Electric potential energy

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.

^\mathbf{r} q\mathbf{E}(\mathbf{r'}) \cdot \mathrm{d} \mathbf{r'}</math>

where E is the electrostatic field and dr' 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:

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.

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

thumb|Electrostatic potential energy of q due to Q₁ and Q₂ 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 q₁, q₂, …, q_N at positions r₁, r₂, …, r_N respectively, is:

,</math>

|

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:

<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.

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

<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 Q₁ as

<math display="block">U_E = \frac{1}{4\pi\varepsilon_0} \frac{q Q_1}{r_1}</math>

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:

<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>

+ \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>

\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_ \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

right|thumb|150x150px|The electric potential energy stored in a [[capacitor is U_E= 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.

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

References