Incomprehensible part in Griffiths' text

[WiFO215]

After deriving the self energy, or the energy to construct a charge configuration\frac{\int_{V} \epsilon E^{2}dV} {2}(where V is the volume over which the E-field of the configuration extends.)

Griffiths goes on to say that the energy required to construct a point charge using the above formula would be infinity. With this statement, I can agree. But then he also adds that this is a big problem with classical EM theory. I cannot comprehend that. I don't quite see the problem.

Remember when we were dealing with the field of an infinite sheet? The field turns out to be a constant,

\frac{\sigma}{2 \epsilon}

where \sigma happens to be uniform charge density over the infinite sheet. In that problem, when we set the zero of the potential at infinity, the problem got very messy. Potentials would always shoot to infinity. That problem was crucial (at least to me) in showing that one ought to be careful in setting the zero of the potential.

If we carry over what we've learnt, then the "problem" of infinities of a point charge disappear. In deriving the above formula for energy of a point charge, he sets the zero at infinity. Why not just set the zero AT the point charge in this problem? Wouldn't that solve the problem?

 

[Vanadium 50]

Why not just set the zero AT the point charge in this problem? Wouldn't that solve the problem?

Please take a look at what is set to zero. Is it the field or is it the potential? What difference does that make? <--- There's a hint in my wording.

 

[Bob S]

Also put a finite charge q on a finite sheet of area A. As the area of the sheet increases, the charge density decreases as σ' = Aσ/A'. So the field drops off with length squared as dimensions increase to ∞.

Bob S