Electrostatic self-energy

In summary: In this approach you do not have to worry about the value of ##\phi## and the factor of cos##\theta## in the expression for ##\phi##.
  • #1
smantics
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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



[itex]\phi[/itex] = U/q

F=-∇U

E=-∇[itex]\phi[/itex]

U = (1/2) [itex]\int[/itex] ρ[itex]\phi[/itex]d3r

The Attempt at a Solution



For an arbitrary continuous charge distribution:

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

For spherically-symmetric distribution of charge:

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

Then solving for U:

U = (1/2) ∫ρ ([itex]\frac{1}{4*pi*ε(0)}[/itex] 2[itex]\pi[/itex] ∫ sin[itex]\theta[/itex]'d[itex]\theta[/itex]' ∫r'2dr' [itex]\frac{ρ(r')}{r''}[/itex] cos[itex]\theta[/itex]) 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[itex]\theta[/itex]'d[itex]\theta[/itex]') is from 0 to pi and the second integral (r'2dr' [itex]\frac{ρ(r')}{r''}[/itex] cos[itex]\theta[/itex]) is from 0 to infinity. These also affect the equation for U, and there is no limits on the first integral in U.
 
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  • #2
smantics said:

Homework Equations



U = (1/2) [itex]\int[/itex] ρ[itex]\phi[/itex]d3r


The Attempt at a Solution



For an arbitrary continuous charge distribution:

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

For spherically-symmetric distribution of charge:

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

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) [itex]\int[/itex] ρ[itex]\phi[/itex]d3r, 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.
 

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1. What is electrostatic self-energy?

Electrostatic self-energy refers to the energy stored in a system of charged particles due to their mutual interactions. It is a measure of the potential energy of the system.

2. How is electrostatic self-energy calculated?

Electrostatic self-energy can be calculated using the formula U = (1/2) ∑i∑j (qi*qj)/4πε₀r, where U is the total self-energy, qi and qj are the charges of the particles, ε₀ is the permittivity of free space, and r is the distance between the particles.

3. What is the significance of electrostatic self-energy?

Electrostatic self-energy plays a crucial role in understanding the behavior of charged particles in a system. It is responsible for phenomena such as attraction and repulsion between particles, as well as the stability of atomic and molecular structures.

4. How does electrostatic self-energy differ from electrostatic potential energy?

Electrostatic self-energy refers to the energy stored in a system of charged particles, while electrostatic potential energy refers to the energy of a single charged particle in an electric field. Self-energy takes into account the interactions between all the particles in a system, while potential energy only considers the interaction between a single particle and the field.

5. Can electrostatic self-energy be negative?

Yes, electrostatic self-energy can be negative. This occurs when the charges in a system are of opposite sign and attract each other, resulting in a decrease in the total energy of the system. However, the magnitude of the self-energy is always positive, as it is a measure of the potential energy of the system.

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