A medium having dielectric constant fills the space between the plates of a parallel plate capacitor. The plates have large area, and the distance between them is . The capacitor is connected to a battery of voltage , as shown in Figure (a). Now, both th

Question

A medium having dielectric constant $K > 1$ fills the space between the plates of a parallel plate capacitor. The plates have large area, and the distance between them is $d$ . The capacitor is connected to a battery of voltage $V$ , as shown in Figure (a). Now, both the plates are moved by a distance of $\frac{d}{2}$ from their original positions, as shown in Figure (b).

In the process of going from the configuration depicted in Figure (a) to that in Figure (b), which of the following statement(s) is(are) correct [2022]

In the process of going from the configuration depicted in Figure (a) to that in Figure (b), which of the following statement(s) is(are) correct? [2022]

Smart Achievers · Accepted Answer

(2)

Case a:

$C_{a} = \frac{K A ε_{0}}{d}$

$q_{a} = C_{a} V = \frac{K A ε_{0}}{d} V$

$E_{a} = \frac{q_{a}}{K A ε_{0}} = \frac{\frac{K A ε_{0} V}{d}}{K A ε_{0}} = \frac{V}{d}$

Case b:

$C_{b} = \frac{A ε_{0}}{2 d + \frac{d}{K} - d} = \frac{A ε_{0}}{d + \frac{d}{K}} = \frac{K A ε_{0}}{d (K + 1)}$

$q_{b} = C_{b} V = \frac{ε_{0} A K V}{d (K + 1)}$

${(E_{b})}_{dielectric} = \frac{{(E_{air})}_{b}}{K} = \frac{q_{b}}{K A ε_{0}} = \frac{ε_{0} A K V}{d (K + 1) K A ε_{0}}$

$= \frac{V}{d (K + 1)}$

So, capacitance decreases by a factor of $\frac{1}{K + 1}$ and the same is true for the electric field. Hence (1) is incorrect and (2) is correct.

$Work done in the process = U_{f} - U_{i}$

$= \frac{1}{2} (C_{f} - C_{i}) V^{2}$

$= \frac{1}{2} V^{2} [\frac{K A ε_{0}}{d} (\frac{1}{K + 1} - 1)]$ $= \frac{1}{2} V^{2} \frac{K A ε_{0}}{d} (- \frac{K}{K + 1})$

$= \frac{1}{2} \frac{ε_{0} A V^{2}}{d} (\frac{- K^{2}}{K + 1})$

So option (4) is incorrect.

As the voltage supply remains constant, there will be no change in voltage across the capacitor. Therefore option (3) is incorrect.

Solution

1 The electric field inside the dielectric material is reduced by a factor of 2K.

2 The capacitance is decreased by a factor of $\frac{1}{K + 1}$ .

3 The voltage between the capacitor plates is increased by a factor of $(K + 1)$ .

4 The work done in the process DOES NOT depend on the presence of the dielectric material.

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