A parallel plate capacitor is made out of two rectangular metal pates of sides `30 cmxx15 cm` and separated by a distance of 2.0 mm. The capcitor is charged in such a way that the charging current has a constant value of 100mA. What must be the rate of change of potential of the charging source to ensure this and what will be the displacement current in the region between the capacitor plates?

Answer

`5xx10^8Vs^-1 0.1A`

Explanation

`I=(dQ)/(dt)=(d(CV))/(dt)=(CdV)/(dt)=((in_0A)/d)(dV)/(dt)`
or `(dV)/(dt)=(Id)/(in_0A)`
or `(dV)/(dt)=(0.1xx(2xx10^-3))/((8.85xx10^(-12))xx(30xx15xx10^-4))`
`=5xx10^8Vs^-1`,
`I_d=I=0.1A`