# Homework Help: Basic Electric Field in Conductor Principles

Tags:
1. Oct 9, 2015

### FallenLeibniz

1. The problem statement, all variables and given/known data
The following is a small concept exercise from an. EM text (Electromagnetism: Pollack and Stump). I believe I have the explanations correct, but am just looking for "peer-review" as they seem "hand-wavy":

Suppose a conductor has a cavity inside it, and there is a point charge q somewhere in the cavity. Prove that the net charge on the wall of the cavity must be -q. If the net charge of the conductor is Q0[\SUB], what is the net charge on its outer surface? Why must every field line in the cavity begin on q and on the cavity wall? (Hint: For the last statement, use the fact that $\oint\vec{E}\cdot d\vec{l}$ )

2. Relevant equations
Other than the one in the statement of the problem, nothing really as it is a concept question more than anything.

3. The attempt at a solution

a)
Let $\vec{E}_{int}$ be the field inside the conductor material (which is of course $\vec{0}$). Let S be a gaussian surface that is large enough to both enclose the cavity and a small portion of the conductive material. Since $\vec{E}_{int}\cdot d\vec{A}$ at S is zero, it follows from Gauss's Law that the enclosed charge is zero. Taken with the fact that a conductor may have no overall charge density inside itself when in equilibrium except at surface boundaries, the wall must have -q.

b) $Q_0$ +q. As before stated, a conductor may only carry charge on its surface boundaries. Since we have established that -q is on one boundary inside the conductor, $Q_0$ +q must be on the outside for the net charge of the conductor to be $Q_0$ only. Note that this also finds validity when considering an empty cavity.

c) To start, a field must have a source. Since the only source for the field in the cavity is the charge, it must start there. Note that the fact that the closed integral of the dot product is zero means that any field that meets the boundaries of a curve must do so perpendicularly. Now the field can not terminate inside the cavity since it can not just arbitrarily terminate in mid air and the field can not travel through or terminate in the conductor since the internal conductor field must be zero. Thus the field must terminate on the cavity boundary.

2. Oct 10, 2015

### rude man

I've rewritten your problem statement slightly for clarity.
I agree with (a) and (b) and am not sure of (c), especially the statement that " ... the fact that the closed integral of the dot product is zero means that any field that meets the boundaries of a curve must do so perpendicularly." That integral is zero anywhere in an electrostatic environment. The reason the E field is perpendicular to the metallic surface is that the metallic surface is an equipotential and the E field is always perpendicular to the equipotential surface. The rest of your statement sounds right. Maybe someone else will comment more on this.