# Electrostatical Anomaly

1. Jun 27, 2008

### aniketp

Hey everyone,
Can you justify this:
The total energy of a thin spherical shell is the sum of its "self" energy and "interaction" energy.By simple calculus for a thin spherical shell,
E(total)= Q^2/8*$$\pi\epsilon$$*R
Here,
Q: total charge
Thus as R$$\rightarrow$$0, i.e the shell becomes a point charge, the total energy tends to infinity.So the analysis of point charge systems becomes impossible from the energy point of view.

2. Jun 27, 2008

### PhysicsDruid

Mmmm.... two things, perhaps.

First, as R->0, I am assuming you have some uniform spread of Q over the surface. Thus the word required to compress all that charge toward a point would grow toward infinity as you tried to compress repelling charges all together into one point. It's similar to the underdivision, undergraduate problems of electrostatic potential, introducing point charges, the work done, etc.

Second, it is often assumed (if you've studied Legendre Polynomials and the solutions to the Laplace equation) that the potential is zero at infinity... however, there are times we reverse this so that our solutions don't diverge and the potential is zero at r = 0, depending on which solutions are being used.

Hope this gives a little insight.

3. Jun 27, 2008

### nicksauce

Griffiths discusses something rather similar in "introduction to electrodynamics":

4. Jun 27, 2008

### aniketp

So is there no explaination for this? And we just "assume" that the energy of a point charge is zero?

5. Jun 27, 2008

### PhysicsDruid

Haha, I think you just gave a very verbose quotation to what I said earlier nicksauce :-)

Its a matter of perspective, aniketp. I suppose, if you really wanted, you could place a reference point at some... say.. 50% of the way to infinity and give yourself some energy. It all depends on where you define your references if you take the point charge by itself... or if you take the view of a collapse of a bunch of charge into a "point", then it would take an infinite amount of energy since you are attempting to push repelling forces all together at exactly into a delta peak.

6. Jun 27, 2008

### aniketp

So is it that bcause we are just interested in the CHANGE of energy it does not matter what our reference point is?