Laplace's equation

We know that $\myv\grad\cdot \myv E(\myv r)=\frac{\rho(\myv r)}{\epsilon_0}$. Therefore, in regions where there is no charge density ($\rho(\myv r)=0$) Laplace's equation for the electric potential must be satisfied: $$\grad^2 V(\myv R)= \frac{\del^2V}{\del x^2}+\frac{\del^2V}{\del y^2}+\frac{\del^2V}{\del z^2} =0$$

Uniqueness

I will not go through the proof. Griffiths shows that:

There is a corollary to this:

The potential in some volume ${\cal V}$ is uniquely determined if $V(\myv r)$ is specified on all boundaries of ${\cal V}$, and the charge density throughout the region is specified.

A second uniqueness theorem that we will also take advantage of is:

The region can either be bounded by another conductor, or completely unbounded.

Read by next class, section 3.15 and 3.16, and make sure you understand the justification for the corollary.

Image solutions

Consider the following problem: A charge of $+q$ is positioned a distance $d$ above the origin, and above an infinite conducting plane in the $xy$ plane. The conducting plane is grounded, so that the potential on the plane is $V(z=0)=0$.

A second condition is that, as usual, we expect $V(r\to\infty)=0$.

What is the electric potential, $V(\myv r)$, in the region above the conducting plane?

First, sketch the field lines you anticipate:

The answer is: $$V(\myv r)=\frac{1}{4\pi \epsilon_0}\left[ \frac{q}{\sqrt{x^2+y^2+(z-d)^2}} -\frac{q}{\sqrt{x^2+y^2+(z+d)^2}} \right].$$ That is...the same as if there were no conducting plane, but there were a single charge $-q$ at $z=-d$ below the origin. (!!)

Again, sketch the field lines...

Do problem 3.6.

Read 3.2, and make sure you understand how to calculate the charge density on the conducting surface using Griffith's Eq (2.48), there's a relation between the electric field and surface charge density: $$\myv E=\frac{\sigma}{\epsilon_0}\uv n.$$