next up previous
Next: Electric Dipoles Up: Electrostatics Previous: Point Charges

Continuous distribution of charge, $\rho _{c}$, per unit volume.

If there is no time dependence, so that $\partial /\partial t = 0$, then we have from (3.2) that $\nabla \times {\bf E} = 0$. The general solution to this equation states that the electric field can be expressed as the gradient of a scalar function, the electrostatic potential, as

\begin{displaymath}
{\bf E} = -\nabla F.
\end{displaymath}

Hence, (3.4)

\begin{displaymath}
\nabla \cdot {\bf E} = {\rho_{c}\over \epsilon_{0}},
\end{displaymath}

becomes
\begin{displaymath}
\nabla^{2} F = -{\rho_{c}\over \epsilon_{0}},
\end{displaymath} (3.9)

where $\rho _{c}$ is the charge density. (3.9) is Poisson's equation. However, when there are no charges so that $\rho_{c}=0$, this reduces to Laplace's equation
\begin{displaymath}
\nabla^{2}F = 0.
\end{displaymath} (3.10)

  1. We will simply state the general solution to (3.9). Firstly, remember that the potential for a point charge is

    \begin{displaymath}
F = {e\over 4\pi \epsilon_{0}\vert{\bf r} - {\bf r_{1}}\vert}.
\end{displaymath}

    If there are more charges we simply add the potentials together. Thus, it is not too surprising that when there is a continuous distribution instead of adding we integrate over the volume containing the charge. Thus,
    \begin{displaymath}
F({\bf r}) = {1\over 4\pi \epsilon_{0}}\int {\rho_{c}
({\bf r_{1}})\over \vert{\bf r} - {\bf r}_{1}\vert} dV,
\end{displaymath} (3.11)

    where $\rho_{c}({\bf r}_{1})= \rho_{c}(x_{1}, y_{1}, z_{1})$ and $dV = dx_{1}
dy_{1} dz_{1}$. Thus,

    \begin{displaymath}
F(x, y, z) = {1\over 4\pi \epsilon_{0}}\int {\rho_{c}
(x_{1}...
...{1})^2 + (y - y_{1})^2 + (z - z_{1})^2}}
dx_{1} dy_{1} dz_{1},
\end{displaymath}

  2. The flux of ${\bf E}$ across a closed surface containing charges is

    \begin{displaymath}
\int {\bf E}\cdot {\bf dS} = \int \nabla \cdot {\bf E} dV,
\end{displaymath}

    by the divergence theorem, and using (3.4) we get

    \begin{displaymath}
\int \nabla \cdot {\bf E} dV = \int {\rho_{c}\over \epsilon} dV =
{1\over \epsilon}\int \rho_{c}dV.
\end{displaymath}

    Hence, we have that the flux of the electric field across the surface is
    \begin{displaymath}
\int {\bf E}\cdot {\bf dS} = {1\over \epsilon} \times \hbox{total
charge inside V}.
\end{displaymath} (3.12)

  3. If $F = F(r)$ and $\rho_{c}=0$, (3.9) becomes in spherical coordinates,

    \begin{displaymath}
{1\over r^{2}}{\partial \over \partial r}\left (r^{2}{\partial
F\over \partial r}\right) = 0.
\end{displaymath}

    Integrating once we obtain

    \begin{displaymath}
r^{2}{\partial F\over \partial r} = - c,
\end{displaymath}

    where $c$ is a constant. Integrating a second time gives the electrostatic potential as

    \begin{displaymath}
F = {c\over r} + d,
\end{displaymath}

    and the radial electric field component is
    \begin{displaymath}
E_{r} = -{\partial F\over \partial r} = {c\over r^{2}}.
\end{displaymath} (3.13)

Thus, there is a possible singularity at $r=0$ and (3.10) fails at this point. If at $r=0$ there is a point charge of strength $e$, what is the value of the contant $c$? We can use (3.12) and draw a small sphere of radius $r$ about the point charge. Thus,

\begin{displaymath}
\int E_{r}dS = {e\over \epsilon},
\end{displaymath}

or substituting for $E_{r} = c/r^{2}$ from (3.13) we get

\begin{displaymath}
{c\over r^{2}}4\pi r^{2} = {e\over \epsilon}.
\end{displaymath}

Thus,

\begin{displaymath}
c = {e\over 4\pi \epsilon},
\end{displaymath}

in agreement with (3.6).
next up previous
Next: Electric Dipoles Up: Electrostatics Previous: Point Charges
Prof. Alan Hood
2000-11-06