A 5.0-cm-radius parallel-plate capacitor has a 5.0 mm gap. What is the displacement current in the capacitor if the potential difference across the capacitor is increasing at 500,000 V/s?

1.389×10^-5A

Explanation:

We can calculate the displacement current in the capacitor using the expresion below

J=Q/t(*)

t= time elapsed

W= stored charge in capacitor

But Q= CV(1)

C = capacitance across given capacitor

V =voltage across given capacitor.

But capacitance in parallel plate capacitor can be expressed vas

C= εA/d(2)

A = area of plates

d= distance between the plates= 5.0 mm gap.= 5×10^-3 m

ε= Vacuum permisivity= 8.85×10^-12F/m

If we input eqn (1) and (2) into eqn (*) we have the expression below

J= εAV/dt(4)

But ratio of V/t=500,000 V/s

But Area=πd^2/2

Where d= diameter= (2×radius)=(2×5cm)=10cm=0.1m

Area=[ π × (0.1)^2]/2

=0.0157m^2

If we substitute all these value into eqn(4) we have

J=[( 8.85×10^-12) × (0.0157)×(500,000)]/5×10^-3

J=1.389×10^-5A

Hence, the displacement current in the capacitor if the potential difference across the capacitor is increasing at 500,000 V/s is 1.389×10^-5A

uio

explanation: