Effect of Dielectric Inside Capacitor Questions in English

(B) When a dielectric is inserted into a capacitor while it remains connected to a battery,the potential difference $V$ remains constant. However,if the capacitor is disconnected from the battery,the charge $Q$ remains constant.
Assuming the capacitor is disconnected from the battery:
$1$. The new capacitance becomes $C' = KC$,where $K > 1$.
$2$. Since the charge $Q$ is constant,the new potential $V' = Q/C' = Q/(KC) = V/K$. Since $K > 1$,$V' < V$,so the potential decreases.
$3$. The new energy $E' = Q^2 / (2C') = Q^2 / (2KC) = E/K$. Since $K > 1$,$E' < E$,so the energy decreases.
Therefore,both $V$ and $E$ decrease.

(A) The initial capacitance of the parallel plate capacitor with air as the medium is $C_0 = \frac{\epsilon_0 A}{d}$.
When a dielectric slab of thickness $t$ and dielectric constant $K$ is introduced,the new capacitance $C'$ is given by $C' = \frac{\epsilon_0 A}{d - t + \frac{t}{K}}$.
Given that the capacity increases by $50 \%$,the new capacitance is $C' = C_0 + 0.5 C_0 = 1.5 C_0 = \frac{3}{2} C_0$.
Substituting the expressions,we get $\frac{\epsilon_0 A}{d - t + \frac{t}{K}} = \frac{3}{2} \frac{\epsilon_0 A}{d}$.
This simplifies to $\frac{1}{d - t + \frac{t}{3}} = \frac{3}{2d}$.
Cross-multiplying gives $2d = 3(d - t + \frac{t}{3}) = 3(d - \frac{2t}{3}) = 3d - 2t$.
Rearranging for $t$,we get $2t = 3d - 2d = d$,which implies $t = \frac{d}{2}$.

(D) Initially,both capacitors have capacity $C$ and are connected in parallel. The initial equivalent capacity is $C_{eq,i} = C + C = 2C$.
After one capacitor is filled with a dielectric of constant $K$,its new capacity becomes $C' = KC$. The other capacitor remains $C$.
The new equivalent capacity of the parallel combination is $C_{eq,f} = KC + C = C(K+1)$.
The change in the effective capacity is $\Delta C = C_{eq,f} - C_{eq,i} = C(K+1) - 2C = CK + C - 2C = C(K-1)$.

(C) Let the area of the plates be $A$ and the distance between them be $d$. The capacitance of the air capacitor is $C_p = \frac{\epsilon_0 A}{d}$.
When the space is filled with two parallel layers of dielectric materials of thickness $d/2$ each,the capacitor acts as two capacitors $C_1$ and $C_2$ in series.
$C_1 = \frac{K_1 \epsilon_0 A}{d/2} = \frac{2 K_1 \epsilon_0 A}{d} = 2 K_1 C_p$.
$C_2 = \frac{K_2 \epsilon_0 A}{d/2} = \frac{2 K_2 \epsilon_0 A}{d} = 2 K_2 C_p$.
Since they are in series,the equivalent capacitance $C_K$ is given by $\frac{1}{C_K} = \frac{1}{C_1} + \frac{1}{C_2}$.
$\frac{1}{C_K} = \frac{1}{2 K_1 C_p} + \frac{1}{2 K_2 C_p} = \frac{1}{2 C_p} \left( \frac{1}{K_1} + \frac{1}{K_2} \right) = \frac{1}{2 C_p} \left( \frac{K_1 + K_2}{K_1 K_2} \right)$.
Therefore,$C_K = \frac{2 C_p K_1 K_2}{K_1 + K_2}$.
The ratio $\frac{C_p}{C_K} = \frac{C_p}{\frac{2 C_p K_1 K_2}{K_1 + K_2}} = \frac{K_1 + K_2}{2 K_1 K_2}$.

(B) The capacitance $C$ of a parallel plate capacitor with a dielectric slab of thickness $t$ and dielectric constant $K$ is given by the formula: $C = \frac{\epsilon_0 A}{d - t + \frac{t}{K}}$.
Given:
Area $A = 50 \ cm^2 = 50 \times 10^{-4} \ m^2 = 5 \times 10^{-3} \ m^2$.
Separation $d = 3 \ mm = 3 \times 10^{-3} \ m$.
Thickness of dielectric $t = 1 \ mm = 1 \times 10^{-3} \ m$.
Dielectric constant $K = 4$.
Substituting the values:
$C = \frac{\epsilon_0 (5 \times 10^{-3})}{3 \times 10^{-3} - 1 \times 10^{-3} + \frac{1 \times 10^{-3}}{4}}$
$C = \frac{\epsilon_0 (5 \times 10^{-3})}{2 \times 10^{-3} + 0.25 \times 10^{-3}}$
$C = \frac{\epsilon_0 (5 \times 10^{-3})}{2.25 \times 10^{-3}}$
$C = \frac{5 \epsilon_0}{2.25} = \frac{5 \epsilon_0}{9/4} = \frac{20 \epsilon_0}{9}$.

(C) The capacitance of a parallel plate capacitor with air is given by $C = \frac{\epsilon_0 A}{d}$.
When a conducting sheet of thickness $t = \frac{2d}{3}$ is inserted between the plates,the effective distance between the plates decreases.
The new capacitance $C_1$ is given by the formula $C_1 = \frac{\epsilon_0 A}{d - t}$.
Substituting the value of $t = \frac{2d}{3}$ into the formula:
$C_1 = \frac{\epsilon_0 A}{d - \frac{2d}{3}} = \frac{\epsilon_0 A}{\frac{d}{3}} = 3 \left( \frac{\epsilon_0 A}{d} \right)$.
Since $C = \frac{\epsilon_0 A}{d}$,we have $C_1 = 3C$.
Therefore,the ratio $\frac{C_1}{C} = 3:1$.

(A) The initial capacity of the air-filled parallel plate capacitor is $C_0 = \frac{\epsilon_0 A}{d}$,where $A$ is the area of the plates and $d$ is the distance between them.
When the space is filled as shown in the figure,the capacitor can be viewed as two capacitors connected in parallel.
One part has an area $A/2$ filled with air,and the other part has an area $A/2$ filled with a dielectric of constant $K$.
The capacity of the air-filled part is $C_1 = \frac{\epsilon_0 (A/2)}{d} = \frac{C_0}{2}$.
The capacity of the dielectric-filled part is $C_2 = \frac{K \epsilon_0 (A/2)}{d} = \frac{K C_0}{2}$.
Since they are in parallel,the new capacity $C_n = C_1 + C_2 = \frac{C_0}{2} + \frac{K C_0}{2} = C_0 \left(\frac{K+1}{2}\right)$.
Therefore,the ratio $\frac{C_n}{C_0} = \frac{K+1}{2}$.

(B) The initial capacitance of the air capacitor is $C_0 = \frac{\epsilon_0 A}{d} = 1 \mu F$.
From the figure,the two dielectrics are placed in parallel,each occupying half the area of the plates $(A_1 = A_2 = A/2)$ and the full distance $d$ between the plates.
The capacitance of the two parts are:
$C_1 = \frac{K_1 \epsilon_0 (A/2)}{d} = K_1 \frac{C_0}{2} = 8 \times \frac{1}{2} = 4 \mu F$
$C_2 = \frac{K_2 \epsilon_0 (A/2)}{d} = K_2 \frac{C_0}{2} = 4 \times \frac{1}{2} = 2 \mu F$
Since they are in parallel,the equivalent capacitance is $C_{eq} = C_1 + C_2 = 4 \mu F + 2 \mu F = 6 \mu F$.

(C) The system can be viewed as two capacitors in series: one with the dielectric and one with the air gap.
$C_1 = \frac{K \varepsilon_0 A}{t} = \frac{5 \varepsilon_0 \times 40 \times 10^{-4}}{1 \times 10^{-3}} = 20 \varepsilon_0 \ F$
$C_2 = \frac{\varepsilon_0 A}{d-t} = \frac{\varepsilon_0 \times 40 \times 10^{-4}}{1 \times 10^{-3}} = 4 \varepsilon_0 \ F$
Since they are in series,the equivalent capacitance $C_{eq}$ is given by:
$\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{1}{20 \varepsilon_0} + \frac{1}{4 \varepsilon_0}$
$\frac{1}{C_{eq}} = \frac{1 + 5}{20 \varepsilon_0} = \frac{6}{20 \varepsilon_0} = \frac{3}{10 \varepsilon_0}$
$C_{eq} = \frac{10}{3} \varepsilon_0 \ F$

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

(A) When a dielectric material is inserted between the plates of a capacitor,it gets polarized. This polarization creates an internal electric field that opposes the external electric field produced by the charges on the plates. As a result,the net electric field $E$ between the plates decreases. Since the potential difference $V$ is related to the electric field by $V = E \cdot d$ (where $d$ is the distance between the plates),a decrease in the electric field leads to a decrease in the potential difference between the plates for a given charge $Q$. Consequently,the capacitance $C = Q/V$ increases.

(A) The capacitance of a parallel plate air capacitor is given by $C = \frac{\varepsilon_0 A}{d} = 1 \mu F$.
From the figure,the area of each plate is divided into two halves,$A/2$ and $A/2$,while the distance $d$ remains the same for both. This configuration represents two capacitors in parallel.
For the first dielectric $(K_1 = 4)$:
$C_1 = \frac{K_1 \varepsilon_0 (A/2)}{d} = K_1 \times \frac{1}{2} \times C = 4 \times 0.5 \times 1 \mu F = 2 \mu F$.
For the second dielectric $(K_2 = 2)$:
$C_2 = \frac{K_2 \varepsilon_0 (A/2)}{d} = K_2 \times \frac{1}{2} \times C = 2 \times 0.5 \times 1 \mu F = 1 \mu F$.
Since the capacitors are in parallel,the effective capacitance is:
$C_{\text{eff}} = C_1 + C_2 = 2 \mu F + 1 \mu F = 3 \mu F$.

(C) Initially,two capacitors of capacity $C$ are in series. The equivalent capacity $C_1$ is given by:
$\frac{1}{C_1} = \frac{1}{C} + \frac{1}{C} = \frac{2}{C} \implies C_1 = \frac{C}{2}$
After filling one capacitor with a dielectric of constant $K$,its new capacity becomes $KC$. The new equivalent capacity $C_2$ is:
$\frac{1}{C_2} = \frac{1}{C} + \frac{1}{KC} = \frac{1}{C} \left(1 + \frac{1}{K}\right) = \frac{1}{C} \left(\frac{K+1}{K}\right) \implies C_2 = \frac{CK}{K+1}$
The change in effective capacity $\Delta C$ is:
$\Delta C = C_2 - C_1 = \frac{CK}{K+1} - \frac{C}{2}$
$\Delta C = C \left[ \frac{2K - (K+1)}{2(K+1)} \right] = \frac{C}{2} \left[ \frac{K-1}{K+1} \right]$

(D) Initially,the capacitors are in series with capacitance $C_A = C$ and $C_B = KC$. The equivalent capacitance is $C_{net} = \frac{C_A C_B}{C_A + C_B} = \frac{C \cdot KC}{C + KC} = \frac{KC}{K+1}$.
The charge on each capacitor in series is $Q_A = Q_B = C_{net}E = \frac{KCE}{K+1}$.
After removing the dielectric,$C_A = C$ and $C_B = C$. The new equivalent capacitance is $C_{net}^{\prime} = \frac{C \cdot C}{C + C} = \frac{C}{2}$.
The new charge on each capacitor is $Q_A^{\prime} = Q_B^{\prime} = C_{net}^{\prime}E = \frac{CE}{2}$.
Calculating the ratio: $\frac{Q_B^{\prime}}{Q_B} = \frac{CE/2}{KCE/(K+1)} = \frac{K+1}{2K}$.

(D) For an air capacitor,$C_1 = \frac{A \varepsilon_0}{d}$.
When two dielectrics are inserted as shown,the arrangement acts as two capacitors in series,each with plate separation $d/2$ and area $A$.
The capacitance of the first part is $C_{a} = \frac{K_1 A \varepsilon_0}{d/2} = \frac{2 K_1 A \varepsilon_0}{d} = 2 K_1 C_1$.
The capacitance of the second part is $C_{b} = \frac{K_2 A \varepsilon_0}{d/2} = \frac{2 K_2 A \varepsilon_0}{d} = 2 K_2 C_1$.
Since they are in series,the equivalent capacitance $C_2$ is given by:
$\frac{1}{C_2} = \frac{1}{C_a} + \frac{1}{C_b} = \frac{1}{2 K_1 C_1} + \frac{1}{2 K_2 C_1} = \frac{1}{2 C_1} \left( \frac{1}{K_1} + \frac{1}{K_2} \right) = \frac{1}{2 C_1} \left( \frac{K_1 + K_2}{K_1 K_2} \right)$.
Therefore,$C_2 = \frac{2 C_1 K_1 K_2}{K_1 + K_2}$.
The ratio $\frac{C_1}{C_2} = \frac{C_1}{\frac{2 C_1 K_1 K_2}{K_1 + K_2}} = \frac{K_1 + K_2}{2 K_1 K_2}$.

(A) The potential energy of a charged capacitor is given by $U_0 = \frac{Q^2}{2C}$,where $Q$ is the charge on the plates and $C$ is the initial capacitance.
When a dielectric slab of constant $K$ is inserted between the plates,the new capacitance becomes $C' = KC$.
Assuming the capacitor is isolated (charge $Q$ remains constant),the new potential energy $U'$ is given by:
$U' = \frac{Q^2}{2C'} = \frac{Q^2}{2(KC)}$
$U' = \frac{1}{K} \left( \frac{Q^2}{2C} \right)$
$U' = \frac{U_0}{K}$

(D) The capacitor can be modeled as two capacitors in series,each with plate area $A$ and dielectric thicknesses $d_1 = \frac{d}{4}$ and $d_2 = \frac{3d}{4}$.
Capacity of the $1^{\text{st}}$ capacitor:
$C_1 = \frac{K_1 \varepsilon_0 A}{d/4} = \frac{4 K_1 \varepsilon_0 A}{d}$
Capacity of the $2^{\text{nd}}$ capacitor:
$C_2 = \frac{K_2 \varepsilon_0 A}{3d/4} = \frac{4 K_2 \varepsilon_0 A}{3d}$
Since they are in series,the equivalent capacitance $C$ is given by:
$\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2}$
$\frac{1}{C} = \frac{d}{4 K_1 \varepsilon_0 A} + \frac{3d}{4 K_2 \varepsilon_0 A}$
$\frac{1}{C} = \frac{d}{4 \varepsilon_0 A} \left[\frac{1}{K_1} + \frac{3}{K_2}\right]$
$\frac{1}{C} = \frac{d}{4 \varepsilon_0 A} \left[\frac{K_2 + 3 K_1}{K_1 K_2}\right]$
Therefore,$C = \frac{4 \varepsilon_0 A}{d} \left[\frac{K_1 K_2}{3 K_1 + K_2}\right]$.

(A) When a battery remains connected to a parallel plate capacitor,the potential difference $V$ between the plates remains constant.
The electric field $E$ between the plates is given by the formula $E = V/d$,where $d$ is the distance between the plates.
Since the battery is connected,$V$ is constant. When the dielectric slab is inserted,the capacitance increases,but the potential difference $V$ across the plates remains equal to the battery voltage.
When the dielectric slab is taken out,the system returns to its original state because the battery maintains the potential difference $V$ at all times.
Therefore,the electric field $E$ remains the same as it was before the insertion of the slab.

(C) The initial capacitance of the air-filled parallel plate capacitor is given by $C_1 = \frac{A \varepsilon_0}{d}$.
When the space is filled with a dielectric of constant $K = 8$ and the distance is changed to $d'$,the new capacitance is $C_2 = \frac{K A \varepsilon_0}{d'} = \frac{8 A \varepsilon_0}{d'}$.
Given that the capacity increases $16$ times,we have $C_2 = 16 C_1$.
Substituting the expressions,we get $\frac{8 A \varepsilon_0}{d'} = 16 \left( \frac{A \varepsilon_0}{d} \right)$.
Simplifying the equation: $\frac{8}{d'} = \frac{16}{d}$.
Solving for $d'$,we get $d' = \frac{8d}{16} = \frac{d}{2}$.

(C) The initial capacitance of the air-filled parallel plate capacitor is given by $C = \frac{\varepsilon_0 A}{d} = 10 \mu F$.
When the area is divided into two equal halves and filled with dielectrics $K_1$ and $K_2$,the two parts act as two capacitors connected in parallel.
The area of each part is $A' = \frac{A}{2}$ and the distance between the plates remains $d$.
The capacitance of the first part is $C_1 = \frac{K_1 \varepsilon_0 A'}{d} = \frac{K_1 \varepsilon_0 A}{2d} = \frac{K_1}{2} C$.
The capacitance of the second part is $C_2 = \frac{K_2 \varepsilon_0 A'}{d} = \frac{K_2 \varepsilon_0 A}{2d} = \frac{K_2}{2} C$.
Since they are in parallel,the equivalent capacitance is $C_{\text{eq}} = C_1 + C_2 = \frac{C}{2} (K_1 + K_2)$.
Substituting the given values: $C_{\text{eq}} = \frac{10 \mu F}{2} (2 + 4) = 5 \mu F \times 6 = 30 \mu F$.

(B) The capacitance of the air-filled parallel plate capacitor is given by:
$C_1 = \frac{\varepsilon_0 A}{d} \quad --- (1)$
When a slab of dielectric constant $K$ and thickness $t$ is introduced between the plates,the new capacitance $C_2$ is given by:
$C_2 = \frac{\varepsilon_0 A}{d - t + \frac{t}{K}}$
For a metal sheet,the dielectric constant $K = \infty$. Given $t = \frac{d}{2}$,we substitute these values:
$C_2 = \frac{\varepsilon_0 A}{d - \frac{d}{2} + \frac{d/2}{\infty}} = \frac{\varepsilon_0 A}{\frac{d}{2} + 0} = \frac{2 \varepsilon_0 A}{d}$
$C_2 = 2 \left( \frac{\varepsilon_0 A}{d} \right) = 2 C_1$
Therefore,the ratio $C_2 : C_1 = 2 : 1$.

(A) When a dielectric slab is inserted into an isolated parallel plate capacitor,the charge $Q$ on the plates remains constant because the system is disconnected from the battery.
As the capacitance increases to $C' = KC$,the potential difference $V' = Q/C' = V/K$ decreases.
The electric field $E' = V'/d = E/K$ also decreases.
The energy stored $U' = Q^2/(2C') = U/K$ decreases.
Therefore,the charge on the capacitor is the only quantity that does not change.

(A) The capacitance of an air-filled parallel plate capacitor is $C_0 = \frac{\varepsilon_0 A}{d}$.
When a dielectric slab of thickness $t$ and dielectric constant $K$ is introduced,the new capacitance $C'$ is given by:
$C' = \frac{\varepsilon_0 A}{d - t + \frac{t}{K}}$.
Given that the new capacitance is one-third of the original value,i.e.,$C' = \frac{C_0}{3}$:
$\frac{\varepsilon_0 A}{d - t + \frac{t}{K}} = \frac{1}{3} \left( \frac{\varepsilon_0 A}{d} \right)$.
Canceling $\varepsilon_0 A$ from both sides:
$\frac{1}{d - t + \frac{t}{K}} = \frac{1}{3d}$.
$3d = d - t + \frac{t}{K}$.
$2d + t = \frac{t}{K}$.
$K = \frac{t}{2d + t}$.

(B) Let $d$ be the initial distance between the plates. The initial capacitance is $C = \frac{A \epsilon_0}{d}$.
When a dielectric plate of thickness $t = 2 \,mm$ is inserted, the new capacitance $C_1$ is given by $C_1 = \frac{A \epsilon_0}{d - t + \frac{t}{k}}$.
When the distance between the plates is increased by $x = 1.6 \,mm$ to maintain the same potential difference (which implies the same capacitance $C_2 = C$), the new distance becomes $d' = d + x$.
The new capacitance is $C_2 = \frac{A \epsilon_0}{d + x - t + \frac{t}{k}}$.
Since $C_2 = C$, we have $d + x - t + \frac{t}{k} = d$.
Simplifying this, we get $x - t + \frac{t}{k} = 0$, or $t - x = \frac{t}{k}$.
Substituting the values $t = 2 \,mm$ and $x = 1.6 \,mm$:
$2 - 1.6 = \frac{2}{k}$
$0.4 = \frac{2}{k}$
$k = \frac{2}{0.4} = 5$.

(D) The capacitance of a parallel plate capacitor filled with a dielectric medium is given by $C = \frac{K \varepsilon_0 A}{d}$.
Given that the dielectric constant $K = 3$,the initial capacitance is $C = \frac{3 \varepsilon_0 A}{d}$.
When the oil (dielectric) is removed,the medium between the plates becomes air (or vacuum),for which the dielectric constant $K' = 1$.
The new capacitance $C'$ is given by $C' = \frac{\varepsilon_0 A}{d}$.
Comparing the two expressions,we find that $C' = \frac{C}{3}$.

(B) The capacitance of a parallel plate capacitor is given by $C = \frac{k A \varepsilon_0}{d}$.
Initially,for an air-filled capacitor,$C_1 = \frac{A \varepsilon_0}{d_1} = 2 \ pF$,where $k_1 = 1$.
Finally,the separation is doubled $(d_2 = 2d_1)$ and the space is filled with a dielectric of constant $k_2 = k$. The new capacitance is $C_2 = \frac{k A \varepsilon_0}{d_2} = 6 \ pF$.
Taking the ratio: $\frac{C_2}{C_1} = \frac{k \cdot A \varepsilon_0 / (2d_1)}{A \varepsilon_0 / d_1} = \frac{k}{2}$.
Given $\frac{C_2}{C_1} = \frac{6}{2} = 3$.
Therefore,$\frac{k}{2} = 3$,which gives $k = 6$.

(C) Initially,both capacitors have capacitance $C$. When one capacitor is filled with a dielectric of constant $K$,its new capacitance becomes $C_1 = KC$,while the other remains $C_2 = C$.
Since the capacitors are connected in series to a battery of emf $V$,the potential difference $V_2$ across the capacitor $C_2$ is given by the voltage divider rule:
$V_2 = V \left( \frac{C_1}{C_1 + C_2} \right)$
Substituting the values:
$V_2 = V \left( \frac{KC}{KC + C} \right) = V \left( \frac{KC}{C(K + 1)} \right)$
$V_2 = \frac{KV}{K + 1}$

(D) When the battery is disconnected,the charge $q$ on the capacitor plates remains constant because there is no path for the charge to flow.
When a dielectric slab of dielectric constant $k$ is introduced between the plates,the capacitance of the capacitor increases from $C$ to $C' = kC$.
The energy stored in a capacitor is given by $U = \frac{q^2}{2C}$.
Since $q$ is constant and $C$ increases,the final energy $U_f = \frac{q^2}{2kC}$ will be less than the initial energy $U_i = \frac{q^2}{2C}$.
Therefore,the energy stored decreases.

(D) The arrangement consists of three capacitors connected in series.
For a parallel plate capacitor with a dielectric slab of thickness $t$ and dielectric constant $k$,the capacitance is $C = \frac{k \varepsilon_0 A}{t}$.
Since the slabs are placed in series,the equivalent capacitance $C_{eq}$ is given by $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$.
Substituting the values,we get $\frac{1}{C_{eq}} = \frac{t_1}{k_1 \varepsilon_0 A} + \frac{t_2}{k_2 \varepsilon_0 A} + \frac{t_3}{k_3 \varepsilon_0 A}$.
Taking $\frac{1}{\varepsilon_0 A}$ as a common factor,$\frac{1}{C_{eq}} = \frac{1}{\varepsilon_0 A} \left[\frac{t_1}{k_1} + \frac{t_2}{k_2} + \frac{t_3}{k_3}\right]$.
Therefore,the resultant capacity is $C_{eq} = \frac{A \varepsilon_0}{\left[\frac{t_1}{k_1} + \frac{t_2}{k_2} + \frac{t_3}{k_3}\right]}$.

(B) Let the initial capacitance of each capacitor be $C$. When one capacitor is filled with a dielectric of constant $K$,its new capacitance becomes $C' = KC$.
The two capacitors are connected in series to a battery of e.m.f. $V$.
Let $C_1 = KC$ be the capacitance of the dielectric-filled capacitor and $C_2 = C$ be the capacitance of the air-filled capacitor.
The potential difference across the air-filled capacitor $(C_2)$ in a series combination is given by the voltage divider rule:
$V_2 = V \times \frac{C_1}{C_1 + C_2}$
Substituting the values:
$V_2 = V \times \frac{KC}{KC + C}$
$V_2 = V \times \frac{KC}{C(K + 1)}$
$V_2 = \frac{KV}{K + 1}$

(C) The capacitance of a parallel plate capacitor is given by $C = \frac{\varepsilon A}{d}$.
For air as the medium,the capacitance is $C_{0} = \frac{\varepsilon_{0} A}{d} = 3 \mu F$.
When a dielectric medium is introduced,the new capacitance is $C = \frac{\varepsilon A}{d} = 15 \mu F$.
Taking the ratio of the two capacitances:
$\frac{C}{C_{0}} = \frac{\varepsilon}{\varepsilon_{0}} = K$ (where $K$ is the dielectric constant).
$\frac{15}{3} = \frac{\varepsilon}{\varepsilon_{0}} \implies 5 = \frac{\varepsilon}{\varepsilon_{0}}$.
Therefore,the permittivity of the medium is $\varepsilon = 5 \times \varepsilon_{0}$.
$\varepsilon = 5 \times 8.85 \times 10^{-12} = 44.25 \times 10^{-12} \text{ F/m}$.
Converting to scientific notation: $\varepsilon = 0.4425 \times 10^{-10} \text{ F/m}$.

(A) The capacitance of a parallel plate air capacitor is given by $C_0 = \frac{\varepsilon_0 A}{d}$.
Here, $\varepsilon_0$ is the permittivity of free space, $A$ is the area of the plates, and $d$ is the distance between the plates.
When the distance between the plates is reduced to $d_1 = \frac{d}{4}$ and the space is filled with a dielectric of constant $K = 2$, the new capacitance $C$ is given by $C = \frac{K \varepsilon_0 A}{d_1}$.
Substituting $d_1 = \frac{d}{4}$ and $K = 2$ into the equation:
$C = \frac{2 \varepsilon_0 A}{d/4} = 8 \left( \frac{\varepsilon_0 A}{d} \right) = 8 C_0$.
Given $C_0 = 4 \mu F$, the new capacitance is $C = 8 \times 4 \mu F = 32 \mu F$.

(A) Initially,two capacitors of capacity $C$ are in series. The equivalent capacitance $C_1$ is given by $\frac{1}{C_1} = \frac{1}{C} + \frac{1}{C} = \frac{2}{C}$,so $C_1 = \frac{C}{2}$.
After filling one capacitor with a dielectric of constant $K$,its new capacity becomes $KC$. The new equivalent capacitance $C_2$ is given by $\frac{1}{C_2} = \frac{1}{C} + \frac{1}{KC} = \frac{1}{C} \left(1 + \frac{1}{K}\right) = \frac{K+1}{KC}$.
Thus,$C_2 = \frac{KC}{K+1}$.
The change in effective capacitance $\Delta C = C_2 - C_1 = \frac{KC}{K+1} - \frac{C}{2}$.
$\Delta C = C \left[ \frac{K}{K+1} - \frac{1}{2} \right] = C \left[ \frac{2K - (K+1)}{2(K+1)} \right] = \frac{C}{2} \left[ \frac{K-1}{K+1} \right]$.

(A) Let the initial capacitance of each capacitor be $C$. When connected in series to a battery of emf $V$,the charge $q$ on both capacitors is the same.
Let $V_1$ be the potential difference across the air capacitor (capacitance $C$) and $V_2$ be the potential difference across the dielectric-filled capacitor (capacitance $KC$).
Since they are in series,$V_1 + V_2 = V$.
Using $q = CV$,we have $V_1 = \frac{q}{C}$ and $V_2 = \frac{q}{KC}$.
Substituting these into the sum equation: $\frac{q}{C} + \frac{q}{KC} = V$.
$\frac{q}{C} (1 + \frac{1}{K}) = V \Rightarrow \frac{q}{C} (\frac{K+1}{K}) = V$.
Therefore,the charge $q = \frac{CKV}{K+1}$.
The potential difference across the air capacitor is $V_1 = \frac{q}{C} = \frac{KV}{K+1}$.

(D) When a dielectric slab of constant $K$ is inserted into a charged parallel plate capacitor after the battery is disconnected,the charge $Q$ on the plates remains constant because there is no path for the charge to flow.
The variations in other physical quantities are as follows:
$1$. Capacitance: $C' = KC$ (increases).
$2$. Charge: $Q' = Q$ (remains constant).
$3$. Potential difference: $V' = V/K$ (decreases).
$4$. Electric field intensity: $E' = E/K$ (decreases).
$5$. Stored energy: $U' = U/K$ (decreases).
Therefore,the charge $Q$ is the quantity that remains constant.

(A) The energy stored in a capacitor is given by $U = \frac{Q^2}{2C}$.
When the battery is removed,the charge $Q$ on the capacitor plates remains constant.
The capacitance of a parallel plate capacitor is given by $C = \frac{\epsilon_0 A}{d}$,where $d$ is the distance between the plates.
If the distance $d$ is increased to $d' = 4d$,the new capacitance becomes $C' = \frac{\epsilon_0 A}{4d} = \frac{C}{4}$.
The new energy stored $U'$ is given by $U' = \frac{Q^2}{2C'} = \frac{Q^2}{2(C/4)} = 4 \times \frac{Q^2}{2C} = 4U$.
Therefore,the energy stored becomes $4U$.

(C) The capacitor is divided into three parts. Let the total area be $A$ and separation be $d$.
$1$. The left part has area $A/2$ and dielectric constant $K_1 = 4$. Its capacitance is $C_1 = \frac{K_1 \epsilon_0 (A/2)}{d} = \frac{4 \epsilon_0 A}{2d} = \frac{2 \epsilon_0 A}{d}$.
$2$. The right side is divided into two parts in series,each with area $A/2$ and separation $d/2$. The dielectric constants are $K_2 = 3$ and $K_3 = 6$.
$3$. Capacitance of the top right part: $C_2 = \frac{K_2 \epsilon_0 (A/2)}{d/2} = \frac{3 \epsilon_0 A}{d}$.
$4$. Capacitance of the bottom right part: $C_3 = \frac{K_3 \epsilon_0 (A/2)}{d/2} = \frac{6 \epsilon_0 A}{d}$.
$5$. $C_2$ and $C_3$ are in series,so their equivalent capacitance $C_{23}$ is given by $\frac{1}{C_{23}} = \frac{1}{C_2} + \frac{1}{C_3} = \frac{d}{3 \epsilon_0 A} + \frac{d}{6 \epsilon_0 A} = \frac{2d + d}{6 \epsilon_0 A} = \frac{3d}{6 \epsilon_0 A} = \frac{d}{2 \epsilon_0 A}$. Thus,$C_{23} = \frac{2 \epsilon_0 A}{d}$.
$6$. $C_1$ and $C_{23}$ are in parallel. Therefore,the total equivalent capacitance $C_{eq} = C_1 + C_{23} = \frac{2 \epsilon_0 A}{d} + \frac{2 \epsilon_0 A}{d} = \frac{4 \epsilon_0 A}{d}$.

(C) The electric field $E_0$ just outside the surface of a charged conductor in vacuum is given by $E_0 = \frac{\sigma}{\epsilon_0}$,where $\sigma$ is the surface charge density.
When the conductor is placed in a medium with dielectric constant $K$,the electric field $E$ is given by $E = \frac{\sigma}{\epsilon_0 K} = \frac{E_0}{K}$.
Given $E_0 = 10^3 \ V/m$ and $K = 4$.
Therefore,$E = \frac{10^3}{4} \ V/m = 250 \ V/m$.

(B) The force between two point charges in a medium is given by $F_{\text{medium}} = \frac{F_{\text{air}}}{K}$,where $F_{\text{air}}$ is the force between the same charges in a vacuum (or air) and $K$ is the dielectric constant of the medium.
Given that the force in the medium is $F$,we have $F = \frac{F_{\text{air}}}{K}$.
To find the force when the material is removed (i.e.,the force in air),we rearrange the equation:
$F_{\text{air}} = F \times K$.
Therefore,the force becomes $FK$.

(D) The initial capacitance of the parallel plate capacitor with air is given by:
$C_0 = \frac{A \varepsilon_0}{d} = 4 \ pF$
When the distance between the plates is reduced by half,the new distance is $d' = \frac{d}{2}$.
When the space is filled with a dielectric of constant $K = 6$,the new capacitance $C$ is given by:
$C = K \frac{A \varepsilon_0}{d'}$
Substituting $d' = \frac{d}{2}$ and $K = 6$:
$C = 6 \times \frac{A \varepsilon_0}{d/2} = 12 \times \frac{A \varepsilon_0}{d}$
Since $\frac{A \varepsilon_0}{d} = C_0 = 4 \ pF$,we have:
$C = 12 \times 4 \ pF = 48 \ pF$.

(C) In the absence of a dielectric,the electric field between the two plates is $E_0 = \frac{V_0}{d}$.
When the dielectric slab is inserted,the electric field inside the dielectric becomes $E = \frac{E_0}{K}$,where $K=3$ is the dielectric constant.
The thickness of the dielectric slab is $t = \frac{3}{4}d$,and the thickness of the air gap is $d - t = d - \frac{3}{4}d = \frac{1}{4}d$.
The new potential difference $V$ is the sum of the potential drops across the air gap and the dielectric slab:
$V = E_0 \left(\frac{1}{4}d\right) + E \left(\frac{3}{4}d\right)$
$V = E_0 \left(\frac{1}{4}d\right) + \frac{E_0}{K} \left(\frac{3}{4}d\right)$
$V = E_0 d \left[ \frac{1}{4} + \frac{3}{4K} \right]$
Since $E_0 d = V_0$ and $K = 3$:
$V = V_0 \left[ \frac{1}{4} + \frac{3}{4 \times 3} \right]$
$V = V_0 \left[ \frac{1}{4} + \frac{1}{4} \right]$
$V = V_0 \left( \frac{2}{4} \right) = \frac{V_0}{2}$

(C) The correct option is $C$.
When the battery is disconnected,the charge $Q$ on the plates remains constant because there is no path for the charge to flow.
The capacitance of a parallel plate capacitor is given by $C = \frac{A \varepsilon_0}{d}$,where $A$ is the area of the plates,$\varepsilon_0$ is the permittivity of free space,and $d$ is the distance between the plates.
When the distance $d$ between the plates is increased,the capacitance $C$ decreases.
Since the charge $Q$ remains constant and $Q = CV$,the potential difference $V = \frac{Q}{C}$ must increase as $C$ decreases.

(D) $1$. When a dielectric slab of constant $K$ is present,the capacitance is $C_1 = \frac{K \varepsilon_0 A}{d}$.
$2$. Since the capacitor remains connected to the battery,the potential difference across the plates remains constant,so $V_1 = V_2 = V$.
$3$. When the dielectric slab is removed,the new capacitance becomes $C_2 = \frac{\varepsilon_0 A}{d}$.
$4$. Since $K > 1$,it is clear that $C_1 > C_2$.
$5$. Because the battery is still connected,the potential difference across the plates does not change,thus $V_1 = V_2$.

(A) When a charged capacitor is disconnected from the battery and a glass slab (dielectric material) is introduced between the plates,the charge $(Q)$ remains constant because the circuit is open.
The capacitance $(C)$ increases as $C = KC_0$,where $K$ is the dielectric constant.
The potential difference $(V)$ decreases as $V = \frac{V_0}{K}$ because $V = \frac{Q}{C}$ and $C$ increases while $Q$ is constant.
The stored energy $(U)$ decreases as $U = \frac{Q^2}{2C} = \frac{U_0}{K}$ because $C$ increases while $Q$ is constant.
Therefore,the potential difference and the stored energy decrease.

(B) Let $t$ be the thickness of the dielectric slab and $K$ be the dielectric constant.
When a dielectric slab of thickness $t$ is introduced,the effective separation between the plates increases by an amount $x = t(1 - 1/K)$ to maintain the same capacitance.
Given,$x = 3.5 \times 10^{-3} \ m$ and $t = 4 \times 10^{-3} \ m$.
Substituting these values into the equation:
$3.5 \times 10^{-3} = 4 \times 10^{-3} \left(1 - \frac{1}{K}\right)$
Dividing both sides by $4 \times 10^{-3}$:
$\frac{3.5}{4} = 1 - \frac{1}{K}$
$0.875 = 1 - \frac{1}{K}$
$\frac{1}{K} = 1 - 0.875 = 0.125$
$K = \frac{1}{0.125} = 8$.
Thus,the dielectric constant of the material is $8$.

(A) The original capacitance of the air-filled parallel plate capacitor is $C = \frac{\varepsilon_0 A}{d}$.
When a dielectric slab of constant $K$ is introduced to fill one-fourth of the area,the capacitor can be considered as two capacitors in parallel.
One capacitor has air as the dielectric with area $\frac{3A}{4}$ and plate separation $d$. Its capacitance is $C_1 = \frac{\varepsilon_0 (3A/4)}{d} = \frac{3}{4} \frac{\varepsilon_0 A}{d} = \frac{3C}{4}$.
The other capacitor has a dielectric $K$ with area $\frac{A}{4}$ and plate separation $d$. Its capacitance is $C_2 = \frac{K \varepsilon_0 (A/4)}{d} = \frac{K}{4} \frac{\varepsilon_0 A}{d} = \frac{KC}{4}$.
Since these two capacitors are in parallel,the equivalent capacitance is $C_{net} = C_1 + C_2$.
$C_{net} = \frac{3C}{4} + \frac{KC}{4} = \frac{C}{4}(K+3)$.

(B) We know that the charge $Q$ on the capacitor remains constant if the battery is disconnected,or the potential difference changes based on the capacitance. Given $Q = CV$,we have $V = Q/C$,which implies $V \propto 1/C$.
The capacitance with air is $C_0 = \epsilon_0 A / d$.
The capacitance with a dielectric slab is $C = K C_0$,where $K$ is the dielectric constant.
Therefore,the ratio of the potentials is given by $V_{dielectric} / V_0 = C_0 / C = C_0 / (K C_0) = 1/K$.
Given $V_0 = 4 \ V$ and $V_{dielectric} = 2 \ V$.
Substituting the values: $2 / 4 = 1/K$.
This gives $1/2 = 1/K$,so $K = 2$.

(D) The initial force between two identical charges $q$ separated by distance $r$ in air is given by $F = \frac{1}{4 \pi \varepsilon_0} \frac{q^2}{r^2} \quad (1)$.
When a dielectric slab of thickness $t$ and dielectric constant $K$ is placed between the charges,the effective separation becomes $r_{eff} = (r - t) + t\sqrt{K}$.
Here,$t = r/2$ and $K = 4$.
Substituting these values,$r_{eff} = (r - r/2) + (r/2)\sqrt{4} = r/2 + (r/2)(2) = r/2 + r = 3r/2$.
The new force $F'$ is given by $F' = \frac{1}{4 \pi \varepsilon_0} \frac{q^2}{(r_{eff})^2} = \frac{1}{4 \pi \varepsilon_0} \frac{q^2}{(3r/2)^2}$.
$F' = \frac{1}{4 \pi \varepsilon_0} \frac{q^2}{(9/4)r^2} = \frac{4}{9} \left( \frac{1}{4 \pi \varepsilon_0} \frac{q^2}{r^2} \right)$.
Using equation $(1)$,we get $F' = \frac{4}{9}F$.