Question

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

Hint:

The displacement current is related to the rate of change of the potential difference by \( I_d = C \frac{dV}{dt} \), where \( C \) is the capacitance.

Answer 1

Correct Answer: 2

The displacement current \( I_d \), proportional to the rate of change of capacitance \( C \) and the potential difference \( V \) across its plates, is defined as: \[ I_d = C \frac{dV}{dt}, \] where \( C \) denotes the capacitance and \( \frac{dV}{dt} \) represents the rate of change of potential difference. Provided values: - \( C = 2.5 \, \mu F = 2.5 \times 10^{-6} \, \text{F} \) - \( I_d = 0.25 \, \text{mA} = 0.25 \times 10^{-3} \, \text{A} \) Calculating the rate of change of potential difference \( \frac{dV}{dt} \): \[ 0.25 \times 10^{-3} = 2.5 \times 10^{-6} \times \frac{dV}{dt}. \] Thus, \[ \frac{dV}{dt} = \frac{0.25 \times 10^{-3}}{2.5 \times 10^{-6}} = 100 \, \text{Vs}^{-1}. \]