= Capacitor energy
{title2=$U=Q^2/(2C)$}
The <capacitor energy> is the <electrostatic energy> stored when <electric charge> is separated. For a linear <capacitor> of fixed positive <capacitance> $C$, plate <electric charge> magnitude $Q$, and terminal <electric potential difference> $V=Q/C$, it is
$$
U=\frac{Q^2}{2C}=\frac12CV^2=\frac12QV.
$$
The zero is the uncharged state. Thus the <capacitor energy> is positive for nonzero $Q$, even though the total <electric charge> of the two plates is zero. https://openstax.org/books/university-physics-volume-2/pages/8-3-energy-stored-in-a-capacitor[OpenStax University Physics 2, §8.3: energy stored in a capacitor]{ref}.
For the <parallel-plate capacitor> example, <capacitance> $C=8.854\times10^{-11}\,\mathrm F$ and terminal <electric potential difference> $V=10\,\mathrm V$ give
$$
U=\frac12(8.854\times10^{-11})(10)^2\,\mathrm J
=4.427\times10^{-9}\,\mathrm J=4.427\,\mathrm{nJ}.
$$
The plate <electric charge> magnitude is $Q=0.8854\,\mathrm{nC}$. Doubling terminal <electric potential difference> with <capacitance> unchanged doubles the plate <electric charge> magnitude and quadruples the <capacitor energy>.
\Image[/capacitor-charge-and-energy.png]
{title=Linear <electric charge> and quadratic <capacitor energy> versus <electric potential difference>}
The <electric charge> is a <linear function> of terminal <electric potential difference>, while <capacitor energy> is a <quadratic function>. The second curve in each plot uses twice the <capacitance>: at a given terminal <electric potential difference>, both plate <electric charge> magnitude and <capacitor energy> double. These relations let us calculate how much <energy> an <electrical circuit> can receive when a <capacitor> discharges.
A brief charging argument explains the factor $1/2$. If the plate <electric charge> magnitude has reached $q$, the terminal <electric potential difference> is $q/C$. Transferring a further amount $dq$ of <electric charge> quasistatically requires <work> $dW=(q/C)\,dq$. With the geometry and <permittivity> fixed, the <integral> is
$$
U=\int_0^Q\frac qC\,dq=\frac{Q^2}{2C}.
$$
Using $Q=CV$ gives the other two forms. This <integral> accounts for the gradually rising terminal <electric potential difference>; it is the <work> stored in the <capacitor>'s <electric field>, rather than a claim that all <energy> drawn from a fixed-voltage <battery> during charging ends up stored.
The field picture gives the same result. In the vacuum gap of a <parallel-plate capacitor>, the <electric field> has <energy density> $u=\epsilon_0 E^2/2$. Multiplying by the gap <volume> $Ad$ gives the <electrostatic energy>
$$
U\simeq\frac12\epsilon_0E^2Ad
=\frac12\frac{\epsilon_0 A}{d}V^2
=\frac12CV^2.
$$
For the worked example, $u=4.427\times10^{-4}\,\mathrm{J/m^3}$ and $Ad=10^{-5}\,\mathrm{m^3}$ reproduce $U=4.427\,\mathrm{nJ}$. The <electrostatic energy> resides in the <electric field> throughout the gap, despite overall <charge neutrality>.
When <capacitance> changes, specify the constraint. At fixed plate <electric charge> magnitude $Q$, increasing <capacitance> decreases <capacitor energy> $Q^2/(2C)$; doubling <capacitance> halves it. At fixed terminal <electric potential difference> $V$, increasing <capacitance> increases <capacitor energy> $CV^2/2$; doubling <capacitance> doubles it. In the latter case a connected <battery> supplies additional <electric charge> and <energy>, and there may also be mechanical <work>. Use the <fixed-voltage capacitor energy balance> to account for all transfers before inferring a <force> from changing <capacitor energy>.
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