(II) It takes 18 J of energy to move a 0.30-mC charge from one plate of a \({\bf{15}}\;{\bf{\mu F}}\) capacitor to the other. How much charge is on each plate?

The potential difference may be defined as the amount of work done in moving a point charge of 1 C from one place to another.

It can be written as,

\(V = \frac{W}{Q}\) … (i)

Here, V is the potential difference, W is the work done and Q is the charge.

The work done is,\(W = 18\;{\rm{J}}\).

The amount of charge is,\(q = 0.30\;{\rm{mC}}\).

The capacitance is,\(C = 15\;{\rm{\mu F}}\).

From equation (i), the work done is given as,

\(\begin{aligned}W &= qV\\W &= q\left( {\frac{Q}{C}} \right)\\Q &= \frac{{C \times W}}{q}\end{aligned}\)

Substitute the values in the above expression.

\(\begin{aligned}Q &= \frac{{\left( {15\;{\rm{\mu F}} \times \frac{{{{10}^{ - 6}}\;{\rm{F}}}}{{1\;{\rm{\mu F}}}}} \right)\left( {18\;{\rm{J}}} \right)}}{{\left( {0.3\;{\rm{mC}} \times \frac{{{{10}^{ - 3}}\;{\rm{C}}}}{{1\;{\rm{mC}}}}} \right)}}\\Q &= 0.90\;{\rm{C}}\end{aligned}\)

Thus, the charge on each plate is \(0.90\;{\rm{C}}\).