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Q.

A parallel plate capacitor having plate area ‘A’ and plate separation ‘d’ is joined to a battery of emf ‘V’ and internal resistance ‘R’ at  t=0 . Consider a plane surface of area A/3,  parallel to the plates and situated symmetrically between them. Find the displacement current through this surface as a function of time. [ The charge on the capacitor at time ‘t’ is given by q=CV1−etτwhere  τ=CR ]

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a

id=V2Re−td∈0AR

b

id=V3Re−td∈0AR

c

id=2V3Re−2td∈0AR

d

id=3V2Re−3td∈0AR

answer is B.

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Detailed Solution

Surface charge density,  σ=qA=CVA1−e−tτ Electric field b/w the plates of capacitor, E=σ∈0=CV∈0A1−e−tτ Electric flux from the given area, ϕE=EA3=CV3∈01−e−tτ Displacement current, id=∈0dϕEdt=∈0ddtCV3∈01−e−tτ=CV3τe−tτ ⇒id=V3Re−tCR ⇒id=V3Re−td∈0AR
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