Suppose that I have a continuous charge distribution that occupies some volume $V$, and my goal is to find the electric field inside of it, One way to approach this is by pretending that $V$ is made of infinitesimal point charges, and sum up all the fields due to these point charges by superposition principle and that's what we call it integration, what I don't understand is that if we want to find the electric field very close to some point charge, does the field approach infinity? if so why if we perform the integral we find a finite value over the whole volume $V$?


1 Answer 1


While there are charges "infinitesimally close" to any point $P$ inside $V$, the amount of charge that is infinitesimally close to $P$ is "infinitesimally small", and the ratio of these quantities in Coloumb's Law does not necessarily diverge.

To see this a bit more quantitatively: the amount of charge in a thin shell of radius $\epsilon$ will be proportional to $\epsilon^2$. So although these charges are an "average distance" of $\epsilon$ from $P$, we would not necessarily expect their contribution to Coloumb's law (of the form $q/r^2$) to diverge as $\epsilon \to 0$, since from the above scaling argument we'd expect the contribution to be proportional to $\epsilon^2 / (\epsilon)^2$.

This is not a rigorous argument, of course, but hopefully it makes it plausible that the process of "integrating over infinitesimally small charges" makes sense even at points where $\rho \neq 0$.


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