Electric potential energy

Units

The SI unit of electric potential energy is joule (named after the English physicist James Prescott Joule). In the CGS system the erg is the unit of energy, being equal to 10⁻⁷ Joules. Also electronvolts may be used, 1 eV = 1.602×10⁻¹⁹ Joules.

Electrostatic potential energy of one point charge

One point charge q in the presence of another point charge Q

The electrostatic potential energy, U_E, of one point charge q at position r in the presence of a point charge Q, taking an infinite separation between the charges as the reference position, is:

$${\displaystyle U_E(\mathbf{r})=\frac{1}{4\pi {\epsilon }_0}\frac{qQ}{r}}$$

where r is the distance between the point charges q and Q, and q and Q are the charges (not the absolute values of the charges—i.e., an electron would have a negative value of charge when placed in the formula). The following outline of proof states the derivation from the definition of electric potential energy and Coulomb's law to this formula.

Outline of proof

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.

$${\displaystyle U_E(r)-U_E(r_{\mathrm{r}\mathrm{e}\mathrm{f}})=-W_{r_{\mathrm{r}\mathrm{e}\mathrm{f}}\to r}=-{\int }_{r_{\mathrm{r}\mathrm{e}\mathrm{f}}}^{r}q\mathbf{E}\cdot \mathrm{d}\mathbf{s}.}$$

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,
• ${\displaystyle W_{r_{\text{ref}}\to r}}$ 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: $${\displaystyle U_E(r_{\mathrm{r}\mathrm{e}\mathrm{f}}=\mathrm{\infty })=0}$$ so $${\displaystyle U_E(r)=-{\int }_{\mathrm{\infty }}^{r}q\mathbf{E}\cdot \mathrm{d}\mathbf{s}}$$

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:

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

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

$${\displaystyle \left|\mathbf{E}\right|=E=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{s^2}}$$

and the integral can be easily evaluated:

$${\displaystyle U_E(r)=-{\int }_{\mathrm{\infty }}^{r}q\mathbf{E}\cdot \mathrm{d}\mathbf{s}=-{\int }_{\mathrm{\infty }}^r\frac{1}{4\pi {\epsilon }_0}\frac{qQ}{s^2}\mathrm{d}s=\frac{1}{4\pi {\epsilon }_0}\frac{qQ}{r}=k_e\frac{qQ}{r}}$$

One point charge q in the presence of n point charges Qi

The electrostatic potential energy, U_E, of one point charge q in the presence of n point charges Q_i, taking an infinite separation between the charges as the reference position, is:

$${\displaystyle U_E(r)=\frac{q}{4\pi {\epsilon }_0}\sum _{i=1}^n\frac{Q_i}{r_i},}$$

where r_i is the distance between the point charges q and Q_i, and q and Q_i are the assigned values of the charges.

Electrostatic potential energy stored in a system of point charges

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:

where, for each i value, V(r_i) is the electrostatic potential due to all point charges except the one at r_i, and is equal to: $${\displaystyle V({\mathbf{r}}_i)=k_e\sum _{\stackrel{j=1}{j\ne i}}^N\frac{q_j}{r_{ij}},}$$ where r_ij is the distance between q_i and q_j.

Outline of proof

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:

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

Energy stored in a system of one point charge

The electrostatic potential energy of a system containing only one point charge is zero, as there are no other sources of electrostatic force against which an external agent must do work in moving the point charge from infinity to its final location.

A common question arises concerning the interaction of a point charge with its own electrostatic potential. Since this interaction doesn't act to move the point charge itself, it doesn't contribute to the stored energy of the system.

Energy stored in a system of two point charges

Consider bringing a point charge, q, into its final position near a point charge, Q₁. The electric potential V(r) due to Q₁ is $${\displaystyle V(\mathbf{r})=k_e\frac{Q_1}{r}}$$

Hence we obtain, the electrostatic potential energy of q in the potential of Q₁ as $${\displaystyle U_E=\frac{1}{4\pi {\epsilon }_0}\frac{q{Q}_1}{r_1}}$$ where r₁ is the separation between the two point charges.

Energy stored in a system of three point charges

The electrostatic potential energy of a system of three charges should not be confused with the electrostatic potential energy of Q₁ due to two charges Q₂ and Q₃, because the latter doesn't include the electrostatic potential energy of the system of the two charges Q₂ and Q₃.

The electrostatic potential energy stored in the system of three charges is: $${\displaystyle U_{\mathrm{E}}=\frac{1}{4\pi {\epsilon }_0}\left[\frac{Q_1{Q}_2}{r_{12}}+\frac{Q_1{Q}_3}{r_{13}}+\frac{Q_2{Q}_3}{r_{23}}\right]}$$

Outline of proof

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:

$${\displaystyle V({\mathbf{r}}_1)=V_2({\mathbf{r}}_1)+V_3({\mathbf{r}}_1)=\frac{1}{4\pi {\epsilon }_0}\frac{Q_2}{r_{12}}+\frac{1}{4\pi {\epsilon }_0}\frac{Q_3}{r_{13}}}$$ $${\displaystyle V({\mathbf{r}}_2)=V_1({\mathbf{r}}_2)+V_3({\mathbf{r}}_2)=\frac{1}{4\pi {\epsilon }_0}\frac{Q_1}{r_{21}}+\frac{1}{4\pi {\epsilon }_0}\frac{Q_3}{r_{23}}}$$ $${\displaystyle V({\mathbf{r}}_3)=V_1({\mathbf{r}}_3)+V_2({\mathbf{r}}_3)=\frac{1}{4\pi {\epsilon }_0}\frac{Q_1}{r_{31}}+\frac{1}{4\pi {\epsilon }_0}\frac{Q_2}{r_{32}}}$$

Where r_ij is the distance between charge Q_i and Q_j.

If we add everything:

$${\displaystyle U_{\mathrm{E}}=\frac{1}{2}\frac{1}{4\pi {\epsilon }_0}\left[\frac{Q_1{Q}_2}{r_{12}}+\frac{Q_1{Q}_3}{r_{13}}+\frac{Q_2{Q}_1}{r_{21}}+\frac{Q_2{Q}_3}{r_{23}}+\frac{Q_3{Q}_1}{r_{31}}+\frac{Q_3{Q}_2}{r_{32}}\right]}$$

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

$${\displaystyle U_{\mathrm{E}}=\frac{1}{4\pi {\epsilon }_0}\left[\frac{Q_1{Q}_2}{r_{12}}+\frac{Q_1{Q}_3}{r_{13}}+\frac{Q_2{Q}_3}{r_{23}}\right]}$$

Energy stored in an electrostatic field distribution in vacuum

The energy density, or energy per unit volume, $\frac{dU}{dV}$, of the electrostatic field of a continuous charge distribution is: $${\displaystyle u_e=\frac{dU}{dV}=\frac{1}{2}{\epsilon }_0{\left|\mathbf{E}\right|}^2.}$$

Outline of proof

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 $${\displaystyle \mathrm{\nabla }\cdot \mathbf{E}=\frac{\rho }{{\epsilon }_0}}$$ 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
• ${\displaystyle \mathrm{\Phi }}$ is the electric potential
• ${\displaystyle {\epsilon }_0}$ is the permittivity of free space,

then, $${\displaystyle \begin{array}{cc}U& =\frac{1}{2}\underset{\text{all space}}{\int }\rho (r)\mathrm{\Phi }(r)dV\\ & =\frac{1}{2}\underset{\text{all space}}{\int }{\epsilon }_0(\mathrm{\nabla }\cdot \mathbf{E})\mathrm{\Phi }dV\end{array}}$$

so, now using the following divergence vector identity

$${\displaystyle \begin{array}{cc}\mathrm{\nabla }\cdot (\mathbf{A}B)& =(\mathrm{\nabla }\cdot \mathbf{A})B+\mathbf{A}\cdot (\mathrm{\nabla }B)\\ \Rightarrow (\mathrm{\nabla }\cdot \mathbf{A})B& =\mathrm{\nabla }\cdot (\mathbf{A}B)-\mathbf{A}\cdot (\mathrm{\nabla }B)\end{array}}$$

we have

$${\displaystyle U=\frac{{\epsilon }_0}{2}\underset{\text{all space}}{\int }\mathrm{\nabla }\cdot (\mathbf{E}\mathrm{\Phi })dV-\frac{{\epsilon }_0}{2}\underset{\text{all space}}{\int }(\mathrm{\nabla }\mathrm{\Phi })\cdot \mathbf{E}dV}$$

using the divergence theorem and taking the area to be at infinity where ${\displaystyle \mathrm{\Phi }(\mathrm{\infty })=0}$, and using ${\displaystyle \mathrm{\nabla }\mathrm{\Phi }=-\mathbf{E}}$

$${\displaystyle \begin{array}{cc}U& =\stackrel{0}{\overbrace{\frac{{\epsilon }_0}{2}\underset{{}_{\text{ of space}}^{\text{boundary}}}{\int }\mathrm{\Phi }\mathbf{E}\cdot d\mathbf{A}}}-\frac{{\epsilon }_0}{2}\underset{\text{all space}}{\int }(-\mathbf{E})\cdot \mathbf{E}dV\\ & =\underset{\text{all space}}{\int }\frac{1}{2}{\epsilon }_0{\left|\mathbf{E}\right|}^{2}dV.\end{array}}$$

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

$${\displaystyle u_e=\frac{1}{2}{\epsilon }_0{\left|\mathbf{E}\right|}^2.}$$

Energy stored in electronic elements

Some elements in a circuit can convert energy from one form to another. For example, a resistor converts electrical energy to heat. This is known as the Joule effect. A capacitor stores it in its electric field. The total electrostatic potential energy stored in a capacitor is given by $${\displaystyle U_E=\frac{1}{2}QV=\frac{1}{2}C{V}^2=\frac{Q^2}{2C}}$$ where C is the capacitance, V is the electric potential difference, and Q the charge stored in the capacitor.

Outline of proof

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

$${\displaystyle W_q=Vdq=\frac{q}{C}dq.}$$

The total work done to fully charge the capacitor in this way is then $${\displaystyle W=\int dW={\int }_0^{Q}Vdq=\frac{1}{C}{\int }_0^{Q}qdq=\frac{Q^2}{2C}.}$$ 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 ${\displaystyle \mathrm{\Delta }q=e}$ 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.

The total electrostatic potential energy may also be expressed in terms of the electric field in the form $${\displaystyle U_E=\frac{1}{2}{\int }_V\mathbf{E}\cdot \mathbf{D}dV}$$

where ${\displaystyle \mathrm{D}}$ is the electric displacement field within a dielectric material and integration is over the entire volume of the dielectric.

The total electrostatic potential energy stored within a charged dielectric may also be expressed in terms of a continuous volume charge, ${\displaystyle \rho }$, $${\displaystyle U_E=\frac{1}{2}{\int }_V\rho \mathrm{\Phi }dV}$$ where integration is over the entire volume of the dielectric.