# Electric fields and energy

## Homework Statement

a. Calculate the energy density of the electric field at a distance r from an electron (presumed to be a particle) at rest.
b. Assume now that the electron is not a point but a sphere of radius R over whose surface the electron charge is uniformly distributed. Determine the energy associated with the external electric field in vacuum of the electron as a function of R.

## Homework Equations

$$u_e = 1/2\epsilon_0E^2$$

## The Attempt at a Solution

a. The electric field of an electron can be assumed to be the same as a point charge, that is
$$E = \frac{q}{4\pi\epsilon_0r^2}$$
Since
$$E^2 = \frac{q^2}{16\pi(\epsilon_0)^2r^4}$$,
$$u_e = \frac{q^2}{32\pi \epsilon_0 r^4}$$

b. We use Gauss's law to find the electric field of this sphere.
$$EA = \frac{\sigma A}{\epsilon_0}$$
So that
$$E = \frac{\sigma}{\epsilon_0}$$, where $$\sigma$$ is the charge per unit area.
So the energy density is
$$u_e = \frac{1}{2}\epsilon_0\frac{\sigma}{(\epsilon_0)^2} = \frac{1}{2} \frac {\sigma^2}{\epsilon_0}$$
The total energy is therefore the energy density multiplied by the volume, so
$$U = \frac{4\sigma^2\pi R^3}{6\epsilon_0}$$

Is this correct?

## Answers and Replies

Orodruin
Staff Emeritus
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The field inside the sphere is zero so this volume does not give any energy. The volume where the electric field is non-zero (and thus has non-zero energy) is outside of the sphere. Also, when you apply Gauss' law, what Gaussian surface are you using? You have an ##A## on both sides of your resulting equation, what area is this? Depending on the Gaussian surface, is it really the area of the sphere?

A hint is that the electric field outside of the sphere is no different from that of a point charge with the same charge, which is likely why (a) is part of this problem.

Thanks. For Gauss's Law, I am using a Gaussian sphere to encompass the sphere. I do know for a fact that the electric field outside of the sphere is the same as a point charge - I guess I used Gauss's law because I thought the problem was testing my knowledge of finding E-fields and using that to find energy.

Anyways, even if E-field is like a point charge - that would still be expressed in terms of r, the distance from the sphere's center to the Gaussian surface right? How would I express it in terms of R? Should I make it so that my Gaussian surface is infinitely close to the sphere's surface, and then r would be the same as R and I could replace that in my equation for (a)? And then to find the total energy, since there is only charge on the surface, do I multiply the energy density by the surface of the sphere, $$4\pi\ r^2$$?

Orodruin
Staff Emeritus