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

$${\bf E} = -\nabla F.$$

Hence, (3.4)

$$\nabla \cdot {\bf E} = {\rho_{c}\over \epsilon_{0}},$$

becomes

$$\nabla^{2} F = -{\rho_{c}\over \epsilon_{0}},$$ (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

$$\nabla^{2}F = 0.$$ (3.10)

1. We will simply state the general solution to (3.9). Firstly, remember that the potential for a point charge is
   
   $$F = {e\over 4\pi \epsilon_{0}\vert{\bf r} - {\bf r_{1}}\vert}.$$
   
   
   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,
   

   $$F({\bf r}) = {1\over 4\pi \epsilon_{0}}\int {\rho_{c} ({\bf r_{1}})\over \vert{\bf r} - {\bf r}_{1}\vert} dV,$$ (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,
   
   $$F(x, y, z) = {1\over 4\pi \epsilon_{0}}\int {\rho_{c} (x_{1}... ...{1})^2 + (y - y_{1})^2 + (z - z_{1})^2}} dx_{1} dy_{1} dz_{1},$$
   
   

2. The flux of ${\bf E}$ across a closed surface containing charges is
   
   $$\int {\bf E}\cdot {\bf dS} = \int \nabla \cdot {\bf E} dV,$$
   
   
   by the divergence theorem, and using (3.4) we get
   
   $$\int \nabla \cdot {\bf E} dV = \int {\rho_{c}\over \epsilon} dV = {1\over \epsilon}\int \rho_{c}dV.$$
   
   
   Hence, we have that the flux of the electric field across the surface is
   

   $$\int {\bf E}\cdot {\bf dS} = {1\over \epsilon} \times \hbox{total charge inside V}.$$ (3.12)

   
   

3. If $F = F(r)$ and $\rho_{c}=0$, (3.9) becomes in spherical coordinates,
   
   $${1\over r^{2}}{\partial \over \partial r}\left (r^{2}{\partial F\over \partial r}\right) = 0.$$
   
   
   Integrating once we obtain
   
   $$r^{2}{\partial F\over \partial r} = - c,$$
   
   
   where $c$ is a constant. Integrating a second time gives the electrostatic potential as
   
   $$F = {c\over r} + d,$$
   
   
   and the radial electric field component is
   

   $$E_{r} = -{\partial F\over \partial r} = {c\over r^{2}}.$$ (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,

$$\int E_{r}dS = {e\over \epsilon},$$

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

$${c\over r^{2}}4\pi r^{2} = {e\over \epsilon}.$$

Thus,

$$c = {e\over 4\pi \epsilon},$$

in agreement with (3.6).