How Does Electron Shape Affect Electric Field Energy?

AI Thread Summary
The discussion focuses on calculating the energy density of the electric field around an electron, initially treated as a point charge, and then as a uniformly charged sphere. For the point charge, the energy density is derived using the formula u_e = 1/2ε₀E², leading to u_e = q²/(32πε₀r⁴). When considering the electron as a sphere, Gauss's law is applied to find the electric field, resulting in the total energy being U = 4σ²πR³/(6ε₀). The participants clarify that the electric field inside the sphere is zero and emphasize the importance of using the correct Gaussian surface for calculations. The discussion concludes with a focus on relating energy density to total energy, highlighting the need to integrate over the volume where the electric field exists.
mintsnapple
Messages
50
Reaction score
0

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?
 
Physics news on Phys.org
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, $$ 4\pi\ r^2 $$?
 
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.)
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top