Physics · Capacitors and R-C Circuits

JEE Main 2025 — 23 January, Evening Shift — Question 70

A time varying potential difference is applied between the plates of a parallel plate capacitor of capacitance $2.5 \mu \mathrm{~F}$ . The dielectric constant of the medium between the capacitor plates is 1 . It produces an instantaneous displacement current of 0.25 mA in the intervening space between the capacitor plates, the magnitude of the rate of change of the potential difference will be $\qquad$ $\mathrm{Vs}^{-1}$ .

Answer: 100

Numerical answer — enter this value.

Step-by-step solution

$\frac{\mathrm{CdV}}{\mathrm{dt}}=\mathrm{I}_{\mathrm{d}}$

$\frac{\mathrm{dV}}{\mathrm{dt}}=\frac{\mathrm{I}_{\mathrm{d}}}{\mathrm{C}}$

$=\frac{0.25 \times 10^{-3}}{2.5 \times 10^{-6}}$

$=100$

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Exam: JEE Main 2025
Paper: 23 January, Evening Shift
Subject: Physics
Chapter: Capacitors and R-C Circuits
Topic: Effect of Dielectrics

At steady state the charge on the capacitor, as shown in the circuit below, is mu C .

Answer: 100

Step-by-step solution

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