JEE MAIN - Physics (2025 - 23rd January Evening Shift - No. 21)
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
100
Explanation
$$ \frac{dV}{dt} = \frac{I}{C} $$
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:
$$ \frac{dV}{dt} = \frac{0.25 \times 10^{-3}}{2.5 \times 10^{-6}} = \frac{0.25}{2.5} \times \frac{10^{-3}}{10^{-6}} = 0.1 \times 10^3 = 100 \, \text{V/s}. $$
Thus, the magnitude of the rate of change of the potential difference is $$ 100 \, \text{V/s} $$.
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