# Electrical energy of a charge distribution and the work required to assemble it

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: 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?

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.