CBSE Class 12-science Answered
the rate of change of potential difference across capacitor is 5×10^2 v/s . if capacitance is 2uF ,then displacement current between it's plate is ?
Asked by pushpakumari291279 | 01 Oct, 2021, 11:08: AM
Expert Answer
Displacement current , Id = εo ( ∂E / ∂t ) A ..........................(1)
where εo is permittivity of free space , ( ∂E / ∂t ) is rate of change of electric field E and A is area of capcitor
if d is the distance between capacitor plates , then ( ∂E / ∂t ) = (1/d) (( ∂V / ∂t ) ...........................(2)
where V is potential difference across plates of capacitor .
By using eqn.(2) , we rewrite eqn.(1) as
Displacement current , Id = [ εo ( A /d ) ] ( ∂V / ∂t ) .................................(3)
Capcitance C = [ εo ( A /d ) ] ...........................(4)
Using eqn.(3) and (4) , Displacement current , Id = C ( ∂V / ∂t )
Hence displacement current id = 2 × 10-6 × 5 × 102 = 10-3 A
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