(III) A \({\bf{2}}{\bf{.50}}\;{\bf{\mu F}}\) capacitor is charged to 746 V and a \({\bf{6}}{\bf{.80}}\;{\bf{\mu F}}\) capacitor is charged to 562 V. These capacitors are then disconnected from their batteries. Next the positive plates are connected to each other and the negative plates are connected to each other. What will be the potential difference across each and the charge on each? [Hint: Charge is conserved.]

A capacitor is a charge storage device. When a capacitor is connected to a battery, the charge stored on each plate is equal in magnitude but opposite in sign.

The capacitance of the first capacitor is,\({C_1} = 2.50\;\mu {\rm{F}}\).

The capacitance of the second capacitor is,\({C_2} = 6.80\;\mu {\rm{F}}\).

The potential across the first capacitor is,\({V_1} = 746\;{\rm{V}}\).

The potential across the second capacitor is,\({V_2} = 562\;{\rm{V}}\).

The initial charge on the first capacitor is,\({Q_1} = {C_1}{V_1}\).

The initial charge on the second capacitor is,\({Q_2} = {C_2}{V_2}\).

After joining the capacitors, their total charge is,

\(\begin{aligned}Q &= {Q_1} + {Q_2}\\ &= {C_1}{V_1} + {C_2}{V_2}\end{aligned}\)

After joining, the net capacitance of the parallel combination is,

\(C = {C_1} + {C_2}\)

The final potential difference across each capacitor is given as:

\(\begin{aligned}V &= \frac{Q}{C}\\ &= \frac{{{C_1}{V_1} + {C_2}{V_2}}}{{{C_1} + {C_2}}}\end{aligned}\)

Substitute the values in the above expression.

\(\begin{aligned}V &= \frac{{\left( {2.50\;\mu {\rm{F}}} \right)\left( {746\;{\rm{V}}} \right) + \left( {562\;{\rm{V}}} \right)6.80\;\mu {\rm{F}}}}{{2.50\;\mu {\rm{F}} + 6.80\;\mu {\rm{F}}}}\\ &= 611\;{\rm{V}}\end{aligned}\)

Thus, the potential difference across each capacitor after connecting them is 611 V.

The final charge on the first capacitor is,

\(\begin{aligned}{{Q'}_1} &= {C_1}V\\ &= \left( {2.50\;\mu {\rm{F}}} \right)\left( {611\;{\rm{V}}} \right)\\ &= 1527.5\;\mu {\rm{C}}\\ &= 1.53 \times {10^{ - 3}}\;{\rm{C}}\end{aligned}\)

The final charge on the second capacitor is,

\(\begin{aligned}{{Q'}_2} &= {C_2}V\\ &= \left( {6.80\;\mu {\rm{F}}} \right)\left( {611\;{\rm{V}}} \right)\\ &= 4154.8\;\mu {\rm{C}}\\ &\approx 4.16 \times {10^{ - 3}}\;{\rm{C}}\end{aligned}\)

Thus, the final charge on each capacitor is \(1.53 \times {10^{ - 3}}\;{\rm{C}}\)and \(4.16 \times {10^{ - 3}}\;{\rm{C}}\).