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

Options

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

Answer

D

Explanation

`I_(d)=1mA=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))=500Vs^(-1)`
Therefore applying a varying potential difference of `500Vs^(-1)` would produce a displacement current of desired value.