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A parallel plate capacitor with plate area Aand 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»

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


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