Electromagnetism / Laplaces Equation

[johnaphun]

Homework Statement

Assuming the Poisson equation d²V = -p/e₀
show that the electric field in a cavity is zero

(Explain clearly each step of your argument)

The Attempt at a Solution

Here's my attempt (i think it's correct, (kind of) I'm just not sure if I've explained the argument clearly enough).

p = 0 inside the cavity as there is no charge contained within the volume. All excess charge resides on the surface therefore d²V = 0

As a result the E field inside the conductor is zero and gradV is also equal to zero meaning V = constant

V in the cavity is given by the solution to Laplace's equation subject to the b.c's that V is constant on the interior surface. Earnshaw's theorem dictates that V can not have either a minima or maxima inside the conductor only on the surface meaning V inside the cavity is constant which means E = 0

 

[mark726]

Hello, thank you for bringing up this interesting topic. I will try to explain the argument step by step.

First, let's start with the Poisson equation: d2V = -p/e0. This equation relates the potential, V, to the charge density, p, and the permittivity of free space, e0. In this case, we are assuming that the charge density, p, is equal to zero inside the cavity. This is because there is no charge contained within the volume of the cavity, and all excess charge resides on the surface.

Next, we can use the fact that d2V = 0 inside the cavity, since p = 0. This means that the electric field, E, is also equal to zero inside the cavity, since E = -gradV. This is because gradV is the gradient of the potential, and if d2V = 0, then there is no change in potential in any direction, resulting in a zero electric field.

Now, let's consider the potential, V, inside the cavity. Since we know that E = 0 inside the cavity, we can use the fact that V = constant to solve for the potential. This is because if E = 0, then there is no change in potential, and therefore V must be constant.

Finally, we can use Earnshaw's theorem to show that V can only have minima or maxima on the surface of the conductor, and not inside the cavity. This is because if there were a minima or maxima inside the cavity, the electric field would not be zero, which contradicts our assumption that E = 0 inside the cavity. Therefore, V must be constant inside the cavity, which means that E = 0.

In conclusion, by assuming a zero charge density inside the cavity and using the Poisson equation, we can show that the electric field inside the cavity is zero. This is because there is no change in potential inside the cavity, and Earnshaw's theorem dictates that V can only have minima or maxima on the surface, resulting in a constant potential and a zero electric field inside the cavity.