# Why is it possible to find electrostatic energy of a conductor by integration of $\int_{space} \vec{E}^2 dV$?

In Griffith's Introduction to Electrostatics, International 4th ed, pg-94 these two equations are given for calculating energy of continous charge distributions:

$$W= \frac12 \int \rho V d \tau \tag{1}$$

And after some simplifications, this other equation is given $$W= \frac{\epsilon_o}{2} \int_{whole space} \vec{E}^2 d \tau \tag{2}$$

And, there are also some exercises to find the energy of conductors using equation (2) but after some careful thought, I realize that in practice, equation (2) may have to integrate over a boundary. This is problem because the component of electric field normal to a surface is discontinuous by an amount $$\frac{\sigma}{\epsilon}$$. Now, since we have a discontinuity at the boundary, how can is the expression integrable?(*)

As far as I learned, discontinuous functions such as $$\frac{1}{x}$$ etc are not integrable on a set containing their discontinuity.

On some reflection, I started realize voltage of a conductor maybe suffering from the same problem because when we integrate from infinity to a point inside a conductor, there is a sudden discontinuity of the electric field at the boundary again. Hence, (1) also suffers the same problem.

A discontinuity is not a problem for the integrability of a function. The reason $$\frac{1}{x}$$ cannot be integrated over $$x=0$$ is because the integral diverges, but this is related to the fact that $$|1/x| \rightarrow \infty$$ as $$x \rightarrow 0$$, not because it is discontinuous. So to answer your question; the integral is in fact well-defined, even though the field may be discontinuous at the boundary.
• In fact a divergent integrand is compatible with a finite integral, e.g. $\int_0^1x^{-1/2}dx=2$.