Electrostatic Self-energy of an arbitrary spherically symmetric charge density

In summary, this homework asks you to find the electrostatic self-energy of an arbitrary spherically symmetric charge density distribution, integrate it to find the potential, and use Gauss's law to find the flux. Unfortunately, the text does not provide enough information to help you understand the problem.
  • #1
harshey
9
0

Homework Statement



Find an expression for the electrostatic self-energy of an arbitrary spherically symmetric charge density distribution p(r). You may not assume that p(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 p(r).


Homework Equations





The Attempt at a Solution



I had no idea how to start this problem because i couldn't figure out what my professor meant by electrostatic self-energy of an arbitrary spherically symmetric charge density distribution.

Does anyone know a professor by the name is Ian Redmount?
 
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  • #2
1. Find E(r) by using Gauss's law.
2. Integrate E to find[tex]\phi(r)[/tex].
3. [tex]U=(1/2)\int\rho\phi d^3 r[/tex] in Gasussian units.
This gives U as a triple integral integral involving rho twice.
What course is this and what is the text/
 
Last edited:
  • #3
The course is Engineering Physics II at Saint Louis University in St. Louis MO.
The text is "Physics" Volume 2 by Halliday, Resnick and Krane.
Our professor doesn't use the text though, he teaches from his own notes which I believe are a few millennium old.

Thanks for the help I'll work on it right now and see if I can do exactly what you suggested but I may have more questions on the method but Thank You, I really appreciate it.
 
  • #4
Meir Achuz said:
1. Find E(r) by using Gauss's law.
2. Integrate E to find[tex]\phi(r)[/tex].
3. [tex]U=(1/2)\int\rho\phi d^3 r[/tex] in Gasussian units.
This gives U as a triple integral integral involving rho twice.
What course is this and what is the text/

Is the electrostatic self-energy the potential? What is phi in your equation? Sorry, I am confused in how you derived that equation.
 
  • #5
  • #6
harshey said:
Is the electrostatic self-energy the potential? What is phi in your equation? Sorry, I am confused in how you derived that equation.
The phi in my equation is the potential, given by integrating E.dr.
The electrostatic self-energy of a point charge is U= q*phi.
For charged sphere, it is given by my third equation.
I think that, on a Haliday and Resnick level,
there are too many simple things not given in the text to understand a problem on this level.
 

1. What is electrostatic self-energy?

Electrostatic self-energy is the energy that a charged particle or system of particles possesses due to its own electric field.

2. What is an arbitrary spherically symmetric charge density?

An arbitrary spherically symmetric charge density refers to a charge distribution that is symmetrical in all directions around a central point, such as a sphere or a point charge.

3. How is the electrostatic self-energy of an arbitrary spherically symmetric charge density calculated?

The electrostatic self-energy of an arbitrary spherically symmetric charge density can be calculated using the formula U = (3/5)kQ^2/R, where U is the self-energy, k is the Coulomb constant, Q is the total charge, and R is the radius of the charge distribution.

4. What is the significance of the electrostatic self-energy of an arbitrary spherically symmetric charge density?

The electrostatic self-energy is important in understanding the stability and behavior of charged systems, as it contributes to the total energy of the system. It is also used in various applications, such as in electrostatics and in the study of atomic and molecular structures.

5. How does the electrostatic self-energy change with distance from the charge distribution?

The electrostatic self-energy decreases with increasing distance from the charge distribution, as the electric field strength weakens with distance. This means that the self-energy is highest at the surface of the charge distribution and decreases as you move further away from it.

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