91Ó°ÊÓ

We have been given that,

Diameter of the capacitor $5cm.$

Gap between parallel plate Capacitor $0.50mm$.

Increasing rate of Potential difference$500000V/s$.

Diameter D = $5cm=0.05m$,

Radius $r$=$D/2$= $\raisebox{1ex}{$0.05m$}\!\left/ \!\raisebox{-1ex}{$2$}\right.=0.025m$,

Gap d between the parallel plates = $0.50mm=0.5x{10}^{-3}m$.

Increasing Potential difference $dV/dt=500000V/s$

Firstly we will Calculate the capacitance $C$, between the parallel plates using the formula :-

$C=\frac{{\mathrm{Î\mu }}_{0}A}{d}$where $A$is area between the parallel plates and $d$is gap between the parallel plates.

$C=\frac{{\mathrm{Î\mu }}_0\mathrm{Ï€}{r}^2}{d}$( Area $A=\mathrm{Ï€}{r}^2$)

$C=\frac{8.854x{10}^{-12}x\mathrm{Ï€}{r}^2}{d}$(${\mathrm{Î\mu }}_0=8.85x{10}^{-12}F/m$)

$C=\frac{8.854x{10}^{-12}x\mathrm{Ï€}x{(0.025)}^2}{0.5x{10}^{-3}}$(substituting the value of $r$)

After solving we get,

$C=6.956x{10}^{-12}F$

Now, For displacement Current we know $I_d=Cx\frac{dV}{dt}$.

$I_d=6.956x{10}^{-12}x5x{10}^5$(substituting the value of $C$and $\frac{dV}{dt}$)

After solving the mathematical part we will get,

$I_d=3.478x{10}^{-6}$$A$

$3.478x{10}^{-6}A$is the displacement current in the capacitor if the potential difference across the capacitor is increasing at $500000$$V/s$.