Effect of Dielectric Inside Capacitor Questions in English

(C) When a dielectric material is inserted between the plates of a capacitor, it gets polarized. The induced electric field inside the dielectric opposes the external electric field applied by 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 plates), a decrease in the electric field leads to a reduction in the effective potential difference between the plates. Because $C = Q/V$, a decrease in $V$ for a constant charge $Q$ results in an increase in the capacitance $C$.

(D) The initial capacitance of the parallel plate capacitor with air as the medium is given by $C = \frac{\epsilon_0 A}{d}$.
When a dielectric slab of thickness $t = \frac{d}{2}$ and dielectric constant $K$ is inserted, the new capacitance $C'$ is given by the formula $C' = \frac{\epsilon_0 A}{d - t + \frac{t}{K}}$.
Substituting the values $t = \frac{d}{2}$ into the formula:
$C' = \frac{\epsilon_0 A}{d - \frac{d}{2} + \frac{d}{2K}}$
$C' = \frac{\epsilon_0 A}{\frac{d}{2} + \frac{d}{2K}}$
$C' = \frac{\epsilon_0 A}{\frac{d}{2} (1 + \frac{1}{K})}$
$C' = \frac{2 \epsilon_0 A}{d (\frac{K + 1}{K})}$
$C' = \frac{2 \epsilon_0 A}{d} \cdot \frac{K}{K + 1}$
Since $C = \frac{\epsilon_0 A}{d}$, we substitute $C$ into the expression:
$C' = \frac{2KC}{K + 1}$.

(A) The capacitance of a parallel plate capacitor without a dielectric is $C_0 = \frac{\epsilon_0 A}{d}$.
When a dielectric slab of thickness $t = (4/5)d$ and dielectric constant $K$ is inserted, the new capacitance $C$ is given by the formula $C = \frac{\epsilon_0 A}{d - t + (t/K)}$.
Substituting $t = (4/5)d$ into the formula:
$C = \frac{\epsilon_0 A}{d - (4/5)d + ((4/5)d/K)}$
$C = \frac{\epsilon_0 A}{(1/5)d + (4d/5K)}$
$C = \frac{\epsilon_0 A}{(d/5) [1 + (4/K)]} = \frac{\epsilon_0 A}{(d/5) [(K + 4)/K]}$
$C = \frac{5K \epsilon_0 A}{d(K + 4)}$
Since $C_0 = \frac{\epsilon_0 A}{d}$, we have $C = C_0 \cdot \frac{5K}{K + 4}$.
Therefore, the ratio $(C/C_0) = \frac{5K}{K + 4}$.

(B) The original capacitance of an air-filled parallel plate capacitor 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 $C' = \frac{1}{3} C_0$, we have $\frac{\epsilon_0 A}{d - t + \frac{t}{K}} = \frac{1}{3} \frac{\epsilon_0 A}{d}$.
This simplifies to $3d = d - t + \frac{t}{K}$.
Rearranging the terms, we get $2d + t = \frac{t}{K}$.
Thus, $K = \frac{t}{2d + t}$.

(D) When a dielectric of dielectric constant $K$ is inserted into a parallel plate capacitor that is disconnected from the battery, the charge $Q$ on the plates remains constant.
$1$. The new capacitance becomes $C' = KC$.
$2$. Since $Q$ is constant, the new potential $V' = Q / C' = Q / (KC) = V / K$. Since $K > 1$, the potential $V$ decreases.
$3$. The new energy $E' = Q^2 / (2C') = Q^2 / (2KC) = E / K$. Since $K > 1$, the energy $E$ also decreases.
Therefore, both $V$ and $E$ decrease.

(A) Let the capacitance of each identical air capacitor be $C$.
Initially, both are in series with a battery of e.m.f. $V$. The equivalent capacitance is $C_{eq} = C/2$.
When one capacitor is filled with a dielectric of constant $K$, its new capacitance becomes $C' = KC$.
The new equivalent capacitance is $C_{eq}' = \frac{C \cdot KC}{C + KC} = \frac{KC}{K + 1}$.
The total charge supplied by the battery is $Q = C_{eq}' V = \frac{KCV}{K + 1}$.
Since the capacitors are in series, the same charge $Q$ flows through both.
The potential difference across the air capacitor (the one not filled with dielectric) is $V_1 = \frac{Q}{C} = \frac{KCV}{C(K + 1)} = \frac{KV}{K + 1}$.

(A) $1$. Initially, the capacitors $2C$ and $C$ are in parallel and connected to a battery of potential $V$. The total charge $Q$ stored is $Q = (2C + C)V = 3CV$.
$2$. When the battery is disconnected, the total charge $Q = 3CV$ remains constant.
$3$. After filling the capacitor $C$ with a dielectric $K$, its new capacitance becomes $C' = KC$. The capacitor $2C$ remains unchanged.
$4$. The capacitors are still in parallel, so they share the same potential difference $V'$.
$5$. The new total capacitance is $C_{eq} = 2C + KC = C(K + 2)$.
$6$. Using the relation $Q = C_{eq}V'$, we get $3CV = C(K + 2)V'$.
$7$. Solving for $V'$, we get $V' = \frac{3CV}{C(K + 2)} = \frac{3V}{(K + 2)}$.

(A) The capacitance of a parallel plate capacitor with air is given by $C_0 = \frac{\epsilon_0 A}{d} = 15 \text{ pF}$.
When the separation is doubled $(d' = 2d)$ and a dielectric medium of constant $K = 3.5$ is introduced, the new capacitance $C'$ is given by $C' = \frac{K \epsilon_0 A}{d'}$.
Substituting the values, $C' = \frac{3.5 \times \epsilon_0 A}{2d} = \frac{3.5}{2} \times C_0$.
$C' = 1.75 \times 15 \text{ pF} = 26.25 \text{ pF}$.
Given that $C' = x/4 \text{ pF}$, we have $26.25 = x/4$.
Therefore, $x = 26.25 \times 4 = 105$.