Question

A parallel plate capacitor with plate area A  and separation between the plates d, is charged by a constant current I. Consider a plane surface of area A/4 parallel to the plates and drawn between the plates. What is the displacement current through this area?

(a) I

(b) \(\frac {I}{4}\)

(c) 4I

(d) \(\frac {I}{2}\)

This question was posed to me in an interview.

My query is from Displacement Current in division Alternating Current of Physics – Class 12

Answer 1

Right option is (b) \(\frac {I}{4}\)

For explanation I would say: Electric field between the plates is given as:

E=\(\frac {q}{A\varepsilon_o}=\frac {It}{A\varepsilon_o}\)

So, the electric flux through the area \(\frac {A}{4}\) is given by:

ΦE=\((\frac {A}{4})\)E=\(\frac {It}{4\varepsilon_o}\)

Then the displacement current will be:

ID = εo\(\frac {d\Phi_E}{dt}\)

ID = εo\(\frac {d}{dt} (\frac {It}{4\epsilon_o})=\frac {I}{4}\)

Therefore, the displacement current through this area is \(\frac {I}{4}\).