A parallel plate capacitor of capacitance 20 | vedclass

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(B) The charge $Q$ on a capacitor is given by $Q = CV$, where $C$ is the capacitance and $V$ is the potential difference.The conduction current $i_c$

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$60 \; \mu A, 60 \; \mu A$

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(B) The charge $Q$ on a capacitor is given by $Q = CV$, where $C$ is the capacitance and $V$ is the potential difference.The conduction current $i_c$ in the connecting wires is the rate of change of charge: $i_c = \frac{dQ}{dt} = C \frac{dV}{dt}$.Given $C = 20 \; \mu F$ and $\frac{dV}{dt} = 3 \; V/s$, we have $i_c = 20 \; \mu F 	imes 3 \; V/s = 60 \; \mu A$.According to Maxwell's modification of Ampere's law, the displacement current $i_d$ between the plates of a capacitor is equal to the conduction current in the connecting wires.Therefore, $i_d = i_c = 60 \; \mu A$.Thus, the conduction current is $60 \; \mu A$ and the displacement current is $60 \; \mu A$.

List-$I$ (Relation)	List-$II$ (Law)
$A$. $\oint \overrightarrow{E} \cdot d\vec{l} = -\frac{d}{dt} \oint \overrightarrow{B} \cdot d\vec{a}$	$I$. Ampere's circuital law
$B$. $\oint \vec{B} \cdot d\vec{l} = \mu_0(I + \epsilon_0 \frac{d\phi_E}{dt})$	$II$. Faraday's laws of electromagnetic induction
$C$. $\oint \overrightarrow{E} \cdot d\vec{a} = \frac{1}{\epsilon_0} \int \rho dv$	$III$. Ampere-Maxwell law
$D$. $\oint \overrightarrow{B} \cdot d\vec{l} = \mu_0 I$	$IV$. Gauss's law of electrostatics

$A$ parallel plate capacitor of capacitance $20 \; \mu F$ 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:

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