A parallel plate capacitor of plate separation `2 mm` is connected in an electric circuit having source voltage `400 V`. If the plate area is `60 cm^(2)`, then the value of displacement current for `10^(-6) sec` will be

Options

A. 1.062 amp.
B. `1.062 xx 10^(-2)` amp
C. `1.062 xx 10^(-3)` amp
D. `1.062 xx 10^(-4)` amp

Answer

B

Explanation

`I_(D) = epsilon_(0) (del phi_(E ))/(del t) = epsilon_(0) (EA)/(t ) = (epsilon_(0) A)/(t ) = (v)/(d)`
`I_(D) = (8.85 xx 10^(-12) xx 400 xx 60 xx 10^(-4))/(2 xx 10^(-3) xx 10^(-6))`
= 1.062 xx 10^(-2)`amp