Electric potential

Remarks

In both cases $E = -\nabla \varphi$, where $E$ is the electric field and $\nabla$ is the gradient. So $\varphi$ is a scalar potential of $-E$, normalised by $\varphi(x) \to 0$ as $\abs{x} \to \infty$. Up to the factor $1/\varepsilon_0$ and a sign convention, the second formula is the Newtonian potential of $\rho$. Example: for one charge $q$ at the origin, $\varphi(x) = q / (4 \pi \varepsilon_0 \abs{x})$, and $-\nabla \varphi(x) = q x / (4 \pi \varepsilon_0 \abs{x}^3)$ is the Coulomb field. Non-example: $\psi(x) = \varphi(x) + 1$ also satisfies $E = -\nabla \psi$, but it is not the electric potential of the definition, since it does not vanish at infinity. The potential difference $\varphi(a) - \varphi(b)$ is the voltage between $a$ and $b$, measured in volts.