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JEE Main 202523 JanuaryEvening Shift
Exam
JEE Main
Year
2025
Date
23 January
Shift
Evening Shift
Subject
Physics
Chapter
Capacitance

JEE Main 2025 · 23 January · Evening Shift · Physics · Capacitance

A time varying potential difference is applied between the plates...

A time varying potential difference is applied between the plates of a parallel plate capacitor of capacitance 2.5μ F2.5 \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 _____ Vs−1\mathrm{Vs}^{-1}.

Correct answer

0

Step-by-step solution

CdVdt=IddVdt=IdC=0.25×10−32.5×10−6=100\begin{aligned} & \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\end{aligned}
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