A rectangular parallel plate capacitor of dimension 5 cm x 4 cm is charging in such a way that the rate of change of electric field between the two plates is `5.65xx10^11 V. m^(-1) .s^(-1)`. Calculate the displacement current for this capacitor .
Given `in_0=8.85xx10^(-12) F.m^(-1)`

Explanation

Area of rectangular plate,
`A=(5xx4) cm^2 =20 cm^2 =0.002 m^2`
If E be the electric field between two plates , then the electric flux,
`phi_E=EA, (dphi_E)/(dt)=A(dE)/(dt)`
Therefore , the displacement current,
`I_d=in_0(dphi_E)/(dt)=in_0A(dE)/(dt)`
`=(8.85xx10^(-12))xx0.002xx(5.65xx10^11)`
=0.01 A = 10 mA