Physics · Capacitors and R-C Circuits

JEE Main 2025 — 7 April, Evening Shift — Question 68

A parallel plate capacitor has charge $5 \times 10^{-6} \mathrm{C}$ .

A dielectric slab is inserted between the plates and almost fills the space between the plates. If the induced charge on one face of the slab is $4 \times 10^{-6} \mathrm{C}$ then the dielectric constant of the slab is $\qquad$ .

Answer: 5

Numerical answer — enter this value.

Step-by-step solution

$E_{\text {net }}=E_{0}-E_{\text {in }}$ $E_{\text {in }}=E_{0}\left(1-\frac{1}{k}\right)$

$Q_{\text {in }}=Q_{0}\left(1-\frac{1}{k}\right)$

$4 \times 10^{-6}=5 \times 10^{-6}\left(1-\frac{1}{k}\right)$ $k=5$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 7 April, Evening Shift
Subject: Physics
Chapter: Capacitors and R-C Circuits
Topic: Effect of Dielectrics