A parallel plate capacitor of capacitance 20 μF is being charged by a voltage source whose potential is changi

Question

A parallel plate capacitor of capacitance 20 μF is being charged by a voltage source whose potential is changing at the rate of 3 V/s. The conduction current through the connecting wires, and the displacement current through the plates of the capacitor would be, respectively:

Accepted Answer

1. **Conduction Current ($I_c$):** The conduction current flowing through the connecting wires is the rate at which charge is supplied to the capacitor plates. Using the relation $q = CV$, we differentiate with respect to time:
   $$ I_c = \frac{dq}{dt} = \frac{d(CV)}{dt} = C \frac{dV}{dt} $$
   Given $C = 20 	ext{ \mu F} = 20 	imes 10^{-6} 	ext{ F}$ and $\frac{dV}{dt} = 3 	ext{ V/s}$:
   $$ I_c = (20 	imes 10^{-6}) 	imes 3 = 60 	imes 10^{-6} 	ext{ A} = 60 	ext{ \mu A} $$
2. **Displacement Current ($I_d$):** According to Maxwell's correction to Ampere's Circuital Law, the changing electric field between the plates constitutes a displacement current. By the property of continuity of current, the displacement current in the gap is always equal to the conduction current in the wires.
   $$ I_d = I_c = 60 	ext{ \mu A} $$
   Alternatively, using the formula $I_d = \varepsilon_0 \frac{d\Phi_E}{dt}$ leads to the same result .

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