Errors in Electrostatic Self-Energy Calculation?

  • Thread starter Thread starter smantics
  • Start date Start date
  • Tags Tags
    Electrostatic
smantics
Messages
11
Reaction score
0

Homework Statement


Find an expression for the electrostatic self-energy of an arbitrary spherically symmetric charge density distribution ρ(r). You may not assume that ρ(r) represents any point charge, or that it is constant, or that it is piecewise constant, or that it does or does not cut off at any finite radius r. Your expression must cover all possibilities. Your expression may include an integral or integrals which cannot be evaluated without knowing the specific form of (r).

Homework Equations



\phi = U/q

F=-∇U

E=-∇\phi

U = (1/2) \int ρ\phid3r

The Attempt at a Solution



For an arbitrary continuous charge distribution:

\phi(r) = \frac{1}{4*pi*ε(0)} \int\frac{ρ(r')}{abs(r - r')}d3r'

For spherically-symmetric distribution of charge:

\phi = (\frac{1}{4*pi*ε(0)} 2\pi ∫ sin\theta'd\theta' ∫ r'2dr' \frac{ρ(r')}{r''} cos\theta)

Then solving for U:

U = (1/2) ∫ρ (\frac{1}{4*pi*ε(0)} 2\pi ∫ sin\theta'd\theta' ∫r'2dr' \frac{ρ(r')}{r''} cos\theta) d3r

Could someone please tell me if I made any errors or assumptions that I should not have made for this situation?

Note: I couldn't figure out how to put limits on the integrals, but for the spherically-symmetric distribution of charge first integral (sin\theta'd\theta') is from 0 to pi and the second integral (r'2dr' \frac{ρ(r')}{r''} cos\theta) is from 0 to infinity. These also affect the equation for U, and there is no limits on the first integral in U.
 
Physics news on Phys.org
smantics said:

Homework Equations



U = (1/2) \int ρ\phid3r


The Attempt at a Solution



For an arbitrary continuous charge distribution:

\phi(r) = \frac{1}{4*pi*ε(0)} \int\frac{ρ(r')}{abs(r - r')}d3r'

For spherically-symmetric distribution of charge:

\phi = (\frac{1}{4*pi*ε(0)} 2\pi ∫ sin\theta'd\theta' ∫ r'2dr' \frac{ρ(r')}{r''} cos\theta)

Hello. I don't see where your factor of cos##\theta## comes from in your last equation above.

If you're not required to use the equation U = (1/2) \int ρ\phid3r, I think there is an easier way.

Imagine bringing in successive thin spherical shells of charge from infinity and building up the charge distribution like layers of an onion. Thus suppose you have currently already built up a spherical distribution of charge of radius r. Find an expression for the change in potential energy dU in bringing in the next layer of charge of thickness dr. Then U = ∫dU.
 

Attachments

  • self energy.png
    self energy.png
    1.7 KB · Views: 528
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top