JEE Advanced 2022 — Paper 1 — PHY-II solution

JEE Advanced2022Paper 1PHY-II

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?

A
The electric field inside the dielectric material is reduced by a factor of $2 K$ .
B
The capacitance is decreased by a factor of $\frac{1}{K + 1}$ .
C
The voltage between the capacitor plates is increased by a factor of $(K + 1)$ .
D
The work done in the process DOES NOT depend on the presence of the dielectric material.

Solution

Configuration (a): $C_{a} = \frac{K ε _{0} A}{d}$ . The battery maintains voltage $V$ , so field in dielectric is $E_{a} = V / d$ .

Configuration (b): Two air gaps of $d /2$ each in series with the dielectric slab of width $d$ (with relative permittivity $K$ ). Equivalent series capacitor:

$\frac{1}{C _{b}} = \frac{d /2}{ε _{0} A} + \frac{d}{K ε _{0} A} + \frac{d /2}{ε _{0} A} \Rightarrow C_{b} = \frac{ε _{0} A K}{d ( K + 1 )}$ .

Ratio $\frac{C _{b}}{C _{a}} = \frac{1}{K + 1}$ . So (B) is correct.

The voltage is still $V$ (battery), so (C) is wrong. The field in the dielectric in (b) is $V / [d (K + 1)]$ , a reduction by factor $K + 1$ , not $2 K$ , so (A) is wrong. The work done involves change in stored energy and work done by the battery, both of which depend on $K$ , so (D) is wrong.

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