Laplace Equation in Electrostatics

• The electrostatic potential created by a static charge distribution at a charge-free region is given by the following Laplace equation:

  $$\nabla^2 V=\frac{\partial^2V}{\partial x^2}+\frac{\partial^2V}{\partial y^2}++\frac{\partial^2V}{\partial z^2}=0$$

  Here, $V(x,y,z)$ is the potential within the region.
• The solution of this problem for particular charge distributions concerns the subject of partial differential equations.
• However, the dimensions of the problem can be reduced if the charge distribution exhibit a spatial symmetry.
• For example, in a system with spherical symmetry, the solution of the problem becomes easier if the partial derivatives in Laplace's equation are expressed in terms of spherical coordinates $(r, \theta, \phi)$:

  $$\nabla^2 V=\frac{1}{r^2} \frac{ \partial}{ \partial r} \left( r^2... ...a}\right) + \frac{1}{r^2 \sin^2\theta} \frac{ \partial^2 V}{ \partial \phi^2}=0$$ (5.4)

• Let the two concentric conductive spherical shells of radii $R_a$ and $R_b$ be held at constant potentials $V_a$ and $V_b$ (Figure 5.10).
• Because of spherical symmetry, the potential distribution in the region between the two spheres ( $R_a <r < R_b$) will be a function of distance $r$ only.
• Accordingly, the derivatives with respect to the variables ( $\theta, \phi$) in Equation 5.4 become zero, and the partial derivative in the remaining term becomes the full derivative:

  $$\frac{1}{r^2} \frac{ \partial}{ \partial r} \left( r^2 \frac{ \pa... ...V}{ \partial r}\right) \rightarrow \frac{d^2V}{dr^2}+\frac{2}{r}\frac{dV}{dr}=0$$

Figure 5.10: The region between two spherical shells of different potential.

The boundary conditions of this differential equation become:

$$V(R_a) = V_a$$

$$V(R_b) = V_b$$

First, we transform this equation into a linear system of equations:

$$V \rightarrow V_1$$

$$\frac{dV}{dr} \rightarrow V_2$$

with these values ( $V_1 and V_2$), the system of equations to be solved

	$$\frac{dV_1}{dr}$$	$$=V_2$$
	$$\frac{dV_2}{dr}$$	$$=-\frac{2}{r}V_2$$

and the boundary conditions are:

$$V(R_a) = 100$$

$$V(R_b) =$$ 0

• Now, we have a set of equations.
• The analytical solution to this spherically symmetric problem would be:

  $$V(R)=\frac{R_a(R_b-r)}{(R_b-R_a)r}V_a$$

We had a set of equations.
We transformed this boundary value problem into a first-order system of equations.

	$$V_1$$	$$=V$$
	$$V_2$$	$$=V'$$
	$$\frac{dV_1}{dr}$$	$$=V_2$$
	$$\frac{dV_2}{dr}$$	$$=-\frac{2}{r}V_2$$

(Example py-file: laplaceequation.py)

Figure 5.11: Solution for the Boundary Value Problem for the ODE: $V''=-\frac {2}{r}V'$.

Figure 5.12: Solution for the Boundary Value Problem for the ODE: $V''=-\frac {2}{r}V'$.