Electric fields and energy

  • #1

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

  • #2
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
2021 Award
17,691
7,644
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.
 
  • #3
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, $$ 4\pi\ r^2 $$?
 
  • #4
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
2021 Award
17,691
7,644
Well, you know the energy density outside of the sphere (inside it is zero since there is no field) because you already solved (a). How do you relate the energy density to the total energy? (The energy density is an energy per volume.)
 

Related Threads on Electric fields and energy

Replies
10
Views
2K
  • Last Post
Replies
1
Views
4K
  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
3
Views
3K
Replies
4
Views
4K
Replies
3
Views
1K
Replies
3
Views
1K
Replies
14
Views
265
Replies
5
Views
13K
  • Last Post
Replies
5
Views
20K
Top