A time varying potential difference is applied between the plates of a parallel plate capacitor of capacitance 2.5 $\mathrm{\mu }$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.

The displacement current $I_d$ in a capacitor is given by the equation:

$$I_d = C \frac{dV}{dt}$$

where:

$I_d$ is the displacement current (0.25 mA = $0.25 \times 10^{-3}$ A),

$C$ is the capacitance of the capacitor (2.5 μF = $2.5 \times 10^{-6}$ F),

$\frac{dV}{dt}$ is the rate of change of potential difference (which we need to find).

Step 1: Solve for $\frac{dV}{dt}$

Rearrange the equation:

$$\frac{dV}{dt} = \frac{I_d}{C}$$

Substituting the given values:

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

 $$\frac{dV}{dt} = \frac{0.25}{2.5} \times 10^3$$ $$\frac{dV}{dt} = 0.1 \times 10^3$$ $$\frac{dV}{dt} = 100 \, \text{Vs}^{-1}$$

Final Answer:

$$100 \, \text{Vs}^{-1}$$