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