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 their separation were doubled while the capacitor remained connected to the battery?

Answer 1

We know that charge on the capacitor, $Q=CV=10\times 12=120\mu C$

The parallel plate capacitance is $C=\frac{A{ϵ}_0}{d}$ or $C\propto \frac{1}{d}$

When the separation d is doubled, the capacitance will be half , i.e $C/2$

Now charges on each plate of capacitor is $Q^{\prime }=\left(C/2\right)V=120/2=60\mu C$