Electric Scalar Potential

Suppose that ${\bf r} = (x,\,y,\,z)$ and ${\bf r'}= (x',\,y',\,z')$ in Cartesian coordinates. The $x$-component of $({\bf r} - {\bf r'})/{\vert{\bf r} - {\bf r'}\vert^3}$ is written

$$\frac{x - x'}{[(x-x')^2+(y-y')^2 + (z-z')^2]^{\,3/2}}.$$ (2.14)

However, it is easily demonstrated that

$$\frac{x - x'}{[(x-x')^2+(y-y')^2 + (z-z')^2]^{\,3/2}} =$$

$$-\frac{\partial}{\partial x}\!\left( \frac{1}{[(x-x')^2+(y-y')^2 + (z-z')^2]^{\,1/2}}\right).$$ (2.15)

Here, $\partial/\partial x$ denotes differentiation with respect to $x$ at constant $y$, $z$, $x'$, $y'$, and $z'$. Because there is nothing special about the $x$-axis, we can write

$$\frac{{\bf r}- {\bf r}' } {\vert{\bf r} - {\bf r}'\vert^3} = -\nabla\!\left(\frac{1}{\vert{\bf r} - {\bf r'}\vert}\right),$$ (2.16)

where $\nabla\equiv (\partial/\partial x,\,\partial/\partial y, \,\partial/\partial z)$ is a differential operator that involves the components of ${\bf r}$, but not those of ${\bf r}'$. (See Section A.19.) It follows from Equation (2.13) that

$${\bf E} = -\nabla \phi,$$ (2.17)

where

$$\phi({\bf r}) = \frac{1}{4\pi\,\epsilon_0} \int_{V'} \frac{ \rho({\bf r}')}{\vert{\bf r} - {\bf r}'\vert} \,dV'.$$ (2.18)

Thus, we conclude that the electric field, ${\bf E}({\bf r})$, generated by a collection of fixed electric charges can be written as minus the gradient of a scalar field, $\phi({\bf r})$—known as the electric scalar potential—and that this scalar field can be expressed as a simple volume integral involving the electric charge distribution.

The scalar potential generated by an electric charge $q$ located at the origin is

$$\phi( r) = \frac{q}{4\pi\,\epsilon_0\,r},$$ (2.19)

where $r$ is a spherical polar coordinate. (See Section A.23.) Moreover, according to Equations (2.11) and (2.16), the scalar potential generated by a set of $N$ discrete charges $q_i$, located at displacements ${\bf r}_i$, is

$$\phi({\bf r}) = \sum_{i=1,N} \phi_i ({\bf r}),$$ (2.20)

where

$$\phi_i({\bf r}) = \frac{q_i}{4\pi\,\epsilon_0\,\vert{\bf r} - {\bf r}_i\vert}.$$ (2.21)

Thus, the net scalar potential is just the sum of the potentials generated by each of the charges taken in isolation.