Electric potential energy

Not to be confused with Electric potential or Electric power.

This article is about the physical magnitude Electric Potential Energy. For electric energy, see Electric energy. For energy sources, see Energy development. For electricity generation, see Electricity generation.

Electric potential energy
Common symbol(s):	UE
SI unit:	joule (J)
Derivations from other quantities:	UE = C · V2 / 2

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Electric potential energy, or electrostatic potential energy, is a potential energy (measured in joules) that results from conservative Coulomb forces and is associated with the configuration of a particular set of point charges within a defined system. An object may have electric potential energy by virtue of two key elements: its own electric charge and its relative position to other electrically charged objects.

The term "electric potential energy" is used to describe the potential energy in systems with time-variant electric fields, while the term "electrostatic potential energy" is used to describe the potential energy in systems with time-invariant electric fields.

Definition

The electrostatic potential energy, U_E, of one point charge q in the presence of an electric field E is defined as the negative of the work W done by the electrostatic force to bring it from the reference position r_ref to some position r.

$U_\mathrm{E}(\mathbf r) = -W_{r_{\rm ref} \rightarrow r } = -\int_{{r}_{\rm ref}}^r \mathbf{F} \cdot \mathrm{d} \mathbf{s} = q \Phi(\mathbf r)$,

where F is the electrostatic force, ds is the displacement vector and Φ is the electrostatic potential generated by the charges, which is a function of position r.

Units

The SI unit of electric potential energy is the 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 J. Also electronvolts may be used, 1 eV = 1.602×10⁻¹⁹ J.

Examples

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

A point charge q in the electric field of another charge Q.

Let's start first with one point charge q in the presence of only one point charge Q. The electrostatic potential energy, U_E, of one point charge q in the presence of a point charge Q, taking an infinite separation between the charges as the reference position, is:

$U_E(r) = k_e\frac{qQ}{r}$,

where $k_e = \frac{1}{4\pi\varepsilon_0}$ is Coulomb's constant, r is the distance between the point charges q & Q, and q & Q are the signed values of the charges (not the modules of the charges. For example, 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 the given formula.

Outline of proof

The small amount of work dW done by a given force over a small distance ds is given by $\mathrm{d} W = \mathbf{F}\cdot\mathrm{d} \mathbf{s}$, The sum of these small amounts of work over the trajectory yields the work, $W_{s_{1} \rightarrow s_{2}} = \int_{s_1}^{s_2}\bold{F} \cdot \mathrm{d} \mathbf{s}$. By definition, the electrostatic potential energy, UE, of one point charge q 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 rref to some position r. $U_E(r) - U_E(r_{\rm ref}) = -W_{r_{\rm ref} \rightarrow r } = -\int_{{r}_{\rm ref}}^r \mathbf{F} \cdot \mathrm{d} \mathbf{s}$. where: r = position in 3d space, using cartesian coordinates r = (x, y, z), taking the position of the Q charge at r = (0,0,0), r = |r| is the norm of the position vector, dr = differential position vector, ds = differential displacement vector, $\scriptstyle W_{r_{\rm ref} \rightarrow r }$ is the work done by the electrostatic force to bring the charge from the reference position rref to r, F = force exerted on charge q by Q, E = electric field due to Q. Usually UE is set to zero when rref is infinity: $U_E (r_{\rm ref}=\infty) = 0 \,\!$ so $U_E(r) = -\int_\infty^r \mathbf{F} \cdot \mathrm{d} \mathbf{s} = -q \int_\infty^r \mathbf{E} \cdot \mathrm{d} \mathbf{s} \,\!$ Since E and therefore F,r, and s, are all radially directed from Q, F and ds must be antiparallel and so $\mathbf{F} \cdot \mathrm{d} \mathbf{s} = |\mathbf{F}| \cdot |\mathrm{d}\mathbf{s}|\cos(\pi) = - F \mathrm{d}s = F \mathrm{d}r \,\!$ using Coulomb's law: $F = \frac{1}{4\pi\varepsilon_0}\frac{qQ}{r^2} \,\!$ The electric field is conservative so the work is path independent allowing the integral to easily be evaluated: $U_E(r) = -\int_\infty^r \mathbf{F} \cdot \mathrm{d} \mathbf{s} = -\int_\infty^r \frac{1}{4\pi\varepsilon_0}\frac{qQ}{r^2}{\rm d}r = \frac{1}{4\pi\varepsilon_0}\frac{qQ}{r} \,\!$ Sometimes the factor ke called Coulomb's constant is used in these expressions. In SI units, the Coulomb constant is given by $k_e = \frac{1}{4 \pi \varepsilon_0}$, in turn $\varepsilon_0$ is the electric constant.

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

$U_E(r) = k_e q \sum_{i=1}^n \frac{Q_i}{r_i}$,

where $k_e = \frac{1}{4\pi\varepsilon_0}$ is Coulomb's constant, r_i is the distance between the point charges q & Q_i, and q & Q_i are the signed values of the charges.

Electrostatic potential energy stored in a system of point charges

In a Two point charges system, Electric potential energy U_E of q in the potential well created by Q₁ is $U_E = \frac{1}{4\pi\varepsilon_0} \frac{q Q_1}{r}$

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 Φ₁, which is a function of position r, then

$U_\mathrm{E} = q_2 \Phi_1(\mathbf r_2).$

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

$U_\mathrm{E} = q_1 \Phi_2(\mathbf r_1).$

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:

$U_\mathrm{E} = \frac{1}{2}\sum_{i=1}^N q_i \Phi(\mathbf{r}_i)$,

(1)

where, for each i value, Φ(r_i) is the electrostatic potential due to all point charges except the one at r_i.

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 potential against which an external agent must do work in moving the point charge from infinity to its final location.

Two point charges

Consider bringing a point charge, q, into its final position in the vicinity of a point charge, Q₁. The electrostatic potential Φ(r) due to Q₁ is

$\Phi(r) = k_e \frac{Q_1}{r}$

Hence we obtain, the electric potential energy of q in the potential of Q₁ as

$U_E = \frac{1}{4\pi\varepsilon_0} \frac{q Q_1}{ r_1 }$

where r₁is the separation between the two point charges.

Three point charges

Electrostatic potential energy of q due to Q₁ and Q₂ charge system, :$U_E = q\frac{1}{4 \pi \varepsilon_0} \left(\frac{Q_1}{r_1} + \frac{Q_2}{r_2} \right)$

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:

$U_\mathrm{E} = \frac{1}{4\pi\varepsilon_0} \left( \frac{Q_1 Q_2}{r_{12}} + \frac{Q_1 Q_3}{r_{13}} + \frac{Q_2 Q_3}{r_{23}} \right)$

Outline of proof

Using the formula given in (1), the electrostatic potential energy of the system of the three charges will then be: $U_\mathrm{E} = \frac{1}{2} ( Q_1 \Phi(\mathbf{r}_1) + Q_2 \Phi(\mathbf{r}_2) + Q_3 \Phi(\mathbf{r}_3) )$ Where $\Phi(\mathbf{r}_1)$ is the electric potential in r1 created by charges Q2 and Q3, $\Phi(\mathbf{r}_2)$ is the electric potential in r2 created by charges Q1 and Q3, and $\Phi(\mathbf{r}_3)$ is the electric potential in r3 created by charges Q1 and Q2. The potentials are: $\Phi(\mathbf{r}_1) = \Phi_2(\mathbf{r}_1) + \Phi_3(\mathbf{r}_1) = \frac{1}{4\pi\varepsilon_0} \frac{Q_2}{r_{12}} + \frac{1}{4\pi\varepsilon_0} \frac{Q_3}{r_{13}}$ $\Phi(\mathbf{r}_2) = \Phi_1(\mathbf{r}_2) + \Phi_3(\mathbf{r}_2) = \frac{1}{4\pi\varepsilon_0} \frac{Q_1}{r_{21}} + \frac{1}{4\pi\varepsilon_0} \frac{Q_3}{r_{23}}$ $\Phi(\mathbf{r}_3) = \Phi_1(\mathbf{r}_3) + \Phi_2(\mathbf{r}_3) = \frac{1}{4\pi\varepsilon_0} \frac{Q_1}{r_{31}} + \frac{1}{4\pi\varepsilon_0} \frac{Q_2}{r_{32}}$ Where rab is the distance between charge Qa and Qb. If we add everything: $U_\mathrm{E} = \frac{1}{2} \frac{1}{4\pi\varepsilon_0} ( \frac{Q_1 Q_2}{r_{12}} + \frac{Q_1 Q_3}{r_{13}} + \frac{Q_2 Q_1}{r_{21}} + \frac{Q_2 Q_3}{r_{23}} + \frac{Q_3 Q_1}{r_{31}} + \frac{Q_3 Q_2}{r_{32}})$ Finally, we get that the electrostatic potential energy stored in the system of three charges: $U_\mathrm{E} = \frac{1}{4\pi\varepsilon_0} ( \frac{Q_1 Q_2}{r_{12}} + \frac{Q_1 Q_3}{r_{13}} + \frac{Q_2 Q_3}{r_{23}})$

Energy stored in an electrostatic field distribution

The energy density, or energy per unit volume, $\frac{dU}{dV}$, of the electrostatic field of a continuous charge distribution is:

$u_e = \frac{dU}{dV} = \frac{1}{2} \varepsilon_0 \left|{\mathbf{E}}\right|^2.$

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' law for electrostatic field in differential form states $\mathbf{\nabla}\cdot\mathbf{E} = \frac{\rho}{\varepsilon_0}$ where $\mathbf{E} \$ is the electric field vector $\rho \$ is the total charge density including dipole charges bound in a material $\varepsilon_0$ is the permittivity of free space, then, $\begin{align} U & = \frac{1}{2}\int \limits_{\text{all space}} \rho(r) \Phi(r) \, dV \\ & = \frac{1}{2}\int \limits_{\text{all space}} \varepsilon_0(\mathbf{\nabla}\cdot{\mathbf{E}})\Phi \, dV \end{align}$ so, now using the following divergence vector identity $\nabla\cdot(\bold{A}{B}) = (\nabla\cdot\bold{A}){B} + \bold{A}\cdot(\nabla{B}) \Rightarrow (\nabla\cdot\bold{A}){B} = \nabla\cdot(\bold{A}{B}) - \bold{A}\cdot(\nabla{B})$ we have $U = \frac{\varepsilon_0}{2}\int \limits_{\text{all space}} \mathbf{\nabla}\cdot(\mathbf{E}\Phi) dV - \frac{\varepsilon_0}{2}\int \limits_{\text{all space}} (\mathbf{\nabla}\Phi)\cdot\mathbf{E} dV$ using the divergence theorem and taking the area to be at infinity where $\Phi(\infty) = 0$ $\begin{align} U & = \overbrace{\frac{\varepsilon_0}{2}\int\limits_{{}^\text{boundary}_\text{ of space}} \Phi\mathbf{E}\cdot d\mathbf A}^{0} - \frac{\varepsilon_0}{2}\int \limits_{\text{all space}} (-\mathbf{E})\cdot\mathbf{E} \, dV \\ & = \int \limits_{\text{all space}} \frac{1}{2}\varepsilon_0\left|{\mathbf{E}}\right|^2 \, dV. \end{align}$ So, the energy density, or energy per unit volume $\frac{dU}{dV}$ of the electrostatic field is: $u_e = \frac{1}{2} \varepsilon_0 \left|{\mathbf{E}}\right|^2.$

Energy 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 electric potential energy stored in a capacitor is given by

$U_E = \frac{1}{2}QV = \frac{1}{2} CV^2 = \frac{Q^2}{2C}$

where C is the capacitance, V is the electric potential difference, and Q the charge stored in the capacitor.