Effect of Dielectric Inside Capacitor Questions in English

(D) The capacitance of a parallel plate capacitor with air as the medium is given by $C = \frac{Q}{V}$.
Given $Q = 10\,\mu C$ and $V = 10\,V$,we have $C = \frac{10\,\mu C}{10\,V} = 1\,\mu F$.
When the space is filled with a dielectric medium of constant $K$,the new capacitance becomes $C' = KC$.
The new charge is $Q' = C'V = KCV = KQ$.
Given $Q' = 100\,\mu C$ and $Q = 10\,\mu C$,we have $100\,\mu C = K \times 10\,\mu C$.
Therefore,$K = \frac{100}{10} = 10$.
The dielectric constant of the oil is $10$.

(C) When a dielectric slab is inserted into a parallel plate capacitor while it remains connected to a battery,the potential difference $V$ across the plates remains constant because it is equal to the $EMF$ of the battery.
The capacitance $C$ of the capacitor increases by a factor of $K$ (dielectric constant),where $C = K C_0$.
The energy stored in the capacitor is given by the formula $U = \frac{1}{2} C V^2$.
Since $V$ is constant and $C$ increases,the stored energy $U$ must increase.

(C) Let the initial distance between the plates be $d$. The capacitance of a parallel plate capacitor is $C = \frac{\varepsilon_0 A}{d}$.
When a dielectric slab of thickness $t = 3 \, mm$ and dielectric constant $K$ is introduced,the new capacitance $C'$ is given by $C' = \frac{\varepsilon_0 A}{d - t + \frac{t}{K}}$.
To maintain the same potential difference,the effective capacitance must remain the same. The problem states that the distance between the plates is increased by $\Delta d = 2.4 \, mm$,so the new distance is $d' = d + 2.4$.
The new capacitance with the slab is $C' = \frac{\varepsilon_0 A}{d' - t + \frac{t}{K}} = \frac{\varepsilon_0 A}{d + 2.4 - 3 + \frac{3}{K}} = \frac{\varepsilon_0 A}{d - 0.6 + \frac{3}{K}}$.
For the capacitance to remain unchanged,we must have $d = d - 0.6 + \frac{3}{K}$.
This simplifies to $0.6 = \frac{3}{K}$.
Therefore,$K = \frac{3}{0.6} = 5$.

(C) The electric force on the slab is given by $F = -\frac{dU}{dx}$.
The energy stored in the capacitor is $U = \frac{Q^2}{2C}$.
The capacitor can be viewed as two capacitors in parallel: one with the dielectric of length $x$ and one without of length $(l-x)$.
$C = C_1 + C_2 = \frac{K \varepsilon_0 w x}{d} + \frac{\varepsilon_0 w (l-x)}{d} = \frac{\varepsilon_0 w}{d} [x(K-1) + l]$.
Substituting $C$ into the energy formula:
$U = \frac{Q^2 d}{2 \varepsilon_0 w [x(K-1) + l]}$.
Now,differentiate with respect to $x$:
$F = -\frac{dU}{dx} = -\frac{Q^2 d}{2 \varepsilon_0 w} \cdot \frac{d}{dx} [x(K-1) + l]^{-1}$.
$F = -\frac{Q^2 d}{2 \varepsilon_0 w} \cdot (-1) [x(K-1) + l]^{-2} \cdot (K-1)$.
$F = \frac{Q^2 d (K-1)}{2 \varepsilon_0 w [x(K-1) + l]^2}$.
At the edge $(x=0)$:
$F = \frac{Q^2 d (K-1)}{2 \varepsilon_0 w l^2}$.

(C) The dielectric constant is given by $K(x) = K_0 + \lambda x$.
Consider a small element of thickness $dx$ at distance $x$. This acts as a capacitor with capacitance $dC = \frac{\epsilon_0 K(x) A}{dx}$.
Since these elements are in series,the total capacitance $C$ is given by $\frac{1}{C} = \int_0^d \frac{1}{dC} = \int_0^d \frac{dx}{\epsilon_0 A (K_0 + \lambda x)}$.
Evaluating the integral: $\frac{1}{C} = \frac{1}{\epsilon_0 A} \int_0^d \frac{dx}{K_0 + \lambda x} = \frac{1}{\epsilon_0 A \lambda} [\ln(K_0 + \lambda x)]_0^d$.
$\frac{1}{C} = \frac{1}{\epsilon_0 A \lambda} \ln \left( \frac{K_0 + \lambda d}{K_0} \right) = \frac{1}{\epsilon_0 A \lambda} \ln \left( 1 + \frac{\lambda d}{K_0} \right)$.
Since vacuum capacitance $C_0 = \frac{\epsilon_0 A}{d}$,we have $\epsilon_0 A = C_0 d$.
Substituting this: $\frac{1}{C} = \frac{1}{C_0 d \lambda} \ln \left( 1 + \frac{\lambda d}{K_0} \right)$.
Therefore,$C = \frac{\lambda d}{\ln(1 + \lambda d / K_0)} C_0$.

(D) Let the permittivity $\varepsilon(x)$ vary linearly with distance $x$ from one plate $(x=0)$ to the other $(x=d)$:
$\varepsilon(x) = \varepsilon_1 + \frac{\varepsilon_2 - \varepsilon_1}{d} x$
Consider a thin elemental slab of thickness $dx$ at distance $x$. This acts as a capacitor with capacitance $dC = \frac{\varepsilon(x) A}{dx}$.
Since these elemental capacitors are in series,the equivalent capacitance $C$ is given by $\frac{1}{C} = \int_0^d \frac{1}{dC} = \int_0^d \frac{dx}{\varepsilon(x) A}$.
$\frac{1}{C} = \frac{1}{A} \int_0^d \frac{dx}{\varepsilon_1 + \frac{\varepsilon_2 - \varepsilon_1}{d} x}$
Let $u = \varepsilon_1 + \frac{\varepsilon_2 - \varepsilon_1}{d} x$,then $du = \frac{\varepsilon_2 - \varepsilon_1}{d} dx$.
$\frac{1}{C} = \frac{1}{A} \cdot \frac{d}{\varepsilon_2 - \varepsilon_1} \int_{\varepsilon_1}^{\varepsilon_2} \frac{du}{u} = \frac{d}{A(\varepsilon_2 - \varepsilon_1)} [\ln u]_{\varepsilon_1}^{\varepsilon_2} = \frac{d \ln(\varepsilon_2/\varepsilon_1)}{A(\varepsilon_2 - \varepsilon_1)}$.
Therefore,$C = \frac{A(\varepsilon_2 - \varepsilon_1)}{d \ln(\varepsilon_2/\varepsilon_1)}$.

(B) The initial capacitance is $C_0 = \frac{K \epsilon_0 A}{d}$.
When the material is replaced,the capacitor acts as two capacitors in parallel.
The first capacitor has area $A_1 = \frac{1}{3}A$,dielectric constant $K_1 = 2K$,and separation $d$. Its capacitance is $C_1 = \frac{K_1 \epsilon_0 A_1}{d} = \frac{(2K) \epsilon_0 (A/3)}{d} = \frac{2}{3} \frac{K \epsilon_0 A}{d} = \frac{2}{3} C_0$.
The second capacitor has area $A_2 = \frac{2}{3}A$,dielectric constant $K_2 = K$,and separation $d$. Its capacitance is $C_2 = \frac{K_2 \epsilon_0 A_2}{d} = \frac{K \epsilon_0 (2A/3)}{d} = \frac{2}{3} \frac{K \epsilon_0 A}{d} = \frac{2}{3} C_0$.
Since they are in parallel,the total capacitance is $C = C_1 + C_2 = \frac{2}{3} C_0 + \frac{2}{3} C_0 = \frac{4}{3} C_0$.
Therefore,$\frac{C}{C_0} = \frac{4}{3}$.

(A) Consider a small strip of width $dx$ at a distance $x$ from the left end. The height of the dielectric part is $y = (d/a)x$.
This strip acts as two capacitors in series: one with air (thickness $d-y$) and one with dielectric (thickness $y$).
The capacitance of the air part is $dC_1 = \frac{\varepsilon_0 a dx}{d-y}$ and the dielectric part is $dC_2 = \frac{K\varepsilon_0 a dx}{y}$.
The equivalent capacitance $dC$ of the strip is given by $\frac{1}{dC} = \frac{1}{dC_1} + \frac{1}{dC_2} = \frac{d-y}{\varepsilon_0 a dx} + \frac{y}{K\varepsilon_0 a dx} = \frac{Kd - Ky + y}{K\varepsilon_0 a dx} = \frac{Kd - (K-1)y}{K\varepsilon_0 a dx}$.
Thus,$dC = \frac{K\varepsilon_0 a dx}{Kd - (K-1)(d/a)x}$.
Integrating from $x=0$ to $x=a$:
$C = \int_0^a \frac{K\varepsilon_0 a dx}{Kd - \frac{(K-1)d}{a}x} = \frac{K\varepsilon_0 a}{d} \int_0^a \frac{dx}{K - \frac{(K-1)}{a}x}$.
Let $u = K - \frac{(K-1)}{a}x$,then $du = -\frac{(K-1)}{a} dx$.
$C = \frac{K\varepsilon_0 a}{d} \left( -\frac{a}{K-1} \right) [\ln(K - \frac{(K-1)}{a}x)]_0^a = \frac{-K\varepsilon_0 a^2}{d(K-1)} [\ln(1) - \ln(K)] = \frac{K\varepsilon_0 a^2}{d(K-1)} \ln K$.

(C) The capacitor can be viewed as two capacitors in series,each of thickness $d/2$.
For the first half (thickness $d/2$),we have two capacitors in parallel with areas $L^2/2$ each:
$C_1 = \frac{\varepsilon_0 K_1 (L^2/2)}{d/2} + \frac{\varepsilon_0 K_3 (L^2/2)}{d/2} = \frac{\varepsilon_0 L^2}{d} (K_1 + K_3)$.
For the second half (thickness $d/2$),we have two capacitors in parallel with areas $L^2/2$ each:
$C_2 = \frac{\varepsilon_0 K_2 (L^2/2)}{d/2} + \frac{\varepsilon_0 K_4 (L^2/2)}{d/2} = \frac{\varepsilon_0 L^2}{d} (K_2 + K_4)$.
Since these two halves are in series,the equivalent capacitance $C$ is given by:
$\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{d}{\varepsilon_0 L^2 (K_1 + K_3)} + \frac{d}{\varepsilon_0 L^2 (K_2 + K_4)}$.
Also,$C = \frac{\varepsilon_0 K L^2}{d}$,so:
$\frac{d}{\varepsilon_0 K L^2} = \frac{d}{\varepsilon_0 L^2} \left( \frac{1}{K_1 + K_3} + \frac{1}{K_2 + K_4} \right)$.
$\frac{1}{K} = \frac{(K_2 + K_4) + (K_1 + K_3)}{(K_1 + K_3)(K_2 + K_4)}$.
Therefore,$K = \frac{(K_1 + K_3)(K_2 + K_4)}{K_1 + K_2 + K_3 + K_4}$.

(C) The capacitor is divided into three parallel capacitors,each with area $A' = A/3$ and the same plate separation $d$.
The equivalent capacitance $C_{net}$ is the sum of the individual capacitances:
$C_{net} = C_1 + C_2 + C_3$
Using the formula for a parallel plate capacitor $C = \frac{K \epsilon_0 A}{d}$,we have:
$\frac{K_{eq} \epsilon_0 A}{d} = \frac{K_1 \epsilon_0 (A/3)}{d} + \frac{K_2 \epsilon_0 (A/3)}{d} + \frac{K_3 \epsilon_0 (A/3)}{d}$
Canceling the common terms $\frac{\epsilon_0 A}{3d}$ from both sides:
$K_{eq} = \frac{K_1 + K_2 + K_3}{3}$
Substituting the given values:
$K_{eq} = \frac{10 + 12 + 14}{3} = \frac{36}{3} = 12$
Thus,the equivalent dielectric constant is $12$.

(B) Initial capacitance $C = 12 \, pF$,Initial potential $V = 10 \, V$.
Initial charge $Q = CV = 12 \, pF \times 10 \, V = 120 \, pC$.
Initial energy $U_i = \frac{Q^2}{2C} = \frac{(120)^2}{2 \times 12} = \frac{14400}{24} = 600 \, pJ$.
When the dielectric slab is inserted,the new capacitance becomes $C' = kC = 6.5 \times 12 = 78 \, pF$.
Since the battery is disconnected,the charge $Q$ remains constant.
Final energy $U_f = \frac{Q^2}{2C'} = \frac{Q^2}{2kC} = \frac{U_i}{k} = \frac{600}{6.5} \approx 92.31 \, pJ$.
The work done by the capacitor on the slab is the decrease in potential energy: $W = U_i - U_f = 600 - 92.31 = 507.69 \, pJ$.
Rounding to the nearest integer,we get $508 \, pJ$.

(D) Given:
Voltage rating $V = 500\,V$
Maximum electric field $E_{\max} = 10^6\,V/m$
Plate area $A = 10^{-4}\,m^2$
Capacitance $C = 15\,pF = 15 \times 10^{-12}\,F$
Permittivity of free space $\epsilon_0 = 8.86 \times 10^{-12}\,C^2/Nm^2$
For a parallel plate capacitor,the maximum voltage $V$ is related to the plate separation $d$ and the maximum electric field $E_{\max}$ by:
$V = E_{\max} \times d$
$d = \frac{V}{E_{\max}} = \frac{500}{10^6} = 5 \times 10^{-4}\,m$
The capacitance of a parallel plate capacitor with a dielectric is given by:
$C = \frac{k \epsilon_0 A}{d}$
Where $k$ is the dielectric constant.
Rearranging for $k$:
$k = \frac{C \times d}{\epsilon_0 \times A}$
Substituting the values:
$k = \frac{15 \times 10^{-12} \times 5 \times 10^{-4}}{8.86 \times 10^{-12} \times 10^{-4}}$
$k = \frac{15 \times 5}{8.86} = \frac{75}{8.86} \approx 8.465$
Rounding to the nearest option,we get $k \approx 8.5$.

(C) $1$. Initially,the capacitors are connected in parallel to a battery of voltage $V$. The total charge $Q_{total}$ stored in the system is given by:
$Q_{total} = C V + n C V = (n + 1) C V$
$2$. After the battery is disconnected,the total charge $Q_{total}$ remains conserved in the system.
$3$. When a dielectric of constant $K$ is inserted into the first capacitor,its new capacitance becomes $C' = K C$. The second capacitor remains unchanged with capacitance $n C$.
$4$. Since the capacitors are in parallel,they share a common potential difference $V_c$. The total capacitance of the system becomes $C_{eq} = K C + n C = (K + n) C$.
$5$. The new potential difference $V_c$ is given by:
$V_c = \frac{Q_{total}}{C_{eq}} = \frac{(n + 1) C V}{(K + n) C} = \frac{(n + 1) V}{K + n}$

(B) For capacitor $(I)$,the dielectrics are in series. The capacitance of each part is $C_i = \frac{3 \varepsilon_0 A K_i}{d}$.
Since they are in series,$\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} = \frac{d}{3 \varepsilon_0 A} (\frac{1}{K_1} + \frac{1}{K_2} + \frac{1}{K_3}) = \frac{d(K_2 K_3 + K_3 K_1 + K_1 K_2)}{3 \varepsilon_0 A K_1 K_2 K_3}$.
Thus,$C_{eq} = \frac{3 \varepsilon_0 A K_1 K_2 K_3}{d(K_1 K_2 + K_2 K_3 + K_3 K_1)}$.
For capacitor $(II)$,the dielectrics are in parallel. The capacitance of each part is $C_i' = \frac{\varepsilon_0 (A/3) K_i}{d} = \frac{\varepsilon_0 A K_i}{3d}$.
Since they are in parallel,$C_{eq}' = C_1' + C_2' + C_3' = \frac{\varepsilon_0 A}{3d} (K_1 + K_2 + K_3)$.
The energy stored is $E = \frac{1}{2} C V^2$. Therefore,$\frac{E_1}{E_2} = \frac{C_{eq}}{C_{eq}'} = \frac{3 \varepsilon_0 A K_1 K_2 K_3}{d(K_1 K_2 + K_2 K_3 + K_3 K_1)} \cdot \frac{3d}{\varepsilon_0 A (K_1 + K_2 + K_3)} = \frac{9 K_1 K_2 K_3}{(K_1 + K_2 + K_3)(K_1 K_2 + K_2 K_3 + K_3 K_1)}$.

(C) Initial charge on the capacitor plates without the dielectric is:
$Q = CV = (5\,\mu F) \times (1\,V) = 5\,\mu C$
The capacitance $C'$ after introducing a dielectric slab of thickness $t$ and dielectric constant $K$ in a capacitor with plate separation $d$ is given by:
$C' = \frac{\varepsilon_0 A}{d - t + \frac{t}{K}} = \frac{C}{1 - \frac{t}{d}(1 - \frac{1}{K})}$
Substituting the given values $(C = 5\,\mu F, d = 6\,cm, t = 4\,cm, K = 4)$:
$C' = \frac{5\,\mu F}{1 - \frac{4}{6}(1 - \frac{1}{4})} = \frac{5\,\mu F}{1 - \frac{2}{3}(\frac{3}{4})} = \frac{5\,\mu F}{1 - \frac{1}{2}} = \frac{5\,\mu F}{0.5} = 10\,\mu F$
The new charge on the capacitor plates is:
$Q' = C'V = (10\,\mu F) \times (1\,V) = 10\,\mu C$
The additional charge that flows from the battery is:
$\Delta Q = Q' - Q = 10\,\mu C - 5\,\mu C = 5\,\mu C$

(A) When the capacitor is disconnected from the source,the charge $Q$ on the plates remains constant because there is no path for the charge to flow.
Inserting a dielectric slab increases the capacitance $C$ of the capacitor,as $C = K C_0$,where $K > 1$ is the dielectric constant.
Since the charge $Q$ is constant and the capacitance $C$ increases,the potential difference $V$ across the plates changes according to the relation $V = Q / C$. As $C$ increases,$V$ decreases.
Therefore,the potential difference $(i)$ and the capacitance $(ii)$ change,while the charge $(iii)$ remains constant.

(D) The capacitor consists of two capacitors in series,each with a dielectric layer.
For a parallel plate capacitor,the capacitance is given by $C = \frac{K \epsilon_0 A}{d}$.
For the first layer (dielectric constant $K_1 = 2$,thickness $d$): $C_1 = \frac{2 \epsilon_0 A}{d}$.
For the second layer (dielectric constant $K_2 = 6$,thickness $2d$): $C_2 = \frac{6 \epsilon_0 A}{2d} = \frac{3 \epsilon_0 A}{d}$.
Since the capacitors are in series,the charge $Q$ on each is the same.
The potential difference across a capacitor is $V = Q/C$.
Therefore,the ratio of potential differences is $\frac{V_1}{V_2} = \frac{Q/C_1}{Q/C_2} = \frac{C_2}{C_1}$.
Substituting the values: $\frac{V_1}{V_2} = \frac{3 \epsilon_0 A / d}{2 \epsilon_0 A / d} = \frac{3}{2} = 1.5$.

(A) Let the area of each plate be $A$ and the distance between them be $d$. The capacity of the capacitor in air is $C = \frac{\varepsilon_0 A}{d}$.
When the space between the plates is filled with a dielectric of constant $\varepsilon_r$ up to half the distance $d/2$,the capacitor can be considered as two capacitors in series.
One capacitor has air as the dielectric with thickness $d_1 = d/2$,so its capacity is $C_1 = \frac{\varepsilon_0 A}{d/2} = 2C$.
The other capacitor has a dielectric material with thickness $d_2 = d/2$,so its capacity is $C_2 = \frac{\varepsilon_r \varepsilon_0 A}{d/2} = 2\varepsilon_r C$.
Since they are in series,the equivalent capacity $C_{eq}$ is given by $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}$.
$\frac{1}{C_{eq}} = \frac{1}{2C} + \frac{1}{2\varepsilon_r C} = \frac{1}{2C} \left(1 + \frac{1}{\varepsilon_r}\right) = \frac{1}{2C} \left(\frac{\varepsilon_r + 1}{\varepsilon_r}\right)$.
Therefore,$C_{eq} = \frac{2\varepsilon_r C}{\varepsilon_r + 1}$.

(D) The capacitor can be divided into three parts. Let the total area be $A$ and distance be $d$.
$1$. The left side consists of two capacitors $C_1$ and $C_2$ in series,each with area $A/2$ and distance $d/2$.
$C_1 = \frac{K_1 \varepsilon_0 (A/2)}{d/2} = \frac{K_1 \varepsilon_0 A}{d}$
$C_2 = \frac{K_2 \varepsilon_0 (A/2)}{d/2} = \frac{K_2 \varepsilon_0 A}{d}$
Since they are in series,the equivalent capacitance $C_{12}$ is given by $\frac{1}{C_{12}} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{d}{\varepsilon_0 A} (\frac{1}{K_1} + \frac{1}{K_2}) = \frac{d}{\varepsilon_0 A} (\frac{K_1 + K_2}{K_1 K_2})$.
So,$C_{12} = \frac{\varepsilon_0 A}{d} (\frac{K_1 K_2}{K_1 + K_2})$.
$2$. The right side is a capacitor $C_3$ with area $A/2$ and distance $d$.
$C_3 = \frac{K_3 \varepsilon_0 (A/2)}{d} = \frac{K_3 \varepsilon_0 A}{2d}$.
$3$. $C_{12}$ and $C_3$ are in parallel. The resultant capacitance $C_{eq} = C_{12} + C_3 = \frac{\varepsilon_0 A}{d} (\frac{K_1 K_2}{K_1 + K_2} + \frac{K_3}{2})$.

(A) Initially,the capacitors are connected in parallel and charged to potential $V$. The total charge stored in the system is $Q_{\text{total}} = (2C)V + (C)V = 3CV$.
When the battery is removed,the total charge $Q_{\text{total}} = 3CV$ remains conserved.
After filling the capacitor $C$ with a dielectric of constant $K$,its new capacitance becomes $C' = KC$.
The new total capacitance of the parallel combination is $C_{\text{eq}} = 2C + KC = C(K + 2)$.
The new potential difference $V'$ across the parallel combination is given by $V' = \frac{Q_{\text{total}}}{C_{\text{eq}}}$.
Substituting the values,$V' = \frac{3CV}{C(K + 2)} = \frac{3V}{K + 2}$.

(C) The initial capacitance of the air-filled parallel plate capacitor is given by $C = \frac{\varepsilon_0 A}{d} = 9 \, pF$.
When the space is filled with two dielectrics of thicknesses $d_1 = d/3$ and $d_2 = 2d/3$,the system acts as two capacitors in series.
The capacitance of the first part is $C_1 = \frac{K_1 \varepsilon_0 A}{d_1} = \frac{3 \varepsilon_0 A}{d/3} = 9 \frac{\varepsilon_0 A}{d} = 9 \times 9 = 81 \, pF$.
The capacitance of the second part is $C_2 = \frac{K_2 \varepsilon_0 A}{d_2} = \frac{6 \varepsilon_0 A}{2d/3} = 9 \frac{\varepsilon_0 A}{d} = 9 \times 9 = 81 \, pF$.
Since they are in series,the equivalent capacitance $C_{eq}$ is given by:
$\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{1}{81} + \frac{1}{81} = \frac{2}{81}$.
Therefore,$C_{eq} = \frac{81}{2} = 40.5 \, pF$.

(D) When a dielectric slab of dielectric constant $\epsilon_r$ is inserted into a capacitor while the cell remains connected,the potential difference $V$ across the plates remains constant because it is fixed by the cell.
The new capacitance $C'$ becomes $C' = \epsilon_r C$. Since $\epsilon_r > 1$,the capacitance increases.
The charge stored on the capacitor is given by $Q = CV$. Since $V$ is constant and $C$ increases,the charge $Q$ stored on the plates will increase $(Q' = \epsilon_r Q)$.

(A) The energy stored in a capacitor is given by $U = Q^2 / (2C)$.
Initially,the capacitance is $C = \epsilon_0 A / d$.
When the battery is disconnected,the charge $Q$ on the plates remains constant.
After introducing a dielectric slab of thickness $d$ and dielectric constant $K$,the new capacitance becomes $C' = KC$.
The new energy stored is $U' = Q^2 / (2C') = Q^2 / (2KC)$.
The ratio of energy before and after is $U / U' = (Q^2 / 2C) / (Q^2 / 2KC) = K$.

(A) The capacitance of a parallel plate capacitor with air is given by $C = \frac{\varepsilon_0 A}{d} = 8\,\mu F$.
When the distance between the plates is reduced to $d' = \frac{d}{2}$ and a dielectric of constant $K = 6$ is introduced,the new capacitance $C'$ is given by:
$C' = \frac{K \varepsilon_0 A}{d'} = \frac{K \varepsilon_0 A}{d/2} = 2K \left( \frac{\varepsilon_0 A}{d} \right)$.
Substituting the given values:
$C' = 2 \times 6 \times C = 12 \times 8\,\mu F = 96\,\mu F$.

(A) The capacitance of a parallel plate capacitor with a dielectric medium is given by the formula: $C = \frac{\varepsilon_{0} \varepsilon_{r} A}{d}$.
Given values:
$C = 2 \times 10^{-4}\,\mu F = 2 \times 10^{-4} \times 10^{-6}\,F = 2 \times 10^{-10}\,F$
$A = 0.01\,m^2$
$d = 2\,mm = 2 \times 10^{-3}\,m$
$\varepsilon_{0} = 8.85 \times 10^{-12}\,F/m$
Rearranging the formula to solve for the dielectric constant $\varepsilon_{r}$:
$\varepsilon_{r} = \frac{C \cdot d}{\varepsilon_{0} \cdot A}$
Substituting the values:
$\varepsilon_{r} = \frac{(2 \times 10^{-10}) \times (2 \times 10^{-3})}{(8.85 \times 10^{-12}) \times (0.01)}$
$\varepsilon_{r} = \frac{4 \times 10^{-13}}{8.85 \times 10^{-14}}$
$\varepsilon_{r} = \frac{40}{8.85} \approx 4.52$.

(A) The initial capacitance of the parallel plate capacitor is given by $C = \frac{\epsilon_0 A}{d}$.
When half the space between the plates is filled with a dielectric of constant $K$ parallel to the plates,the capacitor can be viewed as two capacitors in series,each with plate separation $d/2$.
The capacitance of the part filled with the dielectric is $C_1 = \frac{K \epsilon_0 A}{d/2} = \frac{2K \epsilon_0 A}{d} = 2KC$.
The capacitance of the part filled with air is $C_2 = \frac{\epsilon_0 A}{d/2} = \frac{2 \epsilon_0 A}{d} = 2C$.
Since these two capacitors are in series,the equivalent capacitance $C_{eq}$ is given by:
$\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{1}{2KC} + \frac{1}{2C} = \frac{1}{2C} \left( \frac{1}{K} + 1 \right) = \frac{1}{2C} \left( \frac{1+K}{K} \right)$.
Therefore,$C_{eq} = \frac{2CK}{K+1}$.

(D) Let the initial capacitance be $C_0$ and the initial potential difference be $V_0$. When a dielectric slab of thickness $d$ (equal to the plate separation) is introduced,the new capacitance $C$ becomes $K C_0$,where $K$ is the dielectric constant.
Since the charge $Q$ on the capacitor remains constant $(Q = C_0 V_0 = C V)$,we have $V = Q/C = (C_0 V_0) / (K C_0) = V_0 / K$.
Given that the potential difference reduces to $1/8$ of the original value,we have $V = V_0 / 8$.
Comparing the two expressions,$V_0 / K = V_0 / 8$,which gives $K = 8$.

(B) Initial capacitance $C = \frac{\varepsilon_0 A}{d}$.
When the capacitor is half-filled with a dielectric of dielectric constant $K=5$ in series (as shown in the figure),the system acts as two capacitors in series,each with plate separation $d/2$.
Capacitance of the dielectric part: $C_1 = \frac{K \varepsilon_0 A}{d/2} = \frac{2 K \varepsilon_0 A}{d} = \frac{10 \varepsilon_0 A}{d}$.
Capacitance of the air part: $C_2 = \frac{\varepsilon_0 A}{d/2} = \frac{2 \varepsilon_0 A}{d}$.
The equivalent capacitance $C'$ is given by:
$\frac{1}{C'} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{d}{10 \varepsilon_0 A} + \frac{d}{2 \varepsilon_0 A} = \frac{d}{\varepsilon_0 A} \left( \frac{1}{10} + \frac{1}{2} \right) = \frac{d}{\varepsilon_0 A} \left( \frac{1+5}{10} \right) = \frac{6d}{10 \varepsilon_0 A} = \frac{3d}{5 \varepsilon_0 A}$.
Therefore,$C' = \frac{5}{3} \frac{\varepsilon_0 A}{d} = \frac{5}{3} C$.
Percentage increase in capacitance = $\frac{C' - C}{C} \times 100 = \left( \frac{5/3 C - C}{C} \right) \times 100 = \left( \frac{5}{3} - 1 \right) \times 100 = \frac{2}{3} \times 100 = 66.67\% \approx 66.6\%$.

(B) When a capacitor is charged and then disconnected from the battery,the charge $Q$ on the plates remains constant.
When a dielectric slab of dielectric constant $K$ is introduced between the plates,the capacitance increases to $C' = KC$.
Since $Q$ is constant,the new potential difference $V' = \frac{Q}{C'} = \frac{Q}{KC} = \frac{V}{K}$,which means the potential difference decreases.
The stored energy $U = \frac{Q^2}{2C}$ becomes $U' = \frac{Q^2}{2C'} = \frac{Q^2}{2KC} = \frac{U}{K}$,which means the stored energy decreases.
Therefore,the potential difference and stored energy decrease,while the charge remains unchanged.

(C) The capacitance of a parallel plate capacitor is given by $C = \frac{K \epsilon_0 A}{d}$,where $K$ is the dielectric constant,$A$ is the area,and $d$ is the distance between the plates.
Given that the new distance $d' = \frac{d}{2}$ and the new dielectric constant $K' = 3K$.
The new capacitance $C'$ becomes $C' = \frac{(3K) \epsilon_0 A}{(d/2)} = 6 \left( \frac{K \epsilon_0 A}{d} \right) = 6C$.
Thus,the Assertion is correct.
The capacity of a capacitor depends on the dielectric constant $(K)$ of the material placed between the plates. Therefore,the Reason is incorrect.

(C) When a parallel plate capacitor is connected to a battery,the potential difference $V$ across the plates remains constant $(V = V_0)$.
The energy stored in the capacitor is given by $U = \frac{1}{2} CV^2$.
When a dielectric slab of dielectric constant $K$ is introduced between the plates,the capacitance becomes $C' = KC$.
Therefore,the new energy stored is $U' = \frac{1}{2} (KC) V^2 = K U$. Thus,the energy stored becomes $K$ times.
Since $Q = CV$,the charge on the plates becomes $Q' = KCV = KQ$.
The surface charge density is given by $\sigma = \frac{Q}{A}$. Since $Q$ increases by a factor of $K$,the surface charge density $\sigma' = \frac{KQ}{A} = K\sigma_0$ also increases by a factor of $K$.
Therefore,the Reason is incorrect because the surface charge density does not remain constant.

(C) For an isolated capacitor,the charge $Q$ remains constant.
The force $F$ between the plates is given by $F = \frac{Q^2}{2A\epsilon_0 K}$. Since $K > 1$ for a dielectric,the force $F$ decreases.
Thus,the Assertion is correct.
The electric field $E$ between the plates is given by $E = \frac{\sigma}{K\epsilon_0} = \frac{E_0}{K}$. Since $K > 1$,the electric field $E$ decreases.
Therefore,the Reason is incorrect.
Hence,the correct option is $C$.

(A) Since the dielectric constant $k(x)$ varies with distance $x$,we consider a thin elemental slice of thickness $dx$ at a distance $x$ from one plate.
The capacitance of this elemental slice is $dC = \frac{\varepsilon_0 k(x) A}{dx} = \frac{\varepsilon_0 K(1+\alpha x) A}{dx}$.
Since these elemental capacitors are connected in series,the equivalent capacitance $C$ is given by:
$\frac{1}{C} = \int_0^d \frac{1}{dC} = \int_0^d \frac{dx}{\varepsilon_0 K A (1+\alpha x)}$.
Integrating this,we get:
$\frac{1}{C} = \frac{1}{\varepsilon_0 K A} \left[ \frac{\ln(1+\alpha x)}{\alpha} \right]_0^d = \frac{1}{\alpha \varepsilon_0 K A} \ln(1+\alpha d)$.
Using the Taylor expansion $\ln(1+u) \approx u - \frac{u^2}{2}$ for $u = \alpha d << 1$:
$\frac{1}{C} \approx \frac{1}{\alpha \varepsilon_0 K A} (\alpha d - \frac{(\alpha d)^2}{2}) = \frac{d}{\varepsilon_0 K A} (1 - \frac{\alpha d}{2})$.
Taking the reciprocal to find $C$ and using the binomial approximation $(1-u)^{-1} \approx 1+u$:
$C \approx \frac{\varepsilon_0 K A}{d} (1 - \frac{\alpha d}{2})^{-1} \approx \frac{\varepsilon_0 K A}{d} (1 + \frac{\alpha d}{2})$.

(N/A) Let $E_{0} = V_{0} / d$ be the electric field between the plates when there is no dielectric and the potential difference is $V_{0}$.
If the dielectric is now inserted,the electric field in the dielectric will be $E = E_{0} / K$.
The potential difference $V$ will then be:
$V = E_{0} \left(\frac{1}{4} d\right) + \frac{E_{0}}{K} \left(\frac{3}{4} d\right)$
$V = E_{0} d \left(\frac{1}{4} + \frac{3}{4K}\right) = V_{0} \left(\frac{K+3}{4K}\right)$
The potential difference decreases by the factor $(K+3) / 4K$ while the free charge $Q_{0}$ on the plates remains unchanged.
The capacitance $C$ thus increases as:
$C = \frac{Q_{0}}{V} = \frac{Q_{0}}{V_{0} \left(\frac{K+3}{4K}\right)} = \left(\frac{4K}{K+3}\right) C_{0}$

(C) The initial capacitance of the parallel plate capacitor with air as the medium is given by $C = \frac{\varepsilon_0 A}{d} = 8 \;pF$.
When the distance between the plates is reduced to $d' = \frac{d}{2}$ and the space is filled with a dielectric of constant $k = 6$,the new capacitance $C'$ is given by:
$C' = \frac{k \varepsilon_0 A}{d'} = \frac{6 \varepsilon_0 A}{d/2} = 12 \left( \frac{\varepsilon_0 A}{d} \right)$.
Substituting the initial value $C = 8 \;pF$:
$C' = 12 \times 8 \;pF = 96 \;pF$.
Thus,the new capacitance is $96 \;pF$.

(A) Dielectric constant of the mica sheet,$k = 6$.
If the voltage supply remains connected,the potential difference between the plates remains constant. Supply voltage,$V = 100 \, V$.
Initial capacitance,$C = 17.71 \, pF = 1.771 \times 10^{-11} \, F$.
New capacitance,$C_{1} = k \cdot C = 6 \times 17.71 \, pF = 106.26 \, pF$.
New charge,$q_{1} = C_{1} \cdot V = 106.26 \times 10^{-12} \, F \times 100 \, V = 1.0626 \times 10^{-8} \, C$.
The potential across the plates remains $100 \, V$.
$(b)$ Dielectric constant,$k = 6$.
Initial charge,$q = C \cdot V = 17.71 \times 10^{-12} \, F \times 100 \, V = 1.771 \times 10^{-9} \, C$.
New capacitance,$C_{1} = k \cdot C = 6 \times 17.71 \, pF = 106.26 \, pF$.
If the supply is disconnected,the charge remains constant,$q = 1.771 \times 10^{-9} \, C$.
The new potential across the plates is given by $V_{1} = \frac{q}{C_{1}} = \frac{1.771 \times 10^{-9} \, C}{106.26 \times 10^{-12} \, F} \approx 16.67 \, V$.

(N/A) Consider two large plates each of area $A$ separated by a distance $d$. The charge on the plates is $\pm Q$ corresponding to the surface charge density $\pm \sigma$.
When there is a vacuum between the plates:
$E_{0} = \frac{\sigma}{\epsilon_{0}}$
The potential difference $V_{0}$ is given by:
$V_{0} = E_{0} d$
If $C_{0}$ is the capacitance:
$C_{0} = \frac{Q}{V_{0}} = \frac{\sigma A}{E_{0} d} = \frac{\epsilon_{0} A}{d} \quad \dots(1)$
When a dielectric is inserted between the plates,it becomes polarized by the external field,creating induced surface charge densities $\pm \sigma_{P}$.
The net electric field between the plates becomes:
$E = E_{0} - E_{P} = \frac{\sigma - \sigma_{P}}{\epsilon_{0}}$
The new potential difference is:
$V = E d = \frac{(\sigma - \sigma_{P}) d}{\epsilon_{0}}$
For a linear dielectric,we define the dielectric constant $K$ such that the net field is reduced by a factor $K$:
$E = \frac{E_{0}}{K} \implies \sigma - \sigma_{P} = \frac{\sigma}{K}$
Substituting this into the potential equation:
$V = \frac{\sigma d}{\epsilon_{0} K} = \frac{Q d}{A \epsilon_{0} K}$
The new capacitance $C$ is:
$C = \frac{Q}{V} = \frac{A \epsilon_{0} K}{d} = K C_{0}$
Thus,the dielectric constant $K$ is defined as $K = \frac{C}{C_{0}}$.

(N/A) The capacitance of a parallel plate capacitor in a vacuum is given by $C_0 = \frac{\epsilon_0 A}{d}$,where $\epsilon_0$ is the permittivity of free space,$A$ is the area of the plates,and $d$ is the distance between them.
When a dielectric material with dielectric constant $K$ is introduced between the plates,the new capacitance $C$ becomes $C = K C_0$.
Given that $K = 2$,the formula for the capacitance becomes $C = 2 C_0 = \frac{2 \epsilon_0 A}{d}$.

(A) $1$. Capacitance $(C)$: The capacitance of a capacitor with a dielectric constant $K$ is given by $C = \frac{K \epsilon_0 A}{d}$. Since $K > 1$,removing the dielectric slab (replacing it with air,$K=1$) causes the capacitance $C$ to decrease.
$2$. Charge $(q)$: Since the battery is disconnected before the dielectric is removed,the capacitor is isolated. Therefore,the charge $q$ stored on the plates remains constant.
$3$. Energy stored $(U)$: The energy stored is given by $U = \frac{q^2}{2C}$. Since $q$ is constant and $C$ decreases,$U$ must increase.
$4$. Voltage $(V)$: From the relation $q = CV$,we have $V = \frac{q}{C}$. Since $q$ is constant and $C$ decreases,the voltage $V$ across the plates must increase.
$5$. Electric field $(E)$: The electric field between the plates is given by $E = \frac{V}{d}$. Since $V$ increases and $d$ remains constant,the electric field $E$ must increase.

(C) Let $C_0$ be the capacitance of the capacitor without the dielectric.
The charge on the first capacitor is $Q_0 = C_0 U_0 = 78 C_0$.
When connected in parallel,the final voltage $U$ across both capacitors is the same.
The capacitance of the second capacitor with dielectric is $C = \varepsilon_r C_0 = \varepsilon C_0 = (\alpha U) C_0$.
The total charge is conserved: $Q_0 = C_0 U + C U = C_0 U + (\alpha U) C_0 U = C_0 U (1 + \alpha U)$.
Substituting the values: $78 C_0 = C_0 U (1 + 2U)$.
$78 = U + 2U^2 \implies 2U^2 + U - 78 = 0$.
Solving the quadratic equation using the formula $U = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$U = \frac{-1 \pm \sqrt{1^2 - 4(2)(-78)}}{2(2)} = \frac{-1 \pm \sqrt{1 + 624}}{4} = \frac{-1 \pm \sqrt{625}}{4} = \frac{-1 \pm 25}{4}$.
Since voltage must be positive,$U = \frac{24}{4} = 6 \ V$ is not among options,let's recheck the calculation.
Wait,$78 = U + 2U^2$. If $U = 6$,$6 + 2(36) = 78$. The calculation is correct. Given the options,let's re-evaluate the question parameters. If $U_0 = 78$,$U=6$. If the options are $39, 26, 13, 52$,perhaps $\alpha$ is different or $U_0$ is different. Assuming the logic holds,the result is $6 \ V$.

(C) Initial capacitance $C_i = \frac{\varepsilon_0 A}{d} = \frac{\varepsilon_0 (lw)}{d}$.
Initial energy $U_i = \frac{1}{2} C_i V^2$.
After inserting the slab of length $x$,the capacitor acts as two capacitors in parallel: one with dielectric (length $x$) and one with air (length $l-x$).
$C_f = C_1 + C_2 = \frac{k \varepsilon_0 (xw)}{d} + \frac{\varepsilon_0 ((l-x)w)}{d}$.
Given $U_f = 2 U_i$,and since $V$ is constant,$C_f = 2 C_i$.
$\frac{\varepsilon_0 w}{d} [kx + l - x] = 2 \frac{\varepsilon_0 lw}{d}$.
$kx + l - x = 2l$.
Substitute $k = 4$: $4x + l - x = 2l$.
$3x = l \Rightarrow x = \frac{l}{3}$.

(D) The capacitance of a parallel plate capacitor with air is given by $C_{0} = 6\, \mu F$.
When a dielectric medium is introduced,the new capacitance is $C_{m} = 30\, \mu F$.
The dielectric constant $\epsilon_{r}$ is defined as the ratio of the capacitance with the medium to the capacitance with air:
$\epsilon_{r} = \frac{C_{m}}{C_{0}} = \frac{30}{6} = 5$.
The permittivity of the medium $\epsilon$ is given by $\epsilon = \epsilon_{0} \cdot \epsilon_{r}$.
Substituting the given values:
$\epsilon = 8.85 \times 10^{-12} \times 5 = 44.25 \times 10^{-12} = 0.4425 \times 10^{-10} \approx 0.44 \times 10^{-10} C^{2} N^{-1} m^{-2}$.

(C) The original capacitance of the parallel plate capacitor with air is given by $C_{0} = \frac{\varepsilon_{0} A}{d}$.
When a dielectric slab of thickness $t$ and dielectric constant $K$ is inserted between the plates,the new capacitance $C$ is given by the formula:
$C = \frac{\varepsilon_{0} A}{d - t + \frac{t}{K}}$
Given $t = \frac{d}{2}$ and $K = 4$,we substitute these values into the formula:
$C = \frac{\varepsilon_{0} A}{d - \frac{d}{2} + \frac{d/2}{4}}$
$C = \frac{\varepsilon_{0} A}{\frac{d}{2} + \frac{d}{8}}$
$C = \frac{\varepsilon_{0} A}{\frac{4d + d}{8}} = \frac{\varepsilon_{0} A}{\frac{5d}{8}}$
$C = \frac{8}{5} \frac{\varepsilon_{0} A}{d} = \frac{8}{5} C_{0}$
Therefore,the ratio of the new capacitance to the original capacitance is $\frac{C}{C_{0}} = \frac{8}{5}$.

(B) Let the separation between the plates be $d$. The electric field $E$ is given by $E = \frac{V}{d}$.
Given $E = 20 \, MV/m = 20 \times 10^{6} \, V/m$ and $V = 20 \, V$.
Substituting the values: $20 \times 10^{6} = \frac{20}{d}$,which gives $d = 10^{-6} \, m$.
The capacitance $C$ is given by $C = \frac{\varepsilon_{0} A \varepsilon_{r}}{d}$.
Substituting the values: $9 \times 10^{-9} = \frac{(8.85 \times 10^{-12}) \times A \times 2.4}{10^{-6}}$.
Rearranging for $A$: $A = \frac{9 \times 10^{-9} \times 10^{-6}}{8.85 \times 10^{-12} \times 2.4}$.
$A = \frac{9 \times 10^{-15}}{21.24 \times 10^{-12}} \approx 0.4237 \times 10^{-3} \, m^{2} = 4.237 \times 10^{-4} \, m^{2}$.
Rounding to the nearest option,$A = 4.2 \times 10^{-4} \, m^{2}$.

(A) The expression for capacitance is given by:
$C = \frac{A \varepsilon_{0} \varepsilon_{r}}{d} \quad ......(I)$
The electric field $E$ is related to potential difference $V$ and distance $d$ by:
$E = \frac{V}{d} \implies d = \frac{V}{E}$
Given $V = 30 \, V$ and $E = 30 \times 10^{6} \, V/m$:
$d = \frac{30}{30 \times 10^{6}} = 10^{-6} \, m$
Now,substitute $C = 15 \times 10^{-9} \, F$,$\varepsilon_{0} = 8.85 \times 10^{-12} \, F/m$,$\varepsilon_{r} = 2.5$,and $d = 10^{-6} \, m$ into equation $(I)$:
$15 \times 10^{-9} = \frac{A \times (8.85 \times 10^{-12}) \times 2.5}{10^{-6}}$
$A = \frac{15 \times 10^{-9} \times 10^{-6}}{8.85 \times 10^{-12} \times 2.5}$
$A = \frac{15 \times 10^{-15}}{22.125 \times 10^{-12}}$
$A \approx 0.678 \times 10^{-3} \, m^{2} = 6.78 \times 10^{-4} \, m^{2}$
Rounding to the nearest option,$A = 6.7 \times 10^{-4} \, m^{2}$.

(B) The initial capacitance of the capacitor is $C_{0} = \frac{\varepsilon_{0} A}{d}$.
The initial charge on the capacitor is $q_{0} = C_{0} V = \frac{\varepsilon_{0} A}{d} V$.
When a dielectric slab of dielectric constant $k$ is inserted between the plates while the battery remains connected,the potential difference $V$ across the plates remains constant.
The new capacitance becomes $C = k C_{0} = \frac{k \varepsilon_{0} A}{d}$.
The final charge on the capacitor is $q = C V = \left( \frac{k \varepsilon_{0} A}{d} \right) V = k q_{0}$.

(C) The original capacitance of the parallel plate capacitor is $C_{0} = \frac{\epsilon_{0} A}{d}$.
When a dielectric slab of thickness $t = \frac{3}{4}d$ is inserted,the system can be modeled as two capacitors $C_{1}$ and $C_{2}$ in series.
$C_{1}$ is the capacitance of the part filled with dielectric: $C_{1} = \frac{K \epsilon_{0} A}{3d/4} = \frac{4 K \epsilon_{0} A}{3d}$.
$C_{2}$ is the capacitance of the air gap: $C_{2} = \frac{\epsilon_{0} A}{d - 3d/4} = \frac{\epsilon_{0} A}{d/4} = \frac{4 \epsilon_{0} A}{d}$.
Since they are in series,the equivalent capacitance $C'$ is given by $\frac{1}{C'} = \frac{1}{C_{1}} + \frac{1}{C_{2}}$.
Substituting the values:
$\frac{1}{C'} = \frac{3d}{4 K \epsilon_{0} A} + \frac{d}{4 \epsilon_{0} A}$.
$\frac{1}{C'} = \frac{d}{4 \epsilon_{0} A} \left( \frac{3}{K} + 1 \right) = \frac{d}{4 \epsilon_{0} A} \left( \frac{3+K}{K} \right)$.
Therefore,$C' = \frac{4 K \epsilon_{0} A}{d(3+K)} = \frac{4K}{3+K} C_{0}$.

(D) Initial energy stored in the capacitor is given by $U_i = \frac{1}{2} C V^2$.
Substituting the values: $U_i = \frac{1}{2} \times 14 \, pF \times (12 \, V)^2 = \frac{1}{2} \times 14 \times 144 = 1008 \, pJ$.
When the battery is disconnected,the charge $Q$ on the plates remains constant.
After inserting the dielectric slab of constant $k = 7$,the new capacitance becomes $C' = kC = 7 \times 14 \, pF = 98 \, pF$.
The new energy stored in the capacitor is $U_f = \frac{Q^2}{2C'} = \frac{Q^2}{2(kC)} = \frac{U_i}{k}$.
$U_f = \frac{1008 \, pJ}{7} = 144 \, pJ$.
The loss in electrical potential energy is converted into the mechanical energy of the slab.
Mechanical energy $= U_i - U_f = 1008 \, pJ - 144 \, pJ = 864 \, pJ$.

(B) The capacitor can be modeled as two capacitors in series: one with a dielectric $(C_{1})$ and one with air $(C_{2})$.
Given: $A = 100\, m^{2}$,$d = 10\, m$,$t = 5\, m$,$K = 10$,$\varepsilon_{0} = 8.85 \times 10^{-12} F \cdot m^{-1}$.
The thickness of the dielectric is $t = 5\, m$ and the thickness of the air gap is $d - t = 10 - 5 = 5\, m$.
$C_{1} = \frac{K \varepsilon_{0} A}{t} = \frac{10 \times \varepsilon_{0} \times 100}{5} = 200 \varepsilon_{0}$.
$C_{2} = \frac{\varepsilon_{0} A}{d - t} = \frac{\varepsilon_{0} \times 100}{5} = 20 \varepsilon_{0}$.
Since they are in series,the equivalent capacitance $C_{eq}$ is given by:
$C_{eq} = \frac{C_{1} C_{2}}{C_{1} + C_{2}} = \frac{(200 \varepsilon_{0})(20 \varepsilon_{0})}{200 \varepsilon_{0} + 20 \varepsilon_{0}} = \frac{4000 \varepsilon_{0}^{2}}{220 \varepsilon_{0}} = \frac{400}{22} \varepsilon_{0} = \frac{200}{11} \varepsilon_{0}$.
Substituting $\varepsilon_{0} = 8.85 \times 10^{-12} F \cdot m^{-1}$:
$C_{eq} = \frac{200}{11} \times 8.85 \times 10^{-12} F = 18.1818 \times 8.85 \times 10^{-12} F \approx 160.909 \times 10^{-12} F$.
Since $1\, pF = 10^{-12} F$,$C_{eq} \approx 160.909\, pF$.
Rounding to the nearest integer,$x = 161$.

(C) The capacitance $C$ of a parallel plate capacitor with a dielectric slab of thickness $t$ and dielectric constant $K$ is given by the formula:
$C = \frac{\varepsilon_{0} A}{d - t + \frac{t}{K}}$
Given:
Area $A = 2 \, m^{2}$
Total separation $d = 1 \, m$
Thickness of dielectric $t = 0.5 \, m$
Dielectric constant $K = 3.2$
Substituting the values:
$C = \frac{\varepsilon_{0} \times 2}{1 - 0.5 + \frac{0.5}{3.2}}$
$C = \frac{2 \varepsilon_{0}}{0.5 + 0.15625}$
$C = \frac{2 \varepsilon_{0}}{0.65625}$
$C \approx 3.047 \, \varepsilon_{0}$
Rounding off to the nearest integer,we get $3 \, \varepsilon_{0}$.