Verified Solution • 2025–26 Session moderate

QJEEAdv22Phd3af: JEE Advanced 2022 Physics - Class 12 JEEAdv22Phd3af Solution: A medium having dielectric constant fills the space between the …

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 dd . The capacitor is connected to a battery of voltage VV , as shown in Figure (a). Now, both the plates are moved by a distance of d2\frac{d}{2} from their original positions, as shown in Figure (b).

JEE Advanced 2022 Paper 1 Online Physics - Capacitor Question 4 English

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?

Multiple Choice Options

  • A The electric field inside the dielectric material is reduced by a factor of 2K2 K .
  • B The capacitance is decreased by a factor of 1K+1\frac{1}{K+1} .
  • C The voltage between the capacitor plates is increased by a factor of (K+1)(K+1) .
  • D The work done in the process DOES NOT depend on the presence of the dielectric material.

Step-by-Step Verified Solution

JEE Advanced 2022 Paper 1 Online Physics - Capacitor Question 4 English Explanation 1

Ca=Kε0AdC_a=\frac{K \varepsilon_0 A}{d}

∴\therefore qa=Kε0AdVq_a=\frac{K \varepsilon_0 A}{d} V

Ea=qaKAε0E_a =\frac{q_a}{K A \varepsilon_0}

=Kε0AVdKε0A=\frac{K \varepsilon_0 A V}{d K \varepsilon_0 A}

=Vd=\frac{V}{d}

JEE Advanced 2022 Paper 1 Online Physics - Capacitor Question 4 English Explanation 2

Cbamp;=ε0Ad+(dK)amp;=ε0AKd(K+1)\begin{aligned} C_b & =\frac{\varepsilon_0 A}{d+\left(\frac{d}{K}\right)} \\\\ & =\frac{\varepsilon_0 A K}{d(K+1)}\end{aligned}

qb=ε0AKVd(K+1)q_b=\frac{\varepsilon_0 A K V}{d(K+1)}

(Eb)dielectric \left(E_b\right)_{\text {dielectric }}

=Eair K=\frac{E_{\text {air }}}{K}

=qbKAε0=\frac{q_b}{K A \varepsilon_0}

=ε0AKVd(K+1)(KAε0)=\frac{\varepsilon_0 A K V}{d(K+1)\left(K A \varepsilon_0\right)}

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

amp;∴ Capacitance decrease by a factor of 1K+1amp; Work done in the process =Uf−Uiamp;=12(Cf−Ci)V2amp;=12(ε0AKd(K+1)−Kε0Ad)V2amp;=12V2ε0AKd(1K+1−1)amp;=12ε0AKV2d1−K−1K+1amp;=12ε0AV2d(−K2K+1)\begin{aligned} &\therefore\text { Capacitance decrease by a factor of } \frac{1}{K+1} \\\\ & \text { Work done in the process }=U_f-U_i \\\\ &=\frac{1}{2}\left(C_f-C_i\right) V^2 \\\\ &=\frac{1}{2}\left(\frac{\varepsilon_0 A K}{d(K+1)}-\frac{K \varepsilon_0 A}{d}\right) V^2 \\\\ &=\frac{1}{2} V^2 \frac{\varepsilon_0 A K}{d}\left(\frac{1}{K+1}-1\right) \\\\ &=\frac{1}{2} \frac{\varepsilon_0 A K V^2}{d} \frac{1-K-1}{K+1} \\\\ &=\frac{1}{2} \frac{\varepsilon_0 A V^2}{d}\left(\frac{-K^2}{K+1}\right)\end{aligned}

Related Learning Resources & Exam Preparation

JEE Advanced PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring