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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 mA in the intervening space between the capacitor plates, the magnitude of the rate of change of the potential difference will be _____$$Vs^{-1}$$.
Correct Answer: 100
A parallel plate capacitor with C = 2.5 μF produces a displacement current of 0.25 mA. Find the rate of change of voltage.
We use the displacement current formula: $$I_d = C\frac{dV}{dt}$$
Solving for the rate of change of voltage gives $$\frac{dV}{dt} = \frac{I_d}{C} = \frac{0.25 \times 10^{-3}}{2.5 \times 10^{-6}} = \frac{0.25}{2.5} \times 10^3 = 100$$ V/s
The answer is 100.
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