## Physics Friday 126

Part 6: Work, Electric Potential, and Energy Density

In electrostatics, the electric potential can be interpreted as the potential energy per unit charge of a test charge at the point in question. In particular, if we have a localized charge distribution creating fields with a scalar potential φ, which goes to zero at infinite distance from the distribution; and if we bring (sufficiently slowly) a point charge q from infinity to the point r, when the work done on this charge, and thus the potential energy (relative to zero potential energy at infinite separation), is simply W=(r).
If our potential is due to a charge q2 at point r2, and we bring in from infinity the charge q1 to point r1, then the potential due to q2 is , and so the potential energy is
.
This can also be seen as the charge q2 times the potential due to charge q1 at point r2.
For a set of n charges qi at points ri, the total potential energy is thus
 summed for each distinct pair of charges qi and qj. If we denote by φi(r) the potential due to all charges except qi, then we see that  is the total potential energy; the 1/2 term out front is because the above sum counts each pair of charges twice.
Extending to a localized continuous charge distribution, for a small element d3r at the point r, the charge is , and the potential is ; so, the above sum is replaced with an integral over space:
.
Now, the differential form of Gauss’ law (or Coulomb’s law) is
,
and since, by definition of the potential, we have
,
substituting this latter into the former tells us that
.
Substituting this last into our integral for energy, we obtain
.
Using vector integration by parts (see here), one can see that
.
Now, letting the volume increase, we recall that our charge distribution is localized; examining the surface integral, we note that far from the distribution, the potential φ goes like r-1, and the magnitude of  goes as r-2; their product has a magnitude that goes as r-3. In contrast, the area of the surface goes as r2; thus, as the volume is expanded to encompass all space, the surface integral vanishes, and so
.
Substituting this into our potential energy formula, this leads to the result that the potential energy, the work needed to assemble the charge distribution from infinite separation, is
.
This is the integral over all space of the quantity ; this quantity thus has units of energy density. We identify this integrand as the energy density of the electric field.

### 3 Responses to “Physics Friday 126”

1. Physics Friday 128 « Twisted One 151's Weblog Says:

[…] Friday 128 By twistedone151 Part 8: Work, Induction, and Magnetic Energy Density Previously, we derived the energy density of the electric field by considering the work done in assembling a […]

2. Physics Friday 131 « Twisted One 151's Weblog Says:

[…] in terms of the fields instead. In SI units, we have , so , and so Now, , so , and we recall from here that the term in the parentheses above is the energy density of the electric field UE, so . Next, […]

3. Physics Friday 136 « Twisted One 151's Weblog Says:

[…] time) as , which reduces to the linear or circular case in each of those limits. Also, since the energy density for electric fields is , and the energy density for magnetic fields is . Applying these to our wave, and taking the […]