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 _________ $\mathrm{Vs}^{-1}$.
Answer (integer)
100
Solution
<p>$\frac{dV}{dt} = \frac{I}{C}$</p>
<p>Given that the displacement current is $I = 0.25 \times 10^{-3} \, \text{A}$ and the capacitance is $C = 2.5 \times 10^{-6} \, \text{F},$ the rate of change of the potential difference is calculated as follows:</p>
<p>$$ \frac{dV}{dt} = \frac{0.25 \times 10^{-3}}{2.5 \times 10^{-6}} = \frac{0.25}{2.5} \times \frac{10^{-3}}{10^{-6}} = 0.1 \times 10^3 = 100 \, \text{V/s}. $$</p>
<p>Thus, the magnitude of the rate of change of the potential difference is $100 \, \text{V/s}$.</p>
About this question
Subject: Physics · Chapter: Electromagnetic Waves · Topic: Maxwell's Equations
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