Suppose we have a conducting grounded hollow sphere of radius $R$, and suppose that we have a point charge $q$ located at a distance $d$ from the centre of the sphere. $(d < R)$.

How can the electrostatic energy of this system be expressed?

On the one hand I could sum over the potential-charge product: $$U = \frac{1}{2}\int_{sphere}\varphi QdV + \frac{1}{2}\varphi_q q$$ and then reason that since the potential of there sphere is $0$, I need only consider the second term, for which the potential can be calculated from that induced by the appropriate image charge.

On the other hand, I could argue that since the potential of the sphere is $0$, all of the energy is the work sunk into bringing a point charge $q$ from the surface of the sphere to its destined location, at distance $d$ from the centre.
However, since the image charge occupies the same space as the charge $q$ when the latter is on the sphere, the appropriate integral expressing this work diverges.

Are any of these two approaches correct? Can the arguments in either approach (potential summation and/or work of system assembly) be 'fixed' so that they lead to the correct answer?

For clarity's sake I'll note that in all literature I am familiar with the self-energy of point charges is taken to be $0$ by definition. I am looking for a solution to this problem with that assumption taken into account.


2 Answers 2


Both methods are correct, but they refer to a different model of point charges. That is why they give different answers.

The first method(formula) is valid if no work needs to be done in assembling the point charge from smaller pieces, the only work needed is to move already existing point charge from infinity inside the sphere. Then we can proceed by assuming that all charges, including those in the sphere and in the ground, are also point charges and use, as a starting point, the Coulomb electrostatic energy formula of point charges in the sphere and the one inside:

$$ W =\sum_{[i,k]k\neq i} \frac{1}{4\pi\epsilon_0}\frac{q_iq_k}{r_{ik}} $$

where the summation is over all pairs of particles. Let the particle inside the sphere have index 0. Introducing average potential $\varphi$ and average charge density $\rho$ in the sphere, and introducing potential due to sphere only $\varphi_{S}$, this sum can be approximated by the formula you gave above:

$$ W = \int\frac{1}{2} \varphi \rho dV + \frac{1}{2}\varphi_{S}(\mathbf r_0)q_0 $$

Now if somehow the average potential on the sphere is made to be 0 (grounding), then the integral is zero and we arrive at the result

$$ W = \frac{1}{2}\varphi_{S}(\mathbf r_0)q_0. $$

The second method assumes that the point charge is to be first formed from charge distributed on a sphere and the work necessary is to be accounted for. But assembling finite point charge from smaller ones requires infinite energy, so the total work in this process is infinite.

In my opinion, point particles of finite charge cannot be usefully treated as being composed of smaller charges, so I prefer the first model.

  • $\begingroup$ If we assume that initially the charge $q_0$ lies 'on' the sphere and we then remove it from there, is that then equivalent to constructing it from smaller charges originating from the ground? $\endgroup$
    – Bar Alon
    Apr 18, 2018 at 10:53
  • $\begingroup$ If the charge $q_0$ exists as point particle in the sphere and then we remove it from there, some work will be necessary to get the charge further from the charges in the sphere, but it will be finite. This is very different from concentrating distributed charge on a sphere, for which work required is infinite. $\endgroup$ Apr 18, 2018 at 11:12

One issue with this setup is that the energy associated with a point charge is theoretically infinite.

To correct that, we could change the point charge to a tiny sphere, in which case the problem will be reduced to a problem of finding the capacitance of a non-concentric spherical capacitor (addressed here) and calculating the energy as Q^2/2C.

  • $\begingroup$ Wouldn't this tiny sphere have to be a conductor for us to treat this system as a capacitor? That's quite a different setup from a point charge. $\endgroup$
    – Bar Alon
    Apr 17, 2018 at 10:57
  • $\begingroup$ Yes, it would be a conductor. Again, my understanding is that the energy associated with a point charge is infinity, so if that's what you are interested in, that is possibly your solution. $\endgroup$
    – V.F.
    Apr 17, 2018 at 11:26
  • $\begingroup$ I think it would make more sense to replace the point charge with a tiny uniformly charges insulating sphere. Additionally, in all literature I am familiar with the self-energy of point charges is considered to be $0$. Does this change matters? $\endgroup$
    – Bar Alon
    Apr 17, 2018 at 18:04
  • $\begingroup$ I just go by the definition of a potential and common sense: it would take infinite energy to bring a unit charge to the surface of a point charge, since the force between the charges is inversely proportional to the square of distance. Similarly, if a unit charge is expelled from a point charge, it would develop infinite speed and acquire infinite energy. Hence your system with or without the grounded shell will have infinite energy unless the charge inside is not a point charge. $\endgroup$
    – V.F.
    Apr 17, 2018 at 21:40

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