Question

A parallel-plate capacitor is being charged. The displacement current across an area in the region between the plates and parallel to it is equal to

• A. conduction current in the wire
• B. half of the conduction current in the wire
• C. twice the conduction current in the wire
• D. can't, cannot.

Answer 1

The correct option is A

conduction current in the wire

The electric field between the plates is $E=\frac{Q}{{ϵ}_{0}A}$ where Q is the charge accumulated at the positive plate. The flux of this field through the given area is${\varphi }_E=\frac{Q}{{ϵ}_{0}A}\times A=\frac{Q}{{ϵ}_0}.$
The displacement current is $i_d={ϵ}_0\frac{d{\varphi }_E}{dt}={ϵ}_0\frac{d}{dt}\left(\frac{Q}{{ϵ}_0}\right)=\frac{dQ}{dt}.$
But $\frac{dQ}{dt}$ is the rate at which the charge is carried to the positive plate through the connecting wire. Thus, $i_d=i_c.$