Is energy relative in electrostatics?

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The discussion centers on the total energy of a thin spherical shell in electrostatics, highlighting that as the radius approaches zero, the energy tends to infinity, complicating the analysis of point charge systems. It emphasizes that the energy calculation involves both self-energy and interaction energy, with the work required to compress charges into a point becoming infinite due to repulsion. The conversation also touches on the assumptions regarding potential at infinity versus at the origin, indicating that different equations apply to different scenarios. Griffiths' equations are referenced to illustrate that while both can be correct, they address distinct aspects of charge configurations. Ultimately, the infinite energy of point charges raises challenges in electromagnetic theory, suggesting that the reference point for energy calculations significantly influences outcomes.
aniketp
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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
R: radius of shell
Thus as R\rightarrow0, 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.
 
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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.
 
Griffiths discusses something rather similar in "introduction to electrodynamics":

David Griffiths said:
Equation 2.45, W = \frac{\epsilon_0}{2}\int{E^2d^3r} implies that the energy of a stationary charge distribution is always positive. On the other hand, Equation 2.42, W=\frac{1}{2}\sum_{i=1}^{i=n}q_iV(\vec{r}_i), from which Eq. 2.45 was derived can be positive or negative... Whch equation is correct? The answer is that both equations are correct, but they pertain to slightly different situations. Eq. 2.42 does not take inito account the work necessary to make the point charge in the first place; we started with the point charges and simply found the work required to bring them together. This is a wise policy, since Eq. 2.45 indicates that the energy of a point charge is in fact infinite. Eq. 2.45 is more complete in the sense that it tells you the total energy stored in the charge configuration, but Eq. 2.42 is more appropriate when you're dealing with point charges, because we prefer to leave out that portion of the total energy that is attributable to the fabrication of the point charges themselves. In practice, after all, the point charges (electrons say) are given to us ready-made; all we do is move them around. Since we did not put them together and cannot take them apart it is immaterial how much work the process would involve. Still, the infinite energy of a point charge is a recurring source of embarrassment for electromagnetic theory, afflicting the quantum version as well as the classical.
 
So is there no explanation for this? And we just "assume" that the energy of a point charge is zero?
 
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.
 
So is it that bcause we are just interested in the CHANGE of energy it does not matter what our reference point is?
 
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