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 isA. `(4)/(pi)A`B. `(0.4)/(pi)A`C. `(40)/(pi)A`D. `(1)/(10pi)A`

The area of each plate of parallel plated condenser is `144 cm^(2)`. T

1.	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 isA. `(4)/(pi)A`B. `(0.4)/(pi)A`C. `(40)/(pi)A`D. `(1)/(10pi)A`
Answer» Correct Answer - B `I_(d)=epsilon_(0)A(dE)/(dt)`

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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 isA. `(4)/(pi)A`B. `(0.4)/(pi)A`C. `(40)/(pi)A`D. `(1)/(10pi)A`

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