1.1 \times {{10}^{ - 6}} \times 10 \times 100} \over {110}}\,C = 10\,\mu C$

${d_1} = 4 \times {10^{ - 3}}$

${A_1} = 8 \times 4 \times {10^{ - 4}}\,{m^2}$

$V = 20\,V$

${d_2} = 4 \times {10^{ - 3}},$

${A_2} = 4 \times 1 \times {10^{ - 4}}\,{m^2}$

${C_{eq}} = {{({A_1} + 5{A_2} - {A_2}){\varepsilon _0}} \over d} = {{3(16) \times {{10}^{ - 4}}} \over {4 \times {{10}^{ - 3}}}}{\varepsilon _0}$

$\varepsilon = {1 \over 2}{C_{eq}}{V^2} = {3 \over 2}\left( {{4 \over {10}}} \right)(400){\varepsilon _0} = 240{\varepsilon _0}$

${{{C_1}} \over {{C_2}}} = {{3 \times {t_2}} \over {{t_1} \times 4}} = {3 \over 2}$

${q \over {{C_1}}} = {v_1},\,{q \over {{C_2}}} = {v_2}$

${{{v_1}} \over {{v_2}}} = {{{C_2}} \over {{C_1}}} = {2 \over 3}$

${V_{common}} = {{18CV + 54CV} \over {3C + 9C}} = 6\,V$

Let the charge q is flown in the circuit.

So using Kirchhoff's law,

${q \over {10}} = {{150 - q} \over 5}$

$q = 100\,\mu C$

The circuit is equivalent to 3 capacitors in parallel as shown

${C_{eq}} = {{{\varepsilon _0}A} \over b}\left( {1 + {1 \over 3} + {1 \over 5}} \right) = {{23} \over {15}}{{{\varepsilon _0}A} \over b}$

$ \Rightarrow x = 23$

Electrical energy lost $ = {1 \over 2}\left( {{1 \over 2}C{V^2}} \right)$

$ = {1 \over 2} \times {1 \over 2} \times 50 \times {10^{ - 12}} \times {(100)^2}$

$ = {{500} \over 4}$ nJ

$ = 125$ nJ