Electric potential energy

The electrostatic force F acting on a charge q can be written in terms of the electric field E as $${\displaystyle \mathbf {F} =q\mathbf {E} ,}$$

By definition, the change in electrostatic potential energy, U_E, of a point charge q that has moved from the reference position r_ref to position r in the presence of an electric field E is the negative of the work done by the electrostatic force to bring it from the reference position r_ref to that position r.

$${\displaystyle U_{E}(r)-U_{E}(r_{\rm {ref}})=-W_{r_{\rm {ref}}\rightarrow r}=-\int _{{r}_{\rm {ref}}}^{r}q\mathbf {E} \cdot \mathrm {d} \mathbf {s} .}$$

where:

• r = position in 3d space of the charge q, using cartesian coordinates r = (x, y, z), taking the position of the Q charge at r = (0,0,0), the scalar r = |r| is the norm of the position vector,
• ds = differential displacement vector along a path C going from r_ref to r,
• ${\displaystyle W_{r_{\rm {ref}}\rightarrow r}}$ is the work done by the electrostatic force to bring the charge from the reference position r_ref to r,

Usually U_E is set to zero when r_ref is infinity: $${\displaystyle U_{E}(r_{\rm {ref}}=\infty )=0}$$ so $${\displaystyle U_{E}(r)=-\int _{\infty }^{r}q\mathbf {E} \cdot \mathrm {d} \mathbf {s} }$$

When the curl ∇ × E is zero, the line integral above does not depend on the specific path C chosen but only on its endpoints. This happens in time-invariant electric fields. When talking about electrostatic potential energy, time-invariant electric fields are always assumed so, in this case, the electric field is conservative and Coulomb's law can be used.

Using Coulomb's law, it is known that the electrostatic force F and the electric field E created by a discrete point charge Q are radially directed from Q. By the definition of the position vector r and the displacement vector s, it follows that r and s are also radially directed from Q. So, E and ds must be parallel:

$${\displaystyle \mathbf {E} \cdot \mathrm {d} \mathbf {s} =|\mathbf {E} |\cdot |\mathrm {d} \mathbf {s} |\cos(0)=E\mathrm {d} s}$$

Using Coulomb's law, the electric field is given by

$${\displaystyle |\mathbf {E} |=E={\frac {1}{4\pi \varepsilon _{0}}}{\frac {Q}{s^{2}}}}$$

and the integral can be easily evaluated:

$${\displaystyle U_{E}(r)=-\int _{\infty }^{r}q\mathbf {E} \cdot \mathrm {d} \mathbf {s} =-\int _{\infty }^{r}{\frac {1}{4\pi \varepsilon _{0}}}{\frac {qQ}{s^{2}}}{\rm {d}}s={\frac {1}{4\pi \varepsilon _{0}}}{\frac {qQ}{r}}=k_{e}{\frac {qQ}{r}}}$$

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 V₁, which is a function of position r, then $${\displaystyle U_{\mathrm {E} }=q_{2}V_{1}(\mathbf {r} _{2}).}$$

Doing the same calculation with respect to the other charge, we obtain $${\displaystyle U_{\mathrm {E} }=q_{1}V_{2}(\mathbf {r} _{1}).}$$

The electrostatic potential energy is mutually shared by ${\displaystyle q_{1}}$ and ${\displaystyle q_{2}}$, so the total stored energy is $${\displaystyle U_{E}={\frac {1}{2}}\left[q_{2}V_{1}(\mathbf {r} _{2})+q_{1}V_{2}(\mathbf {r} _{1})\right]}$$

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:

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

Using the formula given in (1), the electrostatic potential energy of the system of the three charges will then be: $${\displaystyle U_{\mathrm {E} }={\frac {1}{2}}\left[Q_{1}V(\mathbf {r} _{1})+Q_{2}V(\mathbf {r} _{2})+Q_{3}V(\mathbf {r} _{3})\right]}$$

Where ${\displaystyle V(\mathbf {r} _{1})}$ is the electric potential in r₁ created by charges Q₂ and Q₃, ${\displaystyle V(\mathbf {r} _{2})}$ is the electric potential in r₂ created by charges Q₁ and Q₃, and ${\displaystyle V(\mathbf {r} _{3})}$ is the electric potential in r₃ created by charges Q₁ and Q₂. The potentials are:

$${\displaystyle V(\mathbf {r} _{1})=V_{2}(\mathbf {r} _{1})+V_{3}(\mathbf {r} _{1})={\frac {1}{4\pi \varepsilon _{0}}}{\frac {Q_{2}}{r_{12}}}+{\frac {1}{4\pi \varepsilon _{0}}}{\frac {Q_{3}}{r_{13}}}}$$ $${\displaystyle V(\mathbf {r} _{2})=V_{1}(\mathbf {r} _{2})+V_{3}(\mathbf {r} _{2})={\frac {1}{4\pi \varepsilon _{0}}}{\frac {Q_{1}}{r_{21}}}+{\frac {1}{4\pi \varepsilon _{0}}}{\frac {Q_{3}}{r_{23}}}}$$ $${\displaystyle V(\mathbf {r} _{3})=V_{1}(\mathbf {r} _{3})+V_{2}(\mathbf {r} _{3})={\frac {1}{4\pi \varepsilon _{0}}}{\frac {Q_{1}}{r_{31}}}+{\frac {1}{4\pi \varepsilon _{0}}}{\frac {Q_{2}}{r_{32}}}}$$

Where r_ij is the distance between charge Q_i and Q_j.

If we add everything:

$${\displaystyle U_{\mathrm {E} }={\frac {1}{2}}{\frac {1}{4\pi \varepsilon _{0}}}\left[{\frac {Q_{1}Q_{2}}{r_{12}}}+{\frac {Q_{1}Q_{3}}{r_{13}}}+{\frac {Q_{2}Q_{1}}{r_{21}}}+{\frac {Q_{2}Q_{3}}{r_{23}}}+{\frac {Q_{3}Q_{1}}{r_{31}}}+{\frac {Q_{3}Q_{2}}{r_{32}}}\right]}$$

Finally, we get that the electrostatic potential energy stored in the system of three charges:

$${\displaystyle U_{\mathrm {E} }={\frac {1}{4\pi \varepsilon _{0}}}\left[{\frac {Q_{1}Q_{2}}{r_{12}}}+{\frac {Q_{1}Q_{3}}{r_{13}}}+{\frac {Q_{2}Q_{3}}{r_{23}}}\right]}$$

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
• ${\displaystyle dV}$ is a volume element
• ${\displaystyle \Phi }$ is the electric potential
• ${\displaystyle \varepsilon _{0}}$ 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 ${\displaystyle \Phi (\infty )=0}$, and using ${\displaystyle \nabla \Phi =-\mathbf {E} }$

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

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