Question

A parallel plate capacitor of capacitance $20\mu F$ is being charged by a voltage source whose potential is changing at the rate of $3V/s$. The conduction current through the connecting wires, and the displacement current through the plates of the capacitor, would be, respectively.
A. zero, $60\mu A$
B. $60\mu A,$ $60\mu A$
C. $60\mu A,$ zero
D. zero, zero

Answer 1

Capacitance is the ratio of the change in electric charge of a system to the corresponding change in electric potential. And current is the rate of flow of charge with respect to time. Find the current, using capacitance, and then use the concept that in case of capacitance, the magnitude of displacement current is equal to the magnitude of conduction current.

Complete step by step answer:
It is given in the question that,
Capacitance of capacitor
$C = 20\mu F$
$= 20 \times {10^{ - 6}}F$
And the rate of change of potential with respect to time is
$\dfrac{{dV}}{{dt}} = 3V/s$
Charge is given by
$q = CV$
By differentiating the above equation with respect to $t$ we get
$\dfrac{{dq}}{{dt}} = C\dfrac{{dV}}{{dt}}$
$\Rightarrow \dfrac{{dq}}{{dt}} = 20 \times {10^{ - 6}} \times 3$
But the rate of change of charge with respect to time is equal to current.
$\Rightarrow \dfrac{{dq}}{{dt}} = {i_c} = 60 \times {10^{ - 6}}$
$\Rightarrow {i_c} = 60\mu A$
As we know that, conduction current is equal to displacement current.
i.e. ${i_d} = {i_c}$
$\Rightarrow {i_d} = {i_c} = 60\mu A$

Therefore, from the above explanation the correct option is (B) $60\mu A,$ $60\mu A$.

Note: Parallel plate capacitors are formed by an arrangement of electrodes and insulting material or dielectric. Displacement current in electromagnet displacement current density in terms of the rate change of $D$.