You are given a $2 \,\mu F$ parallel plate capacitor. How would you establish an instantaneous displacement current of $1 \, mA$ in the space between its plates?

(N/A) The displacement current $I_d$ in a capacitor is given by the formula $I_d = \epsilon_0 \frac{d\Phi_E}{dt}$,which is equivalent to $I_d = C \frac{dV}{dt}$.
Given:
Capacitance $C = 2 \,\mu F = 2 \times 10^{-6} \, F$
Displacement current $I_d = 1 \, mA = 10^{-3} \, A$
We need to find the rate of change of potential difference $\frac{dV}{dt}$.
Using the relation $I_d = C \frac{dV}{dt}$,we get:
$\frac{dV}{dt} = \frac{I_d}{C} = \frac{10^{-3}}{2 \times 10^{-6}} = 0.5 \times 10^3 = 500 \, V/s$.
Therefore,by changing the potential difference across the plates at a rate of $500 \, V/s$,an instantaneous displacement current of $1 \, mA$ can be established.