Imagine we have the first Maxwell Equation:
$$
\nabla \cdot \mathbf E = \frac{\rho}{\varepsilon_0} \\
\mathbf E = -\nabla \phi \\
\nabla \cdot (-\nabla\phi) = \frac{\rho}{\varepsilon_0} \\\nabla^2 \phi = -\frac{\rho}{\varepsilon_0}
$$
We arrive to an equation for finding the potential of an Electric field, given a charge density, $\rho$.
Assuming that: $\phi$ and $\rho$ only depends on the $\mathbf r$, coordinate, we have the following:
$$
\nabla^2 \phi = \frac{\partial^2 \phi}{\partial r^2} + \frac{1}{r} \frac{\partial \phi}{\partial r} = \frac{\rho(r)}{\varepsilon_0}
$$
For zero charge density we have the following:
$$
\frac{\partial \phi}{\partial r} \equiv u(r)\\
0 = u'(r) + \frac{u(r)}{r} \rightarrow u'(r) = – \frac{u(r)}{r} \\
\frac{du}{u} = -\frac{dr}{r} \rightarrow u(r) = \frac{C_1}{r} \\
\phi(r) = C_1\log(r) + \phi_0
$$
But, this mean there exists a potential. I don't quite understand this because there is not charge density, hence, no electric field. But how is there an electric potential, did I derive it wrong?
Best Answer
You can have a non-zero potential -- as long as it is constant in space. This will always generate a zero electric field, since in this case $\mathbf{E} = -\nabla\phi = 0$, which is indeed expected for a zero charge density.
Mathematically, you could notice that since the point $r=0$ is part of your domain, you must have $C_1 = 0$ to avoid divergences in the potential. This generates $\phi = \phi_0$, i.e., a constant potential, as expected.
EDIT: As jensen paull said, you have the wrong Laplacian operator for spherical coordinates, making your final solution wrong. Nevertheless, as the correct solution is $\phi = C_1/r + \phi_0$, the above analysis still hold.