Question

A $10.0\mu F$ parallel-plate capacitor with circular plates is connected to a $12.0V$ battery. How much charge (in $\mu C$) would be on the plates if the capacitor were connected to the $12.0V$ battery after the radius of each plate was doubled without changing their separation?

Answer 1

The magnitude of initial charge on each plate is $Q=CV=10\times 12=120\mu C$

Also, $C=\frac{A{ϵ}_0}{d}=\frac{\left(\pi R^2\right){ϵ}_0}{d}$

As the the quantities are constant except $R$, so $C\propto R^2$

When $R$ is doubled, the capacitance becomes $4C$ and now the charge will be $Q^{\prime }=4CV=4\times 120=480\mu C$