Electric Scalar Potential

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

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

(2.14)

However, it is easily demonstrated that

$\displaystyle \frac{x - x'}{[(x-x')^2+(y-y')^2 + (z-z')^2]^{\,3/2}} =$
$\displaystyle -\frac{\partial}{\partial x}\!\left( \frac{1}{[(x-x')^2+(y-y')^2 + (z-z')^2]^{\,1/2}}\right).$		(2.15)

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

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

(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

$\displaystyle {\bf E} = -\nabla \phi,$

(2.17)

where

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

(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 located at the origin is

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

(2.19)

where 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 discrete charges , located at displacements ${\bf r}_i$ , is

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

(2.20)

where

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

(2.21)

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