Effect of Dielectric Inside Capacitor Questions in English

(D) The initial capacitance of the capacitor is $C_0 = \frac{\varepsilon_0 A}{d}$.
When two dielectric slabs of thickness $d/2$ are placed between the plates,the arrangement acts as two capacitors in series,each with plate separation $d/2$.
The capacitance of the first slab is $C_1 = \frac{K_1 \varepsilon_0 A}{d/2} = \frac{2 K_1 \varepsilon_0 A}{d} = 2 K_1 C_0$.
The capacitance of the second slab is $C_2 = \frac{K_2 \varepsilon_0 A}{d/2} = \frac{2 K_2 \varepsilon_0 A}{d} = 2 K_2 C_0$.
Since they are in series,the equivalent capacitance $C_{eq}$ is given by:
$\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{1}{2 K_1 C_0} + \frac{1}{2 K_2 C_0} = \frac{1}{2 C_0} \left( \frac{1}{K_1} + \frac{1}{K_2} \right) = \frac{1}{2 C_0} \left( \frac{K_1 + K_2}{K_1 K_2} \right)$.
Therefore,$C_{eq} = 2 C_0 \left( \frac{K_1 K_2}{K_1 + K_2} \right)$.

(B) Initial energy stored in the capacitor is $U_i = \frac{1}{2} C V^2$.
Since the capacitor is disconnected from the source,the charge $Q$ on the plates remains constant.
When a dielectric of constant $K$ is inserted,the new capacitance becomes $C' = KC$.
The new energy stored is $U_f = \frac{Q^2}{2C'} = \frac{Q^2}{2KC} = \frac{U_i}{K}$.
Given that the energy decreases by $25 \%$,the final energy is $U_f = U_i - 0.25 U_i = 0.75 U_i = \frac{3}{4} U_i$.
Equating the two expressions for $U_f$: $\frac{U_i}{K} = \frac{3}{4} U_i$.
Solving for $K$,we get $K = 4/3$.

(A) The capacitance of a parallel plate capacitor with a dielectric slab of thickness $t$ and dielectric constant $K$ is given by the formula: $C = \frac{K \epsilon_0 A}{d}$.
Given:
Area $A = 0.4 \pi \,m^2$
Spacing $d = 0.5 \,mm = 0.5 \times 10^{-3} \,m$
Thickness of slab $t = 0.5 \,mm = d$
Dielectric constant $K = 4.5$
Permittivity of free space $\epsilon_0 = 8.854 \times 10^{-12} \,F/m$ or $\frac{1}{36 \pi \times 10^9} \,F/m$.
Since the slab fills the entire space between the plates $(t = d)$, the formula becomes $C = \frac{K \epsilon_0 A}{d}$.
Substituting the values:
$C = \frac{4.5 \times (1 / (36 \pi \times 10^9)) \times 0.4 \pi}{0.5 \times 10^{-3}}$
$C = \frac{4.5 \times 0.4 \pi}{36 \pi \times 10^9 \times 0.5 \times 10^{-3}}$
$C = \frac{1.8}{18 \times 10^6} = 0.1 \times 10^{-6} \,F = 100 \times 10^{-9} \,F = 100 \,nF$.

(C) The capacitance of a parallel plate capacitor is given by $C = \frac{\varepsilon_0 A}{d}$.
When a dielectric slab of thickness $t$ is inserted,the new capacitance $C'$ is given by $C' = \frac{\varepsilon_0 A}{d' - t(1 - 1/K)}$,where $d'$ is the new plate separation.
Given that the potential difference $V$ remains the same and the charge $Q$ is constant (since the battery is disconnected),the capacitance must remain constant,so $C = C'$.
Equating the two expressions: $\frac{\varepsilon_0 A}{d} = \frac{\varepsilon_0 A}{(d + 1.6 \ mm) - 2 \ mm(1 - 1/K)}$.
This simplifies to: $d = d + 1.6 - 2(1 - 1/K)$.
$0 = 1.6 - 2 + 2/K$.
$0 = -0.4 + 2/K$.
$0.4 = 2/K$.
$K = 2 / 0.4 = 5$.

(A) Initial capacitance of the parallel plate capacitor is given by $C_1 = \frac{\varepsilon_0 A}{d}$.
When the distance between the plates is halved $(d' = d/2)$ and a dielectric of relative permittivity $K = 10$ is introduced,the final capacitance $C_2$ is given by $C_2 = \frac{K \varepsilon_0 A}{d'}$.
Substituting the values,we get $C_2 = \frac{10 \varepsilon_0 A}{d/2} = \frac{20 \varepsilon_0 A}{d}$.
Therefore,the ratio of the final to the initial capacitance is $\frac{C_2}{C_1} = \frac{20 \varepsilon_0 A / d}{\varepsilon_0 A / d} = 20$.

(C) The capacitance $C$ of the parallel plate capacitor as it is immersed in oil to a depth $x$ is given by the sum of the capacitances of the part in air and the part in oil:
$C = \frac{\epsilon_0 (A - Ax)}{d} + \frac{K \epsilon_0 Ax}{d} = \frac{\epsilon_0 A}{d} [1 + (K - 1)x/L]$
Given side $L = 1 \,m$, so $A = L^2 = 1 \,m^2$. The depth $x$ increases with time $t$ as $x = vt$, where $v = 0.001 \,m/s$.
$C(t) = \frac{\epsilon_0}{d} [1 + (K - 1)vt]$
Differentiating with respect to time $t$:
$\frac{dC}{dt} = \frac{\epsilon_0}{d} (K - 1) v$
Substituting the values: $\epsilon_0 = 8.85 \times 10^{-12} \,F/m$, $d = 0.01 \,m$, $K = 11$, $v = 0.001 \,m/s$:
$\frac{dC}{dt} = \frac{8.85 \times 10^{-12}}{0.01} \times (11 - 1) \times 0.001 = 8.85 \times 10^{-12} \times 10 \times 0.1 = 8.85 \times 10^{-12} \,F/s$
The current $I$ drawn from the battery is $I = V \frac{dC}{dt}$:
$I = 500 \times 8.85 \times 10^{-12} = 4.425 \times 10^{-9} \,A$.

(B) Given: Thickness of mica $d_1 = 1 \times 10^{-3} \,m$, thickness of fiber $d_2 = 0.5 \times 10^{-3} \,m$. Dielectric constants $K_1 = 8$ (mica) and $K_2 = 2.5$ (fiber). Breakdown electric field of fiber $E_2 = 6.4 \times 10^6 \,V/m$.
Since the sheets are in series, the electric displacement field $D = \epsilon_0 K E$ is constant across the plates, so $K_1 E_1 = K_2 E_2$.
Thus, $E_1 = E_2 (K_2 / K_1) = (6.4 \times 10^6) \times (2.5 / 8) = 2.0 \times 10^6 \,V/m$.
The maximum voltage $V_{max}$ is given by $V_{max} = E_1 d_1 + E_2 d_2$.
$V_{max} = (2.0 \times 10^6 \times 1 \times 10^{-3}) + (6.4 \times 10^6 \times 0.5 \times 10^{-3})$.
$V_{max} = 2000 + 3200 = 5200 \,V$.

(A) The capacitance of a parallel plate capacitor is given by $C = \frac{K A \varepsilon_0}{d}$.
Initially,for air,$K_1 = 1$ and $d_1 = d$,so $C_1 = \frac{A \varepsilon_0}{d} = 12 \mu F$.
When the distance is doubled,$d_2 = 2d$,and the dielectric constant is $K_2 = 4$.
The new capacitance $C_2$ is given by $C_2 = \frac{K_2 A \varepsilon_0}{d_2}$.
Substituting the values,we get $C_2 = \frac{4 A \varepsilon_0}{2d} = 2 \left( \frac{A \varepsilon_0}{d} \right) = 2 \times C_1$.
Therefore,$C_2 = 2 \times 12 \mu F = 24 \mu F$.

(B) Initial capacitance $C = 60 \ \mu F$. Initial potential $V = 250 \ V$. Charge $q = CV = 60 \ \mu F \times 250 \ V = 15000 \ \mu C$.
When a dielectric slab of thickness $t = 3 \ mm$ and dielectric constant $K = 5$ is inserted between plates separated by $d = 6 \ mm$,the new capacitance $C'$ is given by $C' = \frac{\epsilon_0 A}{d - t + \frac{t}{K}}$.
Since $C = \frac{\epsilon_0 A}{d}$,we have $C' = C \left[ \frac{d}{d - t + \frac{t}{K}} \right] = 60 \ \mu F \left[ \frac{6}{6 - 3 + \frac{3}{5}} \right] = 60 \ \mu F \left[ \frac{6}{3 + 0.6} \right] = 60 \ \mu F \left[ \frac{6}{3.6} \right] = 60 \ \mu F \times \frac{60}{36} = 60 \ \mu F \times \frac{5}{3} = 100 \ \mu F$.
Since the charging source is removed,the charge $q$ remains constant. Thus,$q = C'V' \Rightarrow 15000 \ \mu C = 100 \ \mu F \times V' \Rightarrow V' = 150 \ V$.
The change in potential difference is $\Delta V = V - V' = 250 \ V - 150 \ V = 100 \ V$.

(D) The capacitance of a parallel plate capacitor is given by $C = \frac{Q}{V}$,where $Q$ is the charge and $V$ is the potential difference.
When a dielectric material of constant $K$ is inserted between the plates,the charge $Q$ remains constant (if the capacitor is isolated).
The new potential difference $V^{\prime}$ is related to the original potential difference $V$ by the relation $V^{\prime} = \frac{V}{K}$.
Given that the potential difference reduces to $1/8$ of the original value,we have $V^{\prime} = \frac{V}{8}$.
Comparing the two expressions,we get $\frac{V}{K} = \frac{V}{8}$.
Therefore,the dielectric constant $K = 8$.

(D) Key Idea: The capacitance of a parallel plate capacitor is given by the relation,$C = \varepsilon_r \frac{\varepsilon_0 A}{d}$,where $\varepsilon_r$ is the dielectric constant,$A$ is the area of the plates,and $d$ is the distance between the plates.
Given: $d_1 = x$,$\varepsilon_{r1} = 3k$,$d_2 = 2x$,and $\varepsilon_{r2} = 5k$.
The two dielectric layers act as two capacitors $C_1$ and $C_2$ connected in series.
$C_1 = \frac{(3k) \varepsilon_0 A}{x}$ and $C_2 = \frac{(5k) \varepsilon_0 A}{2x}$.
In a series combination,the charge $Q$ stored on each capacitor is the same.
The potential difference across each layer is given by $V = \frac{Q}{C}$.
Thus,$V_1 = \frac{Q}{C_1} = \frac{Qx}{3k \varepsilon_0 A}$ and $V_2 = \frac{Q}{C_2} = \frac{Q(2x)}{5k \varepsilon_0 A}$.
The ratio of the potential differences is $\frac{V_1}{V_2} = \frac{Qx / (3k \varepsilon_0 A)}{2Qx / (5k \varepsilon_0 A)} = \frac{1/3}{2/5} = \frac{1}{3} \times \frac{5}{2} = \frac{5}{6}$.
Therefore,the correct option is $(d)$.

(C) Let the initial distance between the plates be $d$ and the surface charge density be $\sigma$. The initial potential difference is $V = \frac{\sigma d}{\varepsilon_0} = 100 \,V$.
When an insulator of thickness $t = 2 \,mm$ and dielectric constant $k$ is inserted, the new potential difference $V'$ is given by:
$V' = E_{insulator} \cdot t + E_{air} \cdot (d - t) = \frac{\sigma}{\varepsilon_0 k} \cdot 2 + \frac{\sigma}{\varepsilon_0} (d - 2) = \frac{\sigma}{\varepsilon_0} \left( \frac{2}{k} + d - 2 \right)$.
To maintain the same potential difference $V$, the distance is increased by $\Delta d = 1.6 \,mm$. The new potential difference becomes:
$V = \frac{\sigma}{\varepsilon_0 k} \cdot 2 + \frac{\sigma}{\varepsilon_0} (d + 1.6 - 2) = \frac{\sigma}{\varepsilon_0} \left( \frac{2}{k} + d - 0.4 \right)$.
Since the potential difference remains $100 \,V$, we equate the initial and final expressions:
$\frac{\sigma d}{\varepsilon_0} = \frac{\sigma}{\varepsilon_0} \left( \frac{2}{k} + d - 0.4 \right)$.
$d = \frac{2}{k} + d - 0.4$.
$0.4 = \frac{2}{k}$.
$k = \frac{2}{0.4} = 5$.

(C) Let the capacitance of each pair of plates without a dielectric be $C = \frac{\epsilon_0 A}{d}$.
There are three capacitors formed by the plates: $C_1$ between $A$ and $B$,$C_2$ between $B$ and $C$,and $C_3$ between $C$ and $D$.
$C_1 = C$ (air between $A$ and $B$).
$C_2 = K C = 2C$ (dielectric $K=2$ between $B$ and $C$).
$C_3 = K C = 2C$ (dielectric $K=2$ between $C$ and $D$).
Plates $B$ and $D$ are connected together. Thus,$C_2$ and $C_3$ are in parallel between points $B$ and $C$. The equivalent capacitance of this parallel combination is $C_{23} = C_2 + C_3 = 2C + 2C = 4C$.
Now,$C_1$ is in series with the combination $C_{23}$ between points $A$ and $C$. However,looking at the circuit,$A$ is connected to one plate of $C_1$,and the other plate $B$ is connected to $C_2$ and $C_3$. Since $B$ and $D$ are connected,the effective capacitance between $A$ and $C$ is the series combination of $C_1$ and $C_{23}$.
$C_{eq} = \frac{C_1 \times C_{23}}{C_1 + C_{23}} = \frac{C \times 4C}{C + 4C} = \frac{4C^2}{5C} = \frac{4}{5} C$.

(A) Initial capacitance with air is $C = 80 \times 10^{-6} \,F$.
When the dielectric slab of constant $k = 20$ is inserted, the new capacitance becomes $C' = kC = 20 \times 80 \times 10^{-6} = 1600 \times 10^{-6} \,F$.
The capacitor is connected to a battery of $V = 30 \,V$, so the charge on the capacitor is $q' = C'V = (1600 \times 10^{-6}) \times 30 = 48000 \times 10^{-6} \,C = 48 \times 10^{-3} \,C$.
When the dielectric slab is removed, the capacitance returns to $C = 80 \times 10^{-6} \,F$. The new charge on the capacitor is $q = CV = (80 \times 10^{-6}) \times 30 = 2400 \times 10^{-6} \,C = 2.4 \times 10^{-3} \,C$.
The charge that passes through the wire is $\Delta q = q' - q = 48 \times 10^{-3} - 2.4 \times 10^{-3} = 45.6 \times 10^{-3} \,C$.

(D) Initial capacitance with air,$C_{\text{air}} = 80 \times 10^{-6} \ F = 80 \ \mu F$.
Capacitance with dielectric slab,$C_d = K \times C_{\text{air}} = 20 \times 80 \ \mu F = 1600 \ \mu F$.
When connected to a $30 \ V$ battery,the charge stored with the dielectric is $q_d = C_d \times V = 1600 \times 10^{-6} \times 30 = 48000 \ \mu C = 48 \times 10^{-3} \ C$.
When the dielectric is removed,the new capacitance becomes $C_{\text{air}} = 80 \ \mu F$.
The new charge stored is $q_{\text{air}} = C_{\text{air}} \times V = 80 \times 10^{-6} \times 30 = 2400 \ \mu C = 2.4 \times 10^{-3} \ C$.
The charge that passes through the wire is the difference between the initial charge and the final charge: $\Delta q = q_d - q_{\text{air}} = 48 \times 10^{-3} \ C - 2.4 \times 10^{-3} \ C = 45.6 \times 10^{-3} \ C$.

(A) The initial capacitance of the parallel plate capacitor is $C = 0.025 \mu F$.
When a metal plate of thickness $t = d/3$ is introduced,the new capacitance $C'$ becomes $C' = \frac{\varepsilon_0 A}{d - t} = \frac{\varepsilon_0 A}{d - d/3} = \frac{3}{2} \frac{\varepsilon_0 A}{d} = \frac{3}{2} C$.
Since the capacitor remains connected to the $100 \ V$ source,the potential difference $V$ remains constant at $100 \ V$.
The energy stored in the capacitor initially is $U_i = \frac{1}{2} C V^2$.
The energy stored in the capacitor with the metal plate is $U_f = \frac{1}{2} C' V^2 = \frac{1}{2} (\frac{3}{2} C) V^2 = \frac{3}{4} C V^2$.
The work done by the external agent to remove the plate is $W = U_i - U_f$ (since the battery does work to maintain potential).
$W = \frac{1}{2} C V^2 - \frac{3}{4} C V^2 = -\frac{1}{4} C V^2$.
The magnitude of work done by the external agent is $|W| = \frac{1}{4} C V^2$.
$|W| = \frac{1}{4} \times (0.025 \times 10^{-6} \ F) \times (100 \ V)^2 = \frac{0.025 \times 10^{-6} \times 10^4}{4} = \frac{0.025 \times 10^{-2}}{4} = 6.25 \times 10^{-5} \ J = 62.5 \mu J$.

(D) Case $1$: Battery is disconnected. The charge $Q$ remains constant.
$Q = Q_0 = C_0 V_0$
New separation $d' = 2d$. The new capacitance is $C' = \frac{\epsilon_0 A}{d'} = \frac{\epsilon_0 A}{2d} = \frac{C_0}{2}$.
The energy stored is $E_1 = \frac{Q^2}{2C'} = \frac{(C_0 V_0)^2}{2(C_0/2)} = C_0 V_0^2$.
Case $2$: Battery remains connected. The potential $V$ remains constant.
$V = V_0$
New capacitance $C' = \frac{C_0}{2}$.
The energy stored is $E_2 = \frac{1}{2} C' V^2 = \frac{1}{2} \times (\frac{C_0}{2}) \times V_0^2 = \frac{1}{4} C_0 V_0^2$.
Ratio: $\frac{E_2}{E_1} = \frac{\frac{1}{4} C_0 V_0^2}{C_0 V_0^2} = 0.25$.

(C) Let the capacitance of each capacitor be $C$ and the battery voltage be $V$.
Initially,when the switch $S$ is closed,both capacitors are connected in parallel to the battery $V$.
The total electrostatic energy stored initially is $U_1 = \frac{1}{2} CV^2 + \frac{1}{2} CV^2 = CV^2$.
Now,the switch $S$ is opened. Capacitor $A$ remains connected to the battery,so its potential difference remains $V$. Capacitor $B$ is disconnected,so its charge $Q = CV$ remains constant.
After introducing a dielectric of constant $K = 3$ into both capacitors:
For capacitor $A$,the new capacitance is $C' = KC = 3C$. The energy stored is $U_A = \frac{1}{2} C' V^2 = \frac{1}{2} (3C) V^2 = \frac{3}{2} CV^2$.
For capacitor $B$,the new capacitance is $C' = KC = 3C$. Since the charge $Q = CV$ is constant,the energy stored is $U_B = \frac{Q^2}{2C'} = \frac{(CV)^2}{2(3C)} = \frac{CV^2}{6}$.
The total energy after introducing the dielectric is $U_2 = U_A + U_B = \frac{3}{2} CV^2 + \frac{1}{6} CV^2 = \left( \frac{9+1}{6} \right) CV^2 = \frac{10}{6} CV^2 = \frac{5}{3} CV^2$.
The ratio of the total electrostatic energy before and after is $\frac{U_1}{U_2} = \frac{CV^2}{\frac{5}{3} CV^2} = \frac{3}{5}$.

(A) When the switch $S$ is closed,both capacitors are in parallel with the battery of voltage $V$. The equivalent capacitance is $C_{eq} = C + C = 2C$. The total energy stored is $U_1 = \frac{1}{2} C_{eq} V^2 = \frac{1}{2} (2C) V^2 = CV^2$.
When the switch $S$ is opened,the battery is disconnected. The charge on each capacitor remains $Q = CV$. Now,both capacitors are filled with a dielectric of constant $K = 3$. The new capacitance of each capacitor becomes $C' = KC = 3C$.
The total energy of the system is $U_2 = \frac{Q^2}{2C'} + \frac{Q^2}{2C'} = \frac{Q^2}{C'} = \frac{(CV)^2}{3C} = \frac{C^2 V^2}{3C} = \frac{CV^2}{3}$.
Therefore,the ratio $\frac{U_1}{U_2} = \frac{CV^2}{CV^2 / 3} = 3: 1$.

(A) The electric displacement vector $D$ is defined as $D = \epsilon E$,where $\epsilon$ is the permittivity of the medium and $E$ is the electric field.
Given that the dielectric constant $K = 3$,the permittivity of the medium is $\epsilon = K \epsilon_0$,where $\epsilon_0 = 8.854 \times 10^{-12} \ F m^{-1}$.
The electric field $E = 1.5 \times 10^{-9} \pi \ N C^{-1}$.
Substituting these values into the formula:
$D = K \epsilon_0 E$
$D = 3 \times (8.854 \times 10^{-12}) \times (1.5 \times 10^{-9} \pi)$
Using $\epsilon_0 \approx \frac{1}{36\pi} \times 10^{-9} \ F m^{-1}$ for calculation convenience:
$D = 3 \times (\frac{1}{36\pi} \times 10^{-9}) \times (1.5 \times 10^{-9} \pi)$
$D = 3 \times \frac{1}{36} \times 1.5 \times 10^{-18}$
$D = \frac{4.5}{36} \times 10^{-18} = 0.125 \times 10^{-18} \ C m^{-2} = 125 \times 10^{-21} \ C m^{-2}$.
Given the options provided,there appears to be a mismatch in the magnitude of the electric field provided in the question. Assuming the standard form $D = \epsilon_0 E_0$ where $E_0$ is the field in vacuum,the correct calculation leads to $125 \times 10^{-12} \ C m^{-2}$ if $E = 1.5 \pi \times 10^9 \ N C^{-1}$ was intended.

(A) The initial force between two charges $q_1$ and $q_2$ separated by distance $r$ in air is given by $F = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r^2} = 98 \ N$.
When dielectric slabs are introduced,the effective distance $r_{eff}$ between the charges is modified. The effective distance is given by $r_{eff} = (r - t_1 - t_2) + t_1\sqrt{K_1} + t_2\sqrt{K_2}$.
Here,$r = 9 \ cm$,$t_1 = 6 \ cm$,$K_1 = 4$,$t_2 = 3 \ cm$,$K_2 = 9$.
$r_{eff} = (9 - 6 - 3) + 6\sqrt{4} + 3\sqrt{9} = 0 + 6(2) + 3(3) = 12 + 9 = 21 \ cm$.
The new force $F'$ is given by $F' = F \left( \frac{r}{r_{eff}} \right)^2$.
$F' = 98 \times \left( \frac{9}{21} \right)^2 = 98 \times \left( \frac{3}{7} \right)^2 = 98 \times \frac{9}{49} = 2 \times 9 = 18 \ N$.

(D) When a dielectric slab of thickness $t$ is introduced between two point charges separated by distance $r$,the effective force $F$ is given by:
$F = \frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{(r - t + t \sqrt{K})^2}$
Case $1$: $t = r/2$ and $K = 4$.
$F_1 = \frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{(r - r/2 + (r/2) \sqrt{4})^2} = \frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{(r/2 + r)^2} = \frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{(3r/2)^2} = \frac{1}{4 \pi \varepsilon_0} \frac{4 q_1 q_2}{9 r^2} = \frac{q_1 q_2}{9 \pi \varepsilon_0 r^2}$.
Case $2$: $t = r/3$ and $K = 9$.
$F_2 = \frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{(r - r/3 + (r/3) \sqrt{9})^2} = \frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{(2r/3 + r)^2} = \frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{(5r/3)^2} = \frac{1}{4 \pi \varepsilon_0} \frac{9 q_1 q_2}{25 r^2}$.
Ratio $F_1/F_2$:
$\frac{F_1}{F_2} = \left( \frac{q_1 q_2}{9 \pi \varepsilon_0 r^2} \right) / \left( \frac{9 q_1 q_2}{100 \pi \varepsilon_0 r^2} \right) = \frac{1}{9} \times \frac{100}{9} = \frac{100}{81}$.

(B) Area of the square plate,$A = L^2$.
Consider an elemental capacitor at distance $x$ from the top with thickness $dx$. The capacitance of this elemental capacitor is given by:
$dC = \frac{k \varepsilon_0 dA}{d} = \frac{k \varepsilon_0 (L \cdot dx)}{d}$
Since all such elemental capacitors are connected in parallel,the equivalent capacitance $C_d$ is the sum (integral) of all these elemental capacitances:
$C_d = \int_0^L \frac{\varepsilon_0 L}{d} \left[1 + \left(\frac{x}{L}\right)^\alpha\right] dx$
$C_d = \frac{\varepsilon_0 L}{d} \left[ x + \frac{x^{\alpha+1}}{L^\alpha (\alpha+1)} \right]_0^L$
$C_d = \frac{\varepsilon_0 L}{d} \left[ L + \frac{L^{\alpha+1}}{L^\alpha (\alpha+1)} \right] = \frac{\varepsilon_0 L^2}{d} \left( 1 + \frac{1}{\alpha+1} \right) = \frac{\varepsilon_0 L^2}{d} \left( \frac{\alpha+2}{\alpha+1} \right)$
The capacitance in the absence of a dielectric is $C_a = \frac{\varepsilon_0 L^2}{d}$.
Given the ratio $\frac{C_d}{C_a} = \frac{7}{6}$:
$\frac{\frac{\varepsilon_0 L^2}{d} \left( \frac{\alpha+2}{\alpha+1} \right)}{\frac{\varepsilon_0 L^2}{d}} = \frac{7}{6}$
$\frac{\alpha+2}{\alpha+1} = \frac{7}{6}$
$6(\alpha+2) = 7(\alpha+1)$
$6\alpha + 12 = 7\alpha + 7$
$\alpha = 5$

(B) $1$. Initial state: The capacitor has a dielectric of thickness $d/2$ and constant $K$. The capacitance $C_i$ is given by the series combination of two capacitors: one with dielectric $(C_1 = \frac{2K\epsilon_0 A}{d})$ and one with air $(C_2 = \frac{2\epsilon_0 A}{d})$.
$C_i = \frac{C_1 C_2}{C_1 + C_2} = \frac{\frac{2K\epsilon_0 A}{d} \cdot \frac{2\epsilon_0 A}{d}}{\frac{2\epsilon_0 A}{d}(K+1)} = \frac{2K\epsilon_0 A}{d(K+1)}$.
$2$. Charge on the capacitor: $Q = C_i V = \frac{2K\epsilon_0 A V}{d(K+1)}$.
$3$. After disconnecting the battery,the charge $Q$ remains constant.
$4$. Final state: The dielectric is removed,so the capacitor becomes an air-filled parallel plate capacitor with capacitance $C_f = \frac{\epsilon_0 A}{d}$.
$5$. Final potential difference $V_f = \frac{Q}{C_f} = \frac{2K\epsilon_0 A V}{d(K+1)} \cdot \frac{d}{\epsilon_0 A} = V \left( \frac{2K}{K+1} \right)$.

(C) The capacitance of a parallel plate capacitor with a dielectric slab of thickness $t$ and dielectric constant $K$ is given by $C = \frac{\epsilon_0 A}{d - t + \frac{t}{K}}$.
For the first case,$t = \frac{d}{2}$ and $K = 4$:
$C_1 = \frac{\epsilon_0 A}{d - \frac{d}{2} + \frac{d/2}{4}} = \frac{\epsilon_0 A}{\frac{d}{2} + \frac{d}{8}} = \frac{\epsilon_0 A}{\frac{5d}{8}} = \frac{8 \epsilon_0 A}{5d}$.
For the second case,$t = \frac{d}{3}$ and $K = 4$:
$C_2 = \frac{\epsilon_0 A}{d - \frac{d}{3} + \frac{d/3}{4}} = \frac{\epsilon_0 A}{\frac{2d}{3} + \frac{d}{12}} = \frac{\epsilon_0 A}{\frac{9d}{12}} = \frac{12 \epsilon_0 A}{9d} = \frac{4 \epsilon_0 A}{3d}$.
Taking the ratio $C_1 : C_2$:
$\frac{C_1}{C_2} = \frac{8 \epsilon_0 A}{5d} \times \frac{3d}{4 \epsilon_0 A} = \frac{24}{20} = \frac{6}{5}$.
Thus,$C_1 : C_2 = 6: 5$.

(C) Initial capacitance $C = \frac{\varepsilon_0 A}{d} = \frac{\varepsilon_0 \times 10 \times 10^{-4}}{3 \times 10^{-3}} = \frac{\varepsilon_0}{3} \text{ F}$.
Initial charge $Q = CV = \frac{\varepsilon_0}{3} \times 12 = 4 \varepsilon_0 \text{ C}$.
Initial energy $U_i = \frac{Q^2}{2C} = \frac{(4 \varepsilon_0)^2}{2(\varepsilon_0/3)} = \frac{16 \varepsilon_0^2}{2 \varepsilon_0 / 3} = 24 \varepsilon_0 \text{ J}$.
After inserting the dielectric,the new capacitance $C' = KC = 3 \times \frac{\varepsilon_0}{3} = \varepsilon_0 \text{ F}$.
Since the battery is disconnected,the charge $Q$ remains constant.
Final energy $U_f = \frac{Q^2}{2C'} = \frac{(4 \varepsilon_0)^2}{2 \varepsilon_0} = \frac{16 \varepsilon_0^2}{2 \varepsilon_0} = 8 \varepsilon_0 \text{ J}$.
Work done $W = U_f - U_i = 8 \varepsilon_0 - 24 \varepsilon_0 = -16 \varepsilon_0 \text{ J}$.
The magnitude of work done on the system is $16 \varepsilon_0$. Thus,$\alpha = 16$.

(C) When a metal plate of thickness $t = 2 \,mm$ is inserted into a parallel plate capacitor with plate spacing $d = 6 \,mm$,the air gaps on either side of the metal plate are $d_1 = 3 \,mm$ and $d_2 = d - d_1 - t = 6 - 3 - 2 = 1 \,mm$.
This arrangement is equivalent to two capacitors in series,each with area $A = 36 \pi \,cm^2 = 36 \pi \times 10^{-4} \,m^2$.
The equivalent capacitance $C_{\text{eq}}$ is given by $\frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{d_1}{\varepsilon_0 A} + \frac{d_2}{\varepsilon_0 A} = \frac{d_1 + d_2}{\varepsilon_0 A}$.
Substituting the values: $d_1 + d_2 = 3 \,mm + 1 \,mm = 4 \,mm = 4 \times 10^{-3} \,m$.
$C_{\text{eq}} = \frac{\varepsilon_0 A}{d_1 + d_2} = \frac{1}{4 \pi \times 9 \times 10^9} \times \frac{36 \pi \times 10^{-4}}{4 \times 10^{-3}}$.
$C_{\text{eq}} = \frac{1}{36 \times 10^9} \times \frac{36 \times 10^{-4}}{4 \times 10^{-3}} = \frac{10^{-4}}{4 \times 10^{-3} \times 10^9} = \frac{1}{4} \times 10^{-10} \,F = 0.25 \times 10^{-10} \,F = 25 \times 10^{-12} \,F = 25 \,pF$.

(B) The initial capacitance of the parallel plate capacitor with air as the dielectric is given by $C = \frac{\varepsilon_0 A}{d}$.
When a metal plate of thickness $t = \frac{d}{2}$ is inserted between the plates,the effective distance between the plates becomes $d_{eff} = d - t = d - \frac{d}{2} = \frac{d}{2}$.
The new capacitance $C^{\prime}$ is given by $C^{\prime} = \frac{\varepsilon_0 A}{d - t} = \frac{\varepsilon_0 A}{d/2} = \frac{2 \varepsilon_0 A}{d}$.
Substituting the expression for $C$,we get $C^{\prime} = 2C$.
Therefore,the ratio $\frac{C^{\prime}}{C} = 2$.

(A) Initial capacitance with air is $C = 80 \times 10^{-6} \ F$.
When the dielectric slab of constant $K = 20$ is inserted,the new capacitance becomes $C' = K \times C = 20 \times 80 \times 10^{-6} = 1600 \times 10^{-6} \ F$.
The charge on the capacitor with the dielectric is $q_1 = C' V = (1600 \times 10^{-6}) \times 30 = 48000 \times 10^{-6} \ C = 48 \times 10^{-3} \ C$.
After removing the dielectric,the capacitance returns to $C = 80 \times 10^{-6} \ F$.
The new charge on the capacitor is $q_2 = C V = (80 \times 10^{-6}) \times 30 = 2400 \times 10^{-6} \ C = 2.4 \times 10^{-3} \ C$.
The charge that flows through the wire is $\Delta q = q_1 - q_2 = 48 \times 10^{-3} - 2.4 \times 10^{-3} = 45.6 \times 10^{-3} \ C$.

(A) Let the initial capacitance of each condenser be $C = \frac{A \varepsilon_0}{d}$.
For condenser $M$,the space is filled with a dielectric of constant $K = 8$. The new capacitance is $C_M = K C = 8C$.
For condenser $N$,a copper plate of thickness $t = d/2$ is introduced. The capacitance of a capacitor with a metal plate of thickness $t$ is $C_N = \frac{A \varepsilon_0}{d - t}$.
Substituting $t = d/2$,we get $C_N = \frac{A \varepsilon_0}{d - d/2} = \frac{A \varepsilon_0}{d/2} = 2 \left( \frac{A \varepsilon_0}{d} \right) = 2C$.
Since the capacitors are connected in series,the charge $Q$ on both is the same.
The potential difference $V$ is given by $V = Q/C$,so $V \propto 1/C$.
Therefore,the ratio of potential differences is $V_M : V_N = \frac{1}{C_M} : \frac{1}{C_N} = \frac{1}{8C} : \frac{1}{2C} = \frac{1}{8} : \frac{1}{2} = 2 : 8 = 1 : 4$.

(C) The capacitance of a parallel plate capacitor with air is $C = \frac{\varepsilon_0 A}{d}$.
When a dielectric slab of thickness $t$ and dielectric constant $K$ is inserted,the new capacitance is $C' = \frac{\varepsilon_0 A}{d - t + \frac{t}{K}}$.
Since the potential difference $V$ is maintained constant and the charge $Q$ remains constant (as the battery is disconnected),the capacitance must remain the same: $C = C'$.
Therefore,$\frac{\varepsilon_0 A}{d} = \frac{\varepsilon_0 A}{d' - t + \frac{t}{K}}$,where $d'$ is the new distance between the plates.
This simplifies to $d = d' - t + \frac{t}{K}$,or $d' - d = t(1 - \frac{1}{K})$.
Given $d' - d = 3.2 \ mm$ and $t = 4 \ mm$,we have:
$3.2 = 4(1 - \frac{1}{K})$
$\frac{3.2}{4} = 1 - \frac{1}{K}$
$0.8 = 1 - \frac{1}{K}$
$\frac{1}{K} = 1 - 0.8 = 0.2$
$K = \frac{1}{0.2} = 5$.

(C) When a capacitor is fully charged and then disconnected from the battery, the total charge $Q$ on the plates remains constant because there is no path for the charge to flow.
When a dielectric is inserted between the plates, the electric field inside the capacitor causes the dielectric to polarize.
This polarization results in the formation of induced charges $(q')$ on the surfaces of the dielectric. As the dielectric material is introduced, these induced charges appear, meaning the charge on the surface of the dielectric increases from zero.
Since the capacitor is isolated, the total charge on the outer surfaces of the metal plates remains constant $(Q = \text{constant})$.
Therefore, the charge on the surface of the dielectric increases, and the charge on the outer surface of the plates remains unchanged.

(D) Initial capacitance with air $C_0 = 4 \mu F$.
After filling with dielectric $(K = 5)$,the new capacitance is $C = K C_0 = 5 \times 4 \mu F = 20 \mu F$.
The charge on the capacitor is $Q = C V = 20 \mu F \times 100 \ V = 2000 \mu C = 2 \times 10^{-3} \ C$.
Since the capacitor is disconnected from the battery,the charge $Q$ remains constant.
Initial energy stored with dielectric: $U_i = \frac{Q^2}{2C} = \frac{(2 \times 10^{-3})^2}{2 \times 20 \times 10^{-6}} = \frac{4 \times 10^{-6}}{40 \times 10^{-6}} = 0.1 \ J$.
Final capacitance after removing the dielectric: $C_f = C_0 = 4 \mu F$.
Final energy stored: $U_f = \frac{Q^2}{2C_f} = \frac{(2 \times 10^{-3})^2}{2 \times 4 \times 10^{-6}} = \frac{4 \times 10^{-6}}{8 \times 10^{-6}} = 0.5 \ J$.
Work done by external agent $W = U_f - U_i = 0.5 \ J - 0.1 \ J = 0.4 \ J$.

(C) Let the distance between the two charges be $d$. The force in air is $F = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{d^2}$.
When a medium of thickness $t = 0.3d$ and dielectric constant $K$ is introduced,the effective distance $d_{eff}$ becomes $d_{eff} = (d - t) + t\sqrt{K} = (0.7d) + 0.3d\sqrt{K} = d(0.7 + 0.3\sqrt{K})$.
The new force is $F' = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{d_{eff}^2} = \frac{F}{2.56}$.
Thus,$d_{eff}^2 = 2.56 d^2$,which implies $d_{eff} = 1.6d$.
Equating the expressions: $0.7 + 0.3\sqrt{K} = 1.6$.
$0.3\sqrt{K} = 0.9$.
$\sqrt{K} = 3$.
$K = 9$.

(B, C, D) The capacitor can be viewed as two capacitors in parallel,each with area $A/2$ and plate separation $d$.
The capacitance of the dielectric-filled part is $C_1 = \frac{K \varepsilon_0 (A/2)}{d} = \frac{K C_0}{2}$ and the air-filled part is $C_2 = \frac{\varepsilon_0 (A/2)}{d} = \frac{C_0}{2}$,where $C_0 = \frac{\varepsilon_0 A}{d}$.
The equivalent capacitance is $C_{eq} = C_1 + C_2 = \frac{C_0}{2}(K+1)$. Thus,option $(d)$ is correct.
Since the capacitors are in parallel,the potential difference $V$ across both is the same. The charges are $Q_1 = C_1 V$ and $Q_2 = C_2 V$.
The ratio of charges is $\frac{Q_1}{Q_2} = \frac{C_1}{C_2} = K$. Since $Q_1 + Q_2 = Q$,we have $Q_2 = \frac{Q}{K+1}$ and $Q_1 = \frac{KQ}{K+1}$. Thus,option $(c)$ is correct.
The charge densities are $\sigma_1 = Q_1 / (A/2)$ and $\sigma_2 = Q_2 / (A/2)$. Since $Q_1 \neq Q_2$,the charge densities are unequal. Thus,option $(b)$ is correct.
The electric fields are $E_1 = \frac{\sigma_1}{K \varepsilon_0}$ and $E_2 = \frac{\sigma_2}{\varepsilon_0}$. Substituting $\sigma_1 = K \sigma_2$,we get $E_1 = \frac{K \sigma_2}{K \varepsilon_0} = \frac{\sigma_2}{\varepsilon_0} = E_2$. The fields are equal,so $(a)$ is incorrect.

(B) Initial capacitance of the capacitor is $C = \frac{\epsilon_0 A}{d} = \frac{\epsilon_0 A}{5 \times 10^{-3}}$.
Initial charge $Q_1 = CV$.
When a mica sheet of thickness $t = 2 \text{ mm}$ is introduced,the capacitor acts as two capacitors in series: one with air of thickness $(d-t) = 3 \text{ mm}$ and one with mica of thickness $t = 2 \text{ mm}$.
$C_1 = \frac{\epsilon_0 A}{d-t} = \frac{\epsilon_0 A}{3 \times 10^{-3}}$ and $C_2 = \frac{K \epsilon_0 A}{t} = \frac{K \epsilon_0 A}{2 \times 10^{-3}}$.
The equivalent capacitance is $C_{eq} = \frac{C_1 C_2}{C_1 + C_2} = \frac{(\frac{\epsilon_0 A}{3 \times 10^{-3}}) (\frac{K \epsilon_0 A}{2 \times 10^{-3}})}{\frac{\epsilon_0 A}{3 \times 10^{-3}} + \frac{K \epsilon_0 A}{2 \times 10^{-3}}} = \frac{K \epsilon_0 A}{2 \times 10^{-3} + 3 \times 10^{-3} K} = \frac{K \epsilon_0 A}{10^{-3}(2 + 3K)}$.
Since the battery remains connected,the new charge is $Q_2 = C_{eq} V$. Given $Q_2 = 1.25 Q_1$,we have $C_{eq} = 1.25 C$.
$\frac{K \epsilon_0 A}{10^{-3}(2 + 3K)} = 1.25 \frac{\epsilon_0 A}{5 \times 10^{-3}}$.
$\frac{K}{2 + 3K} = \frac{1.25}{5} = 0.25 = \frac{1}{4}$.
$4K = 2 + 3K \Rightarrow K = 2$.

(D) Let $C_0 = \frac{\varepsilon_0 A}{d}$.
For capacitor $A$:
The top half has dielectric $k_1$ and the bottom half is split into two parallel parts with dielectrics $k_1$ and $k_2$. The equivalent capacitance is $C_A = \left( \frac{1}{C_{k1}} + \frac{1}{C_{parallel}} \right)^{-1} = \left( \frac{1}{2k_1 C_0} + \frac{1}{k_1 C_0/2 + k_2 C_0/2} \right)^{-1} = \frac{2k_1(k_1+k_2)C_0}{3k_1+k_2}$.
For capacitor $B$:
The top half has dielectric $k_2$ and the bottom half is split into two parallel parts with dielectrics $k_1$ and $k_2$. Similarly,$C_B = \frac{2k_2(k_1+k_2)C_0}{k_1+3k_2}$.
For capacitor $C$:
It consists of two parallel branches,each having two dielectrics in series. $C_C = \frac{k_1 k_2}{k_1+k_2} C_0 + \frac{k_2 k_1}{k_2+k_1} C_0 = \frac{2k_1 k_2}{k_1+k_2} C_0$.
Given $k_1 > k_2$,comparing the expressions leads to $C_A > C_C > C_B$.

\(C _1=\frac{5 \in_0 A / 2}{d / 2}=\frac{5 \in_0 A}{ d }=5 C\)
\(C _2=\frac{2 \in_0 A / 2}{d / 2}=\frac{2 \in_0 A}{ d }=2 C\)
\(C _1 \& C _2\) in series.
\(C^{\prime}=\frac{C_1 C_2}{C_1+C_2}=\frac{(5 C)(2 C)}{7 C}=\frac{10}{7} C\)
\(C _3=\frac{3 \in_0 A / 2}{d / 2}=3 C\)
\(C _4=\frac{2 \epsilon_0 A / 2}{d / 2}=2 C\)
\(C _4 \& C _3\) in series; \(C ^{\prime \prime}=\frac{(2 C )(3 C )}{5 C }=\frac{6}{5} C\)
\(C ^{\prime} \& C ^{\prime \prime}\) in parallel;
So, \(C _{ eq }= C \left(\frac{6}{5}+\frac{10}{7}\right)= C \left(\frac{42+50}{35}\right)=\left(\frac{92}{35}\right) C\)
\(\frac{92}{35} C =\frac{ nC }{3}\)
\(n =\frac{92 \times 3}{35}=7.9 \Rightarrow n \simeq 8\)

(A) The initial capacitance of the parallel plate capacitor with vacuum is $C = \frac{\epsilon_0 A}{d}$.
When a dielectric slab of thickness $t = d/3$ and dielectric constant $K$ is introduced,the system can be modeled as two capacitors in series: one with air of thickness $d - t = 2d/3$ and one with dielectric of thickness $t = d/3$.
The capacitance of the air part is $C_1 = \frac{\epsilon_0 A}{2d/3} = \frac{3}{2} \frac{\epsilon_0 A}{d} = \frac{3}{2} C$.
The capacitance of the dielectric part is $C_2 = \frac{K \epsilon_0 A}{d/3} = 3K \frac{\epsilon_0 A}{d} = 3KC$.
Since these are in series,the equivalent capacitance $C_{eq}$ is given by:
$C_{eq} = \frac{C_1 C_2}{C_1 + C_2} = \frac{(\frac{3}{2} C) (3KC)}{\frac{3}{2} C + 3KC} = \frac{\frac{9}{2} KC^2}{\frac{3}{2} C (1 + 2K)} = \frac{3KC}{2K + 1}$.

(D) The initial capacitance of the parallel plate capacitor with air is given by $C_0 = \frac{\epsilon_0 A}{d} = 1.0 \text{ pF}$.
When the distance between the plates is doubled $(d' = 2d)$ and the space is filled with a dielectric of constant $K$,the new capacitance is $C = \frac{K \epsilon_0 A}{d'}$.
Substituting $d' = 2d$ into the equation,we get $C = \frac{K \epsilon_0 A}{2d} = \frac{K}{2} \times C_0$.
Given that $C = 2.0 \text{ pF}$ and $C_0 = 1.0 \text{ pF}$,we have $2.0 = \frac{K}{2} \times 1.0$.
Solving for $K$,we get $K = 2.0 \times 2 = 4.0$.

(B) The initial capacitance of the air capacitor is $C = \frac{\epsilon_0 A}{d}$.
When the plate separation $d$ is half-filled with a dielectric of constant $K=5$,it acts as two capacitors in series.
The first capacitor (air) has a plate separation of $d/2$,so $C_1 = \frac{\epsilon_0 A}{d/2} = \frac{2\epsilon_0 A}{d} = 2C$.
The second capacitor (dielectric) has a plate separation of $d/2$,so $C_2 = \frac{K\epsilon_0 A}{d/2} = \frac{2K\epsilon_0 A}{d} = 2KC = 2(5)C = 10C$.
Since they are in series,the new capacitance $C_{new}$ is given by:
$C_{new} = \frac{C_1 C_2}{C_1 + C_2} = \frac{(2C)(10C)}{2C + 10C} = \frac{20C^2}{12C} = \frac{5}{3}C$.
The increase in capacitance is $\Delta C = C_{new} - C = \frac{5}{3}C - C = \frac{2}{3}C$.
The percentage increase is $\frac{\Delta C}{C} \times 100 = \frac{2/3 C}{C} \times 100 = \frac{2}{3} \times 100 \approx 66.67\%$.

(B) The capacitance of a parallel plate capacitor is given by $C = \frac{K \epsilon_0 A}{d}$.
The resistance of the dielectric material is given by $R = \frac{\rho d}{A}$.
Multiplying these two expressions, we get $RC = \left( \frac{K \epsilon_0 A}{d} \right) \left( \frac{\rho d}{A} \right) = K \epsilon_0 \rho$.
Given values are $R = 17.7 \times 10^{14}$ $\Omega$, $C = 1 \times 10^{-6}$ $F$, $\rho = 10^{13}$ $\Omega$m, and $\epsilon_0 = 8.85 \times 10^{-12}$ $F$/m.
Substituting these values into the equation: $(17.7 \times 10^{14}) \times (1 \times 10^{-6}) = K \times (8.85 \times 10^{-12}) \times (10^{13})$.
$17.7 \times 10^8 = K \times 8.85 \times 10^1$.
$17.7 \times 10^8 = 88.5 K$.
$K = \frac{17.7 \times 10^8}{88.5} = 0.2 \times 10^8 = 2 \times 10^7$.
Given that the relative permittivity $K = \alpha \times 10^7$, we find $\alpha = 2$.