Electric potential energy

{{Short description|Potential energy that results from conservative Coulomb forces}} {{Distinguish|Electric potential|Electric power}} {{About|the physical magnitude Electric Potential Energy|electrical energy as a resource and commodity|Electrical energy|energy sources|Energy development|electricity generation|Electricity generation}} {{More citations needed|date=April 2026}} {{Infobox Physical quantity |bgcolour={default} |name=Electric potential energy |image= |caption= |unit=joule (J) |symbols=U_E |derivations=U_E = C · V² / 2 }} {{Electromagnetism|cTopic=Electrostatics}}

'''Electric 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 be said to have electric potential energy by virtue of either its own electric charge or 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 electric potential energy of a system of point charges is defined as the work required to assemble this system of charges by bringing them close together, as in the system from an infinite distance. Alternatively, the electric potential energy of any given charge or system of charges is termed as the total work done by an external agent in bringing the charge or the system of charges from infinity to the present configuration '''without''' undergoing any acceleration. {{block indent|em=1.2|text=The electrostatic potential energy, ''U_E'', of one point charge ''q'' at position '''r''' 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<ref group="note">The reference zero is usually taken to be a state in which the individual point charges are very well separated ("are at infinite separation") and are at rest.</ref> to that position '''r'''.<ref>Electromagnetism (2nd edition), I.S. Grant, W.R. Phillips, Manchester Physics Series, 2008 {{ISBN|0-471-92712-0}}</ref><ref name="HRW1997">{{cite book |last=Halliday |first=David |author2=Resnick, Robert |author3=Walker, Jearl |title=Fundamentals of Physics |url=https://archive.org/details/fundamentalsofp000davi |url-access=registration |edition=5th |year=1997 |publisher=John Wiley & Sons |chapter=Electric Potential |isbn=0-471-10559-7}}</ref>{{rp|§25-1}}

<math display="block">U_\mathrm{E}(\mathbf r) = -W_{r_\text{ref} \to r } = -\int_{\mathbf{r}_\text{ref}}^\mathbf{r} q\mathbf{E}(\mathbf{r'}) \cdot \mathrm{d} \mathbf{r'}</math>

where '''E''' is the electrostatic field and d'''r'''' is the displacement vector in a curve from the reference position '''r'''_ref to the final position '''r'''.}}

The electrostatic potential energy can also be defined from the electric potential as follows:

{{block indent|em=1.2|text=The electrostatic potential energy, ''U_E'', of one point charge ''q'' at position '''r''' in the presence of an electric potential <math>V</math> is defined as the product of the charge and the electric potential.

<math display="block">U_\mathrm{E}(\mathbf r) = qV(\mathbf r)</math>

where <math>V</math> is the electric potential generated by the charges, which is a function of position '''r'''.}}

==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''===

right|A point charge q in the electric field of another charge Q.|thumb|434px

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:

<math display="block"> U_E(\mathbf r) = \frac{1}{4\pi \varepsilon_0} \frac{qQ}{r}</math>

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.

{{math proof |title=Outline of proof |proof= The electrostatic force '''F''' acting on a charge ''q'' can be written in terms of the electric field '''E''' as <math display="block"> \mathbf{F} = q\mathbf{E} ,</math>

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'''.

<math display="block"> 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} .</math>

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, *d'''s''' = differential displacement vector along a path ''C'' going from '''r'''_ref to '''r''', *<math> W_{r_\text{ref} \to r } </math> 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: <math display="block"> U_E (r_{\rm ref}=\infty) = 0 </math> so <math display="block"> U_E(r) = - \int_\infty^r q\mathbf{E} \cdot \mathrm{d} \mathbf{s} </math>

When the curl {{nowrap|'''∇''' × '''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 d'''s''' must be parallel:

<math display="block"> \mathbf{E} \cdot \mathrm{d} \mathbf{s} = \left|\mathbf{E}\right| \cdot \left|\mathrm{d}\mathbf{s}\right| \cos(0) = E \, \mathrm{d}s </math>

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

<math display="block"> \left|\mathbf{E}\right| = E = \frac{1}{4\pi\varepsilon_0}\frac{Q}{s^2} </math>

and the integral can be easily evaluated:

<math display="block"> 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} \, \mathrm{d}s = \frac{1}{4\pi\varepsilon_0}\frac{qQ}{r} = k_e\frac{qQ}{r} </math>

}}

===One point charge ''q'' in the presence of ''n'' point charges ''Q_i''=== thumb|Electrostatic potential energy of ''q'' due to ''Q''₁ and ''Q''₂ charge system:<math>U_E = \frac{q}{4\pi\varepsilon_0} \left(\frac{Q_1}{r_1} + \frac{Q_2}{r_2} \right) </math> 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:

<math display="block"> U_E(r) = \frac{q}{4\pi\varepsilon_0} \sum_{i=1}^n \frac{Q_i}{r_i},</math>

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 ''q''₁, ''q''₂, …, ''q''_''N'' at positions '''r'''₁, '''r'''₂, …, '''r'''_''N'' respectively, is: {{NumBlk||<math display="block">U_\mathrm{E} = \frac{1}{2} \sum_{i=1}^N q_i V(\mathbf{r}_i) = \frac{1}{2} k_e\sum_{i=1}^N q_i \sum_\stackrel{j=1}{j \ne i}^N \frac{q_j}{r_{ij}},</math> |{{EquationRef|1}}}}

where, for each ''i'' value, V('''r'''_''i'') is the electrostatic potential due to all point charges except the one at '''r'''_''i'',<ref group="note">The factor of one half accounts for the 'double counting' of charge pairs. For example, consider the case of just two charges.</ref> and is equal to: <math display="block">V(\mathbf{r}_i) = k_e\sum_\stackrel{j=1}{j \ne i}^N \frac{q_j}{r_{ij}},</math> where ''r''_''ij'' is the distance between ''q''_''i'' and ''q''_''j''.

{{math proof |title=Outline of proof |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 <math display="block">U_\mathrm{E} = q_2 V_1(\mathbf r_2).</math>

Doing the same calculation with respect to the other charge, we obtain <math display="block">U_\mathrm{E} = q_1 V_2(\mathbf r_1).</math>

The electrostatic potential energy is mutually shared by <math>q_1</math> and <math>q_2</math>, so the total stored energy is <math display="block">U_E = \frac{1}{2}\left[q_2 V_1(\mathbf r_2) + q_1 V_2(\mathbf r_1)\right]</math>

This can be generalized to say that the electrostatic potential energy ''U''_E stored in a system of ''n'' charges ''q''₁, ''q''₂, …, ''q_n'' at positions '''r'''₁, '''r'''₂, …, '''r'''_''n'' respectively, is:

<math display="block">U_\mathrm{E} = \frac{1}{2}\sum_{i=1}^n q_i V(\mathbf{r}_i).</math> }}

===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, ''Q''₁. The electric potential ''V''('''r''') due to ''Q''₁ is <math display="block"> V(\mathbf r) = k_e \frac{Q_1}{r} </math>

Hence we obtain, the electrostatic potential energy of ''q'' in the potential of ''Q''₁ as <math display="block">U_E = \frac{1}{4\pi\varepsilon_0} \frac{q Q_1}{r_1}</math> where ''r''₁ 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 ''Q''₁ due to two charges ''Q''₂ and ''Q''₃, because the latter doesn't include the electrostatic potential energy of the system of the two charges ''Q''₂ and ''Q''₃.

The electrostatic potential energy stored in the system of three charges is: <math display="block">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]</math>

{{math proof |title=Outline of proof |proof= Using the formula given in ({{EquationNote|1}}), the electrostatic potential energy of the system of the three charges will then be: <math display="block">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]</math>

Where <math>V(\mathbf{r}_1)</math> is the electric potential in '''r'''₁ created by charges ''Q''₂ and ''Q''₃, <math>V(\mathbf{r}_2)</math> is the electric potential in '''r'''₂ created by charges ''Q''₁ and ''Q''₃, and <math>V(\mathbf{r}_3)</math> is the electric potential in '''r'''₃ created by charges ''Q''₁ and ''Q''₂. The potentials are:

<math display="block">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}}</math> <math display="block">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}}</math> <math display="block">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}}</math>

Where ''r''_''ij'' is the distance between charge ''Q''_''i'' and ''Q''_''j''.

If we add everything:

<math display="block">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]</math>

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

<math display="block">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]</math>

}}

== Energy stored in an electrostatic field distribution in vacuum ==

The energy density, or energy per unit volume, <math display="inline">\frac{dU}{dV}</math>, of the electrostatic field of a continuous charge distribution is: <math display="block"> u_e = \frac{dU}{dV} = \frac{1}{2} \varepsilon_0 \left|{\mathbf{E}}\right|^2.</math>

{{math proof |title=Outline of proof |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 <math display="block">\mathbf{\nabla}\cdot\mathbf{E} = \frac{\rho}{\varepsilon_0}</math> where * <math>\mathbf{E} </math> is the electric field vector * <math>\rho </math> is the total charge density including dipole charges bound in a material * <math> dV </math> is a volume element * <math>\Phi </math> is the electric potential * <math>\varepsilon_0 </math> is the permittivity of free space, then, <math display="block"> \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} </math>

so, now using the following divergence vector identity

<math display="block">\begin{align} \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}) \end{align}</math>

we have

<math display="block"> 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</math>

using the divergence theorem and taking the area to be at infinity where <math>\Phi(\infty) = 0</math>, and using <math> \nabla \Phi = -\mathbf{E} </math>

<math display="block">\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}</math>

So, the energy density, or energy per unit volume <math display="inline">\frac{dU}{dV}</math> of the electrostatic field is:

<math display="block"> u_e = \frac{1}{2} \varepsilon_0 \left|{\mathbf{E}}\right|^2.</math> }}

== Energy stored in electronic elements== [[File:Electronic component electrolytic capacitors.jpg|right|thumb|150x150px|The electric potential energy stored in a capacitor is U_E={{sfrac|1|2}} CV²]] 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 <math display="block"> U_E = \frac{1}{2}QV = \frac{1}{2} CV^2 = \frac{Q^2}{2C}</math> where ''C'' is the capacitance, ''V'' is the electric potential difference, and ''Q'' the charge stored in the capacitor.

{{math proof |title=Outline of proof |proof= One may assemble charges to a capacitor in infinitesimal increments, <math>dq \to 0</math>, such that the amount of work done to assemble each increment to its final location may be expressed as

<math display="block"> W_q = V \, dq = \frac{q}{C}dq.</math>

The total work done to fully charge the capacitor in this way is then <math display="block"> W = \int dW = \int_0^Q V \, dq = \frac{1}{C} \int_0^Q q \, dq = \frac{Q^2}{2C}.</math> where <math>Q</math> is the total charge on the capacitor. This work is stored as electrostatic potential energy, hence, <math display="block"> W = U_E = \frac{Q^2}{2C}.</math>

Notably, this expression is only valid if <math>dq \to 0</math>, 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 <math display="block"> U_E = \frac{Q^2}{2C}</math> which is obtained by a method of charge assembly utilizing the smallest physical charge increment <math>\Delta q = e</math> where <math>e</math> is the elementary unit of charge and <math>Q = Ne</math> where <math>N</math> 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 <math display="block">U_E = \frac{1}{2} \int_V \mathbf{E} \cdot \mathbf{D} \, dV</math>

where <math>\mathrm{D}</math> 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, <math>\rho</math>, <math display="block">U_E = \frac{1}{2} \int_V \rho \Phi \, dV</math> where integration is over the entire volume of the dielectric.

==Notes== {{reflist|group=note|refs=.}}

== References == <references> . </references>

==External links== *{{Commons category inline}}

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