Question 451: Electrostatics & Capacitance

Question

The voltage between the plates of a parallel plate capacitor of capacity $1 \mu \mathrm{~F}$ is changing at the rate of $4 \mathrm{~V} / \mathrm{s}$. the displacement current in the capacitor is

Accepted Answer

Given:
Capacitance, $C = 1~\mu\mathrm{F} = 1 	imes 10^{-6}~\mathrm{F}$
Rate of change of voltage, $\dfrac{dV}{dt} = 4~\mathrm{V/s}$
Displacement current
$(I_d)$ in a capacitor is given by:
$ I_d = C \dfrac{dV}{dt} $
Substitute the given values:
$ I_d = 1 	imes 10^{-6}~\mathrm{F} 	imes 4~\mathrm{V/s} $
$ I_d = 4 	imes 10^{-6}~\mathrm{A} $
$ I_d = 4~\mu\mathrm{A} $
Answer:
Option A: $4~\mu\mathrm{A}$

Answer

$4 \mu \mathrm{~A}$

Answer

$3 \mu \mathrm{~A}$

Answer

$1 \mu \mathrm{~A}$

Answer

$6 \mu \mathrm{~A}$

Electrostatics & Capacitance

The voltage between the plates of a parallel plate capacitor of capacity 1 \mu \mathrm{~F} is changing at the rate of 4 \mathrm{~V} / \mathrm{s}. the displacement current in the capacitor is

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