The electrical energy of a charge distribution is equal to the integral of $\,E^{2}dv/8\pi$ over all space, and if I have a charge distribution of finite extent, then the energy is finite.

Moreover, $\int E^{2}dv/8\pi$ equals the work required to assemble the charge distibution itself. Suppose we have the following charge distribution:

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Since, it's finite in extent, then so does its potential energy, and, ultimately, the work required to assemble it is finite too.

Consider the assembling process, suppose we now arrived at a step were we need to bring a charge $dq_{2}$ near $dq_{1}$, Coulomb's law tells us that the nearer we get from $dq_{1}$, the larger the amount of energy we would need to spend in order to make $dq_{2}$ even nearer, thus, the amount of work required to bring $dq_{2}$ to an infinitely small distance from $dq_{1}$ is infinite. This is in contradiction with the fact that the work required to assemble our charge distribution is finite and equals $\int E^{2}dv/8\pi$. What is the reason behind this discrepancy?


1 Answer 1


The separation of $dq_1$ and $dq_2$ may be "infinitesimally small," but so are the charges $dq_1$ and $dq_2$ themselves, to which the energy associated with this pair is proportional.

For a continuous charge distribution, you need to properly evaluate the total energy as an integral over the charge distribution, possibly as a Riemann integral where you divide (or "partition") the charge distribution into tiny pieces. You can then take the charge of each piece to be localized at a point in the interior of this piece. In the limit where the volume of each such piece goes to zero, while the charges in this limit get closer and closer together, the charged pieces also shrink, so the charge they carry decreases. The total energy of this configuration thus approaches a finite number and there is no contradiction.


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