Electric potential energy

The electrostatic potential energy U_E stored in a system of two charges is equal to the electrostatic potential energy of a charge in the electrostatic potential generated by the other. That is to say, if charge q₁ generates an electrostatic potential Φ₁, which is a function of position r, then

$${\displaystyle U_{\mathrm {E} }=q_{2}\Phi _{1}(\mathbf {r} _{2}).}$$

Doing the same calculation with respect to the other charge, we obtain

$${\displaystyle U_{\mathrm {E} }=q_{1}\Phi _{2}(\mathbf {r} _{1}).}$$

The electrostatic potential energy is mutually shared by ${\displaystyle q_{1}}$ and ${\displaystyle q_{2}}$, so the total stored energy is

$${\displaystyle U_{E}={\frac {1}{2}}\left[q_{2}\Phi _{1}(\mathbf {r} _{2})+q_{1}\Phi _{2}(\mathbf {r} _{1})\right]}$$

This can be generalized to say that the electrostatic potential energy U_E stored in a system of N charges q₁, q₂, …, q_N at positions r₁, r₂, …, r_N respectively, is:

$${\displaystyle U_{\mathrm {E} }={\frac {1}{2}}\sum _{i=1}^{N}q_{i}\Phi (\mathbf {r} _{i}).}$$

One may take the equation for the electrostatic potential energy of a continuous charge distribution and put it in terms of the electrostatic field.

Since Gauss's law for electrostatic field in differential form states

$${\displaystyle \mathbf {\nabla } \cdot \mathbf {E} ={\frac {\rho }{\varepsilon _{0}}}}$$

where

• ${\displaystyle \mathbf {E} }$ is the electric field vector
• ${\displaystyle \rho }$ is the total charge density including dipole charges bound in a material
• ${\displaystyle \varepsilon _{0}}$ is the permittivity of free space,

then,

$${\displaystyle {\begin{aligned}U&={\frac {1}{2}}\int \limits _{\text{all space}}\rho (r)\Phi (r)\,dV\\&={\frac {1}{2}}\int \limits _{\text{all space}}\varepsilon _{0}(\mathbf {\nabla } \cdot {\mathbf {E} })\Phi \,dV\end{aligned}}}$$

so, now using the following divergence vector identity

$${\displaystyle \nabla \cdot (\mathbf {A} {B})=(\nabla \cdot \mathbf {A} ){B}+\mathbf {A} \cdot (\nabla {B})\Rightarrow (\nabla \cdot \mathbf {A} ){B}=\nabla \cdot (\mathbf {A} {B})-\mathbf {A} \cdot (\nabla {B})}$$

we have

$${\displaystyle U={\frac {\varepsilon _{0}}{2}}\int \limits _{\text{all space}}\mathbf {\nabla } \cdot (\mathbf {E} \Phi )dV-{\frac {\varepsilon _{0}}{2}}\int \limits _{\text{all space}}(\mathbf {\nabla } \Phi )\cdot \mathbf {E} dV}$$

using the divergence theorem and taking the area to be at infinity where ${\displaystyle \Phi (\infty )=0}$

$${\displaystyle {\begin{aligned}U&=\overbrace {{\frac {\varepsilon _{0}}{2}}\int \limits _{{}_{\text{ of space}}^{\text{boundary}}}\Phi \mathbf {E} \cdot d\mathbf {A} } ^{0}-{\frac {\varepsilon _{0}}{2}}\int \limits _{\text{all space}}(-\mathbf {E} )\cdot \mathbf {E} \,dV\\&=\int \limits _{\text{all space}}{\frac {1}{2}}\varepsilon _{0}\left|{\mathbf {E} }\right|^{2}\,dV.\end{aligned}}}$$

So, the energy density, or energy per unit volume ${\textstyle {\frac {dU}{dV}}}$ of the electrostatic field is:

$${\displaystyle u_{e}={\frac {1}{2}}\varepsilon _{0}\left|{\mathbf {E} }\right|^{2}.}$$

One may assemble charges to a capacitor in infinitesimal increments, ${\displaystyle dq\to 0}$, such that the amount of work done to assemble each increment to its final location may be expressed as

$${\displaystyle W_{q}=V\,dq={\frac {q}{C}}dq.}$$

The total work done to fully charge the capacitor in this way is then

$${\displaystyle W=\int dW=\int _{0}^{Q}V\,dq={\frac {1}{C}}\int _{0}^{Q}q\,dq={\frac {Q^{2}}{2C}}.}$$

where ${\displaystyle Q}$ is the total charge on the capacitor. This work is stored as electrostatic potential energy, hence,

$${\displaystyle W=U_{E}={\frac {Q^{2}}{2C}}.}$$

Notably, this expression is only valid if ${\displaystyle dq\to 0}$, which holds for many-charge systems such as large capacitors having metallic electrodes. For few-charge systems the discrete nature of charge is important. The total energy stored in a few-charge capacitor is

$${\displaystyle U_{E}={\frac {Q^{2}}{C}}}$$

which is obtained by a method of charge assembly utilizing the smallest physical charge increment ${\displaystyle \Delta q=e}$ where ${\displaystyle e}$ is the elementary unit of charge and ${\displaystyle Q=Ne}$ where ${\displaystyle N}$ is the total number of charges in the capacitor.