Physics with Calculus/Electromagnetism/Field Energy

From Wikibooks, open books for an open world
< Physics with Calculus
Jump to: navigation, search

Taken from: https://en.wikipedia.org/wiki/Electrostatics#Electrostatic_energy


A single test particle's potential energy,   U_\mathrm{E}^{\text{single}}, can be calculated from a line integral of the work, q_n\vec E\cdot\mathrm d\vec\ell . We integrate from a point at infinity, and assume a collection of N particles of charge Q_n, are already situated at the points \vec r_i. This potential energy (in Joules) is:

  U_\mathrm{E}^{\text{single}}=q\phi(\vec r)=\frac{q }{4\pi \varepsilon_0}\sum_{i=1}^N \frac{Q_i}{\left \|\mathfrak\vec r_i \right \|}

where  \vec\mathfrak r_i =  \vec r - \vec r_i , is the distance of each charge Q_i from the test charge q, which situated at the point \vec r , and  \phi(\vec r) is the electric potential that would be at \vec r if the test charge were not present. If only two charges are present, the potential energy is k_eQ_1Q_2/r. The total electric potential energy due a collection of N charges is calculating by assembling these particles one at a time:

U_\mathrm{E}^{\text{total}} = \frac{1 }{4\pi \varepsilon _0}\sum_{j=1}^N Q_j \sum_{i=1}^{j-1} \frac{Q_i}{r_{ij}}= \frac{1}{2}\sum_{i=1}^N Q_i\phi_i ,

where the following sum from, j = 1 to N, excludes i = j:

\phi_i = \frac{1}{4\pi \varepsilon _0}\sum_{j=1 (j\ne i)}^N \frac{Q_j}{4\pi \varepsilon _0 r_{ij}}.

This electric potential, \phi_i is what would be measured at \vec r_i if the charge Q_i were missing. This formula obviously excludes the (infinite) energy that would be required to assemble each point charge from a disperse cloud of charge. The sum over charges can be converted into an integral over charge density using the prescription \sum (\cdots) \rightarrow \int(\cdots)\rho\mathrm d^3r:

U_\mathrm{E}^{\text{total}} = \frac{1}{2} \int\rho(\vec{r})\phi(\vec{r}) \operatorname{d}^3 r = \frac{\varepsilon_0 }{2} \int  \left|{\mathbf{E}}\right|^2 \operatorname{d}^3 r,

This second expression for electrostatic energy uses the fact that the electric field is the negative gradient of the electric potential, as well as vector calculus identities in a way that resembles integration by parts. These two integrals for electric field energy seem to indicate two mutually exclusive formulas for electrostatic energy density, namely \frac{1}{2}\rho\phi and \frac{\varepsilon_0 }{2}E^2; they yield equal values for the total electrostatic energy only if both are integrated over all space.