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I was trying to get Poisson's equation for a discrete charge system. But my calculation is giving result. I used the scalar potential $\phi(r)=\frac{kq}{r}$, where r=$\sqrt{x^2+y^2+z^2}$ to find $\nabla^2\phi$. And i got $$\nabla^2\phi=\frac{q}{4\pi\epsilon_o}.\frac{3r^2-3x^2-3y^2-3z^2}{r^{\frac{5}{2}}}.$$ At origin,$\nabla^2\phi=\frac{0}{0}$ This result is indeterminate at origin, and also when $r\to0$, but according to Poisson's equation, $$\nabla^2\phi=-\frac{\rho(r)}{\epsilon_o}$$ And for a point charge, $\rho(r)\to\infty$. $$\nabla^2\phi=\infty$$ at origin. Why my result and this result vary?

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The potential that you consider is that of a point charge, which has charge density $$\rho(\mathbf{r})=q\delta(x)\delta(y)\delta(z).$$ If you are not used to delta-function, we can state this in words: the charge density is zero everywhere, except a single point (the one where the charge is located.) That is, anywhere outside this point, the potential satisfies the Laplace equation: $$ \nabla^2\phi(\mathbf{r})=0.$$ Indeed, this is the result obtained in the OP (if we do not forget that $r^2=x^2+y^2+z^2$.) It is not valid at the point where the charge is located, where the denominator becomes zero as well. The way to go around is either use delta-function or build an infinitesimally small Gaussian surface around the charge, convert the Poisson equation to integral form, and show that the integral does not vanish (but rather takes value proportional to $q$.) as the surface radius is reduced to zero. That $\mathbf{r}=0$ is a special point is seen already from the form of the potential, which is singular at this point.

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  • $\begingroup$ Ok. So it means that 0/0 can take up the form of $-\frac{\rho(r)}{\epsilon_o}$. Where $\rho(r)=q\delta^3(r)$. $\endgroup$
    – Alv
    Jun 9, 2023 at 8:47
  • $\begingroup$ $0/0$ is mathematically poorly defined. We can speak only of the limit or of generalized function/singularity (which seems to bypass the problem, but is actually grounded in more mathematical theory.) You could try considering potential of a small charged ball, which will not have singularity, but cutoff at the ball surface. $\endgroup$
    – Roger V.
    Jun 9, 2023 at 9:07

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