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The area of each plate of parallel plate...

The area of each plate of parallel plated condenser is `144 cm^(2)`. The electrical field in the gap between the plates changes at the rate of `10^(12)Vm^(-1)s^(-1)`. The displacement current is

A

`(4)/(pi)A `

B

`(0.4)/(pi)`

C

`(40)/(pi)`

D

`(1)/(10pi)A`

Text Solution

Verified by Experts

The correct Answer is:
B
`I_(d)=epsilon_(0)A (dE)/(dt)`
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