In order to establish an instantaneous displacement current of `1 mA` in the space between the plates of `2muF` parallel plate capacitor, the potential difference need to apply is

Options

A. `100 V s^(-1)`
B. `200V s^(-1)`
C. `300Vs^(-1)`
D. `500 V s^(-1)`

Answer

D

Explanation

`I_(D)="1 mA" =10^(-3)A,C=2muF=2xx10^(-6)F`
`I_(D)=I_(C)=(d)/(dt)(CV)=C(dV)/(dt)`
Therefore, `(dV)/(dt)=(I_(D))/(C)=(10^(-3))/(2xx10^(-6))=500"V s"^(-1)`
Therefore, applying a varying potential difference of 500 V `s^(-1)` would produce a displacement current of desired value.