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Physics · Capacitors and R-C Circuits

JEE Advanced 2022 — Paper 1 — Question 27

A medium having dielectric constant K>1K>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 d/2\mathrm{d} / 2 of 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?

Question figure
  1. Option A:

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

  2. Option B:

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

    Correct
  3. Option C:

    The voltage between the capacitor plates is increased by a factor of (K+1)(\mathrm{K}+1).

  4. Option D:

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

Answer: B

Step-by-step solution

B\mathbf{B} Refer to figure (a) V=E1 d\mathrm{V}=\mathrm{E}_{1} \mathrm{~d}

C1=kε0 A d\mathrm{C}_{1}=\frac{\mathrm{k} \varepsilon_{0} \mathrm{~A}}{\mathrm{~d}} Figure(a) Refer to figure (b)

V=E3 d2+E2 d2+E3 d2\mathrm{V}=\mathrm{E}_{3} \frac{\mathrm{~d}}{2}+\mathrm{E}_{2} \frac{\mathrm{~d}}{2}+\mathrm{E}_{3} \frac{\mathrm{~d}}{2}

Also E3=E2KE_{3}=E_{2} K V=E2 d+KE2 d\mathrm{V}=\mathrm{E}_{2} \mathrm{~d}+\mathrm{KE}_{2} \mathrm{~d}

V=(K+1)E2dV=(K+1) E_{2} d C2=ε0Ad+dK=KK+1ε0Ad\mathrm{C}_{2}=\frac{\varepsilon_{0} A}{d+\frac{d}{K}}=\frac{K}{K+1} \frac{\varepsilon_{0} A}{d}

Now E1E2=K+1\frac{E_{1}}{E_{2}}=K+1 C1C2=(K+1)\frac{\mathrm{C}_{1}}{\mathrm{C}_{2}}=(\mathrm{K}+1)

Wext +Wbattery =ΔU\mathrm{W}_{\text {ext }}+\mathrm{W}_{\text {battery }}=\Delta \mathrm{U}

Wext+(C2−C1)V2=(C2−C1)V22\mathrm{W}_{\mathrm{ext}}+\left(\mathrm{C}_{2}-\mathrm{C}_{1}\right) \mathrm{V}^{2}=\left(\mathrm{C}_{2}-\mathrm{C}_{1}\right) \frac{\mathrm{V}^{2}}{2}

Wext =(C1−C2)V22=Kε0 A d(1−1 K+1)V22=K2ε0AV22 d( K+1)\mathrm{W}_{\text {ext }}=\left(\mathrm{C}_{1}-\mathrm{C}_{2}\right) \frac{\mathrm{V}^{2}}{2}=\frac{\mathrm{K} \varepsilon_{0} \mathrm{~A}}{\mathrm{~d}}\left(1-\frac{1}{\mathrm{~K}+1}\right) \frac{\mathrm{V}^{2}}{2}=\frac{\mathrm{K}^{2} \varepsilon_{0} \mathrm{AV}^{2}}{2 \mathrm{~d}(\mathrm{~K}+1)}

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2022
Paper
Paper 1
Subject
Physics
Chapter
Capacitors and R-C Circuits
Topic
Effect of Dielectrics

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