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*=*qφ*(**r**).

If our potential is due to a charge *q*_{2} at point **r**_{2}, and we bring in from infinity the charge *q*_{1} to point **r**_{1}, then the potential due to *q*_{2} is , and so the potential energy is

.

This can also be seen as the charge *q*_{2} times the potential due to charge *q*_{1} at point **r**_{2}.

For a set of *n* charges *q _{i}* at points

**r**

*, the total potential energy is thus*

_{i}summed for each distinct pair of charges

*q*and

_{i}*q*. If we denote by

_{j}*φ*(

_{i}**r**) the potential due to all charges

*except*

*q*, 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.

_{i}Extending to a localized continuous charge distribution, for a small element

*d*

^{3}

*r*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

*r*

^{2}; 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.

Tags: Electricity, Electrostatics, Friday Physics, Integration by Parts, physics, Potential Energy, work

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