# Calculating the point charge potential via Gauss's law

I want to calculate the potential of a point charge via Gauss's law
$$\vec{\nabla}\cdot\vec{E}=4\pi\rho(\vec{r})$$
My Idea:
we can find an expression for $$4\pi$$ that has the form of an integral,
by using spherical coordinates we can derive:
$$\int_{V}\vec{\nabla}^2 \left( \frac{1}{r} \right)dV =\int_V \vec{\nabla}\cdot\left( \vec{\nabla} \frac{1}{r} \right)dV$$
by using the Divergence theorem and applying spherical coordinates we obtain: $$\oint_S \vec{\nabla} \left(\frac{1}{r}\right) \cdot \vec{n}dS = \oint_S \partial_r \left( \frac{1}{r} \right)r^2\sin(\theta) d\theta dr = -4\pi$$
$$\Rightarrow \vec{\nabla}\cdot \vec{E}=-\int_V \vec{\nabla}^2 \frac{\rho(\vec{r})}{|\vec{r}-\vec{r}'|}dV = -\vec{\nabla}^2\int_V \frac{\rho(\vec{r})}{|\vec{r}-\vec{r}'|}dV$$
since following expression is known: $$\vec{\nabla}\cdot \vec{E}=-\vec{\nabla}^2\phi$$
we can deduce that:
$$\Rightarrow \phi(\vec{r})=\int_V \frac{\rho(\vec{r}')}{|\vec{r}-\vec{r}'|}dV$$
My problem is that I'm not sure how to solve this integral. I'm aware that the charge density for discrete charges is:
$$\rho(\vec{r})=\sum_i^nq_i \delta(\vec{r}-\vec{r}')$$
and since we are talking about a single charge at the origin ($$\vec{r}'=0$$) this sum will equal $$q$$.
$$\Rightarrow \phi(\vec{r})=\int_V \frac{q}{r}dV=q\int_0^{2\pi}\int_0^\pi \int_0^R \frac{1}{r}r^2\sin(\theta) \space d\theta d\varphi dr$$
However, this delivers:
$$\phi(\vec{r})=2q\pi r^2$$
Which is of course wrong. Any help is appreciated

Your reasoning is correct until your equation $$\rho(\vec{r})=\sum_i^nq_i \delta(\vec{r}-\vec{r}'_i)$$ (I've added an index $$i$$ here for clarity, because $$\vec{r}'_i$$ denotes the place of charge $$q_i$$)
But then, for the special case of a single charge $$q$$ at the origin ($$\vec{r}' = \vec{0}$$) this sum will be equal $$\rho(\vec{r})=q \delta(\vec{r})$$
From this it follows by integration \begin{align} \phi(\vec{r})&=\int_V \frac{\rho(\vec{r}')}{|\vec{r}-\vec{r}'|}dV \\ &=\int_V \frac{q\delta(\vec{r}')}{|\vec{r}-\vec{r}'|}dV \\ &=\frac{q}{|\vec{r}|} \end{align} (The last step here used the definition of the $$\delta$$ function)
• If I am not mistaken we should have $\int_V \frac{q}{r} \delta(\vec{r})dV$ at $\vec{r}'=0$. what happened to the integral? It seems like $\int_V \frac{q}{r} \delta (\vec{r})dV = \frac{q}{r}$ but that doesn't seem right Nov 4, 2019 at 21:52
Note that $$\mathbf{\nabla}\cdot\mathbf{E}=4\pi\rho(\mathbf{r})$$ is equivalent to \begin{align} \int_V \mathbf{\nabla}\cdot\mathbf{E}\, d\tau & = 4\pi\int_V \rho\,d\tau \\ \oint_S \mathbf{E}\cdot d\mathbf{a}&=4\pi \,Q_{enc} \end{align} by the divergence theorem. Define a sphere around the point. Then, $$\mathbf{E}$$ is spherically symmetric, constant, and comes out of the integral. If we say the point has charge $$q$$, then we have \begin{align} \oint_S |\mathbf{E}|\,da &= 4\pi q \\ |\mathbf{E}|\oint da &= 4\pi q \\ |\mathbf{E}|\cdot 4\pi r^2 &= 4\pi q \\ |\mathbf{E}| & = \frac{q}{r^2} \\ \mathbf{E} & = \frac{q}{r^2}\hat{\mathbf{r}} \end{align} Now, we have $$V(\mathbf{r})=-\int^\mathbf{r}_\infty \mathbf{E}\cdot d\mathbf{l}$$ Therefore, \begin{align} V(\mathbf{r})&=-\int^\mathbf{r}_\infty \frac{q}{r'^2}\, dr' \\ &= \frac{q}{r'}\bigg|_\infty^r \\ &= \frac{q}{r} \end{align}