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Question

A parallel-plate capacitor having plate area A and plate separation d is joined to a battery of emf E and internal resistance R at t=0. Consider a plane surface of area A/2, parallel to the plates and situated symmetrically between them. If the displacement current through this surface as a function of time is given as:
id=EnRetdε0AR
Where, ε0 is the permittivity of free space.

The value of n is :


A
2.0
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B
2
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C
2.00
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Solution

Let the charge on the capacitor at time t is Q.

Then, the electric field between the plates of the capacitor is

E=Qε0A

The flux through the area considered is

ϕE=Qε0A×A2=Q2ε0

Therefore, the displacement current flowing through area A2 is

id=ε0dϕEdt

id=ε0×(12ε0)dQdt=12dQdt

We know that when a capacitor is connected to a DC voltage, the charge on the plates of the capacitor at any instant is given by

Q=E C[1et/RC]

Here, C= Capacitance of the capacitor

Therefore,
id=12d[E C(1et/RC)]dt

id=12×EC×(1RC)×et/RC

id=E2Ret/RC

Here, capacitance is given by

C=ε0Ad

id=E2Retdε0AR

Hence, n=2.

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