When capacitors are connected in series across a potential difference, there is a loss in their stored energy because:

Answer Details

The energy stored in a capacitor charged to a potential difference \(V\) is

For a fixed supply voltage the stored energy therefore depends only on the capacitance of the combination, so that is the quantity to examine.

For capacitors in series the effective capacitance obeys

which always gives a value smaller than the smallest individual capacitance. For two \(4\,\mu\mathrm{F}\) capacitors, for instance, the series value is \(2\,\mu\mathrm{F}\), so across a \(10\,\mathrm{V}\) supply the pair stores \(\tfrac{1}{2}(2\times10^{-6})(10)^{2} = 1.0\times10^{-4}\,\mathrm{J}\), whereas one of them alone across the same supply would store \(2.0\times10^{-4}\,\mathrm{J}\). The fall in stored energy therefore traces directly to the fall in overall capacitance produced by the series connection. Physically, the applied p.d. is shared among the capacitors, so no single capacitor receives the full \(V\), and each stores less than it would on its own.

The suggestion that unequal charges are deposited is the misconception worth clearing up: in a series chain the charge on every capacitor is the same, because the plates between neighbouring capacitors are isolated and can only separate charge, not create it. What differs between unequal capacitors in series is the voltage each carries, from \(V = Q/C\). Internal resistance of the source affects how quickly charging happens and causes heating in the wires, but it is not the reason the fully charged combination holds less energy. In the examination, tie any energy comparison for capacitors back to \(E = \tfrac{1}{2}CV^{2}\) and ask what has changed, \(C\) or \(V\).