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Electric potential energy

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From Wikipedia, the free encyclopedia

Potential energy that results from conservative Coulomb forces

Not to be confused withElectric potential orElectric power.

This article is about the physical magnitude Electric Potential Energy. For electrical energy as a resource and commodity, seeElectrical energy. For energy sources, seeEnergy development. For electricity generation, seeElectricity generation.

Electric potential energy
Common symbols	U_E
SI unit	joule (J)
Derivations from other quantities	U_E =C ·V² / 2

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Electric potential energy is apotential energy (measured injoules) that results fromconservative Coulomb forces and is associated with the configuration of a particular set of pointcharges within a definedsystem. Anobject may be said to have electric potential energy by virtue of either its own electric charge or its relative position to other electrically chargedobjects.

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

Definition

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The electric potential energy of a system of point charges is defined as thework 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 configurationwithout undergoing any acceleration.

The electrostatic potential energy,U_E, of onepoint chargeq at positionr in the presence of anelectric fieldE is defined as the negative of theworkW done by theelectrostatic force to bring it from the reference positionr_ref^{[note 1]} to that positionr.^[1]^[2]^: §25-1

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

whereE is the electrostatic field and dr' is the displacement vector in a curve from the reference positionr_ref to the final positionr.

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

The electrostatic potential energy,U_E, of one point chargeq at positionr in the presence of anelectric potential

V {\displaystyle V}

is defined as the product of the charge and the electric potential.

$U_{\mathrm {E} }(\mathbf {r} )=qV(\mathbf {r} )$

where

V {\displaystyle V}

is theelectric potential generated by the charges, which is a function of positionr.

Units

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TheSI unit of electric potential energy isjoule (named after the English physicistJames Prescott Joule). In theCGS system theerg is the unit of energy, being equal to 10⁻⁷ Joules. Alsoelectronvolts may be used, 1 eV = 1.602×10⁻¹⁹ Joules.

Electrostatic potential energy of one point charge

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One point chargeq in the presence of another point chargeQ

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

The electrostatic potential energy,U_E, of one point chargeq at positionr in the presence of a point chargeQ, taking an infinite separation between the charges as the reference position, is:

$U_{E}(\mathbf {r} )={\frac {1}{4\pi \varepsilon _{0}}}{\frac {qQ}{r}}$

wherer is the distance between the point chargesq andQ, andq andQ are the charges (not the absolute values of the charges—i.e., anelectron 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 andCoulomb's law to this formula.

Outline of proof

The electrostatic forceF acting on a chargeq can be written in terms of the electric fieldE as $\mathbf {F} =q\mathbf {E} ,$

By definition, the change in electrostatic potential energy,U_E, of a point chargeq that has moved from the reference positionr_ref to positionr in the presence of an electric fieldE is the negative of the work done by theelectrostatic force to bring it from the reference positionr_ref to that positionr.

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

where:

r = position in 3d space of the chargeq, using cartesian coordinatesr = (x,y,z), taking the position of theQ charge atr = (0,0,0), the scalarr = |r| is thenorm of the position vector,
ds = differentialdisplacement vector along a pathC going fromr_ref tor,
$W_{r_{\rm {ref}}\rightarrow r}$ is the work done by the electrostatic force to bring the charge from the reference positionr_ref tor,

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

When thecurl∇ ×E is zero, the line integral above does not depend on the specific pathC 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 isconservative and Coulomb's law can be used.

UsingCoulomb's law, it is known that the electrostatic forceF and the electric fieldE created by a discrete point chargeQ are radially directed fromQ. By the definition of the position vectorr and the displacement vectors, it follows thatr ands are also radially directed fromQ. So,E and ds must be parallel:

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

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

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

and the integral can be easily evaluated:

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

One point chargeq in the presence ofn point chargesQ_i

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Electrostatic potential energy ofq due toQ₁ andQ₂ charge system: $U_{E}={\frac {q}{4\pi \varepsilon _{0}}}\left({\frac {Q_{1}}{r_{1}}}+{\frac {Q_{2}}{r_{2}}}\right)$

{\displaystyle U_{E}={\frac {q}{4\pi \varepsilon _{0}}}\left({\frac {Q_{1}}{r_{1}}}+{\frac {Q_{2}}{r_{2}}}\right)} — Electrostatic potential energy ofq due toQ₁ andQ₂ charge system: $U_{E}={\frac {q}{4\pi \varepsilon _{0}}}\left({\frac {Q_{1}}{r_{1}}}+{\frac {Q_{2}}{r_{2}}}\right)$

The electrostatic potential energy,U_E, of one point chargeq in the presence ofn point chargesQ_i, taking an infinite separation between the charges as the reference position, is:

$U_{E}(r)={\frac {q}{4\pi \varepsilon _{0}}}\sum _{i=1}^{n}{\frac {Q_{i}}{r_{i}}},$

wherer_i is the distance between the point chargesq andQ_i, andq andQ_i are the assigned values of the charges.

Electrostatic potential energy stored in a system of point charges

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The electrostatic potential energyU_E stored in a system ofN chargesq₁,q₂, …,q_N at positionsr₁,r₂, …,r_N respectively, is:

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\neq i}}^{N}{\frac {q_{j}}{r_{ij}}},

where, for eachi value, V(r_i) is the electrostatic potential due to all point charges except the one atr_i,^{[note 2]} and is equal to: $V(\mathbf {r} _{i})=k_{e}\sum _{\stackrel {j=1}{j\neq i}}^{N}{\frac {q_{j}}{r_{ij}}},$ wherer_ij is the distance betweenq_i andq_j.

Outline of proof

The electrostatic potential energyU_E stored in a system of two charges is equal to the electrostatic potential energy of a charge in theelectrostatic potential generated by the other. That is to say, if charge q₁ generates an electrostatic potential V₁, which is a function of positionr, then $U_{\mathrm {E} }=q_{2}V_{1}(\mathbf {r} _{2}).$

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

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

This can be generalized to say that the electrostatic potential energyU_E stored in a system ofn chargesq₁,q₂, …,q_n at positionsr₁,r₂, …,r_n respectively, is:

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

Energy stored in a system of one point charge

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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

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Consider bringing a point charge,q, into its final position near a point charge,Q₁. The electric potential V(r) due toQ₁ is $V(\mathbf {r} )=k_{e}{\frac {Q_{1}}{r}}$

Hence we obtain, the electrostatic potential energy ofq in the potential ofQ₁ as $U_{E}={\frac {1}{4\pi \varepsilon _{0}}}{\frac {qQ_{1}}{r_{1}}}$ wherer₁ is the separation between the two point charges.

Energy stored in a system of three point charges

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The electrostatic potential energy of a system of three charges should not be confused with the electrostatic potential energy ofQ₁ due to two chargesQ₂ andQ₃, because the latter doesn't include the electrostatic potential energy of the system of the two chargesQ₂ andQ₃.

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: $U_{\mathrm {E} }={\frac {1}{2}}\left[Q_{1}V(\mathbf {r} _{1})+Q_{2}V(\mathbf {r} _{2})+Q_{3}V(\mathbf {r} _{3})\right]$

Where $V(\mathbf {r} _{1})$ is the electric potential inr₁ created by chargesQ₂ andQ₃, $V(\mathbf {r} _{2})$ is the electric potential inr₂ created by chargesQ₁ andQ₃, and $V(\mathbf {r} _{3})$ is the electric potential inr₃ created by chargesQ₁ andQ₂. The potentials are:

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

Wherer_ij is the distance between chargeQ_i andQ_j.

If we add everything:

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

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

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

Energy stored in an electrostatic field distribution in vacuum

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The energy density, or energy per unit volume, ${\textstyle {\frac {dU}{dV}}}$ , of theelectrostatic 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 electrostaticpotential energy of a continuous charge distribution and put it in terms of theelectrostatic field.

SinceGauss's 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 totalcharge density includingdipole chargesbound in a material
$d V {\displaystyle dV}$ is avolume element
$\Phi$ is theelectric potential
$\varepsilon _{0}$ is thepermittivity of free space,

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

so, now using the following divergence vector identity

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

we have

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

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

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

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

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

Energy stored in electronic elements

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The electric potential energy stored in acapacitor is U_E=⁠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 theJoule effect. Acapacitor stores it in its electric field. The total electrostatic potential energy stored in a capacitor is given by $U_{E}={\frac {1}{2}}QV={\frac {1}{2}}CV^{2}={\frac {Q^{2}}{2C}}$ whereC is thecapacitance,V is theelectric potential difference, andQ thecharge stored in the capacitor.

Outline of proof

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

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

The total work done to fully charge the capacitor in this way is then $W=\int dW=\int _{0}^{Q}V\,dq={\frac {1}{C}}\int _{0}^{Q}q\,dq={\frac {Q^{2}}{2C}}.$ where $Q {\displaystyle Q}$ is the total charge on the capacitor. This work is stored as electrostatic potential energy, hence, $W=U_{E}={\frac {Q^{2}}{2C}}.$

Notably, this expression is only valid if $dq\to 0$ , 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 $U_{E}={\frac {Q^{2}}{2C}}$ which is obtained by a method of charge assembly utilizing the smallest physical charge increment $\Delta q=e$ where $e {\displaystyle e}$ is theelementary unit of charge and $Q=Ne$ where $N {\displaystyle N}$ 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 $U_{E}={\frac {1}{2}}\int _{V}\mathrm {E} \cdot \mathrm {D} \,dV$

where $\mathrm {D}$ is theelectric 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, $\rho$ , $U_{E}={\frac {1}{2}}\int _{V}\rho \Phi \,dV$ where integration is over the entire volume of the dielectric.

These latter two expressions are valid only for cases when the smallest increment of charge is zero ( $dq\to 0$ ) such as dielectrics in the presence of metallic electrodes or dielectrics containing many charges.

Note that a virtual experiment based on the energy transfer between capacitor plates reveals that an additional term should be taken into account when dealing with semiconductors for instance.^[3] While this extra energy cancels when dealing with insulators, the derivation predicts that it cannot be ignored as it may exceed the polarization energy.

Notes

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^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.
^The factor of one half accounts for the 'double counting' of charge pairs. For example, consider the case of just two charges.

References

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^Electromagnetism (2nd edition), I.S. Grant, W.R. Phillips, Manchester Physics Series, 2008ISBN 0-471-92712-0
^Halliday, David; Resnick, Robert; Walker, Jearl (1997). "Electric Potential".Fundamentals of Physics (5th ed.).John Wiley & Sons.ISBN 0-471-10559-7.
^Sallese (2016-06-01)."A new constituent of electrostatic energy in semiconductors".The European Physical Journal B.89 (6): 136.arXiv:1510.06708.doi:10.1140/epjb/e2016-60865-4.ISSN 1434-6036.S2CID 120731496.

External links

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Media related toElectric potential energy at Wikimedia Commons

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