# Electric Field Around a Biased Material

I feel like this is a fairly straight forward question, but I cant seem to come to a good understanding.

Imagine I have a very large metal box. Inside that box I have a suspended metal plate. I have some voltage generator in which I attach the hot lead to the suspended plate and the ground lead to the large metal box. If I set the voltage generator to 100V, what is the electric field around the plate? Also how can you set the surface of a material to a set voltage like that? My understanding was as you got closer and closer to point charges your potential diverges?

I think the confusion I have is my picture and understanding of potentials and electric fields comes from point charges, and I am not sure how to draw a corollary to larger systems like charged plates.

By imposing $V=0$ for some very large metal box, you're basically just imposing $V=0$ at infinity, which is what we normally do for most finite charge distributions. Since the suspended plate is a finite charge distribution, the very large box can be ignored.

The electric field from a finite charged plate of dimensions $a\times b$ can be found through direct integration of

$$\mathbf{E}(\mathbf{r})=\int\frac{\rho(\mathbf{r'})(\mathbf{r}-\mathbf{r'})}{4\pi\epsilon_0|\mathbf{r}-\mathbf{r'}|^3}d\mathbf{r}'$$

where, assuming the plate is in the $xy$ plane,

$$\rho(\mathbf{r})=\begin{cases}\sigma\delta(z)\;\mathrm{ for }\;|x|<a,|y|<b\\ 0\;\mathrm{otherwise}\end{cases}$$

where $\delta(z)$ is the Dirac delta function. The result of this calculation is quite nasty for points off the $z$-axis, but we can still get some intuition from its form on the $z$-axis:

$$\mathbf{E}(z)=\frac{\sigma}{\pi\epsilon_0}\arctan\left(\frac{a}{2z\sqrt{(z/b)^2+(a/b)^2+1}}\right)\hat{\mathbf{z}}$$

As $z\to0$, we have that $\arctan(1/z)\to\frac{\pi}{2}$, so $\mathbf{E}\to\frac{\sigma}{2\epsilon_0}\hat{\mathbf{z}}$, which is the field of an infinite sheet. So, when you get close to a finite sheet, it looks infinite.

As $z\to\infty$, we have that

$$\arctan\left(\frac{a}{2z\sqrt{(z/b)^2+(a/b)^2+1}}\right)\to\frac{ab}{2z^2}$$

by Taylor expansion in $1/z$, so $\mathbf{E}\to\frac{\sigma ab}{4\pi\epsilon_0z^2}\hat{\mathbf{z}}$, which is exactly the formula for the field of a point charge $\sigma a b$ (which is the total charge on the plate). So the plate looks like a point if you go far enough away.

There's still one undetermined parameter left: $\sigma$, the charge density on the plate. This is related to the self-capacitance $C$ of the plate, which determines the total charge you can put on the plate at potential $V$ through the relation:

$$\sigma ab=CV$$

The potential $V$ on the plate can be found by integrating $\mathbf{E}$ along the $z$-axis from zero to infinity. This integral is pretty intractable, but from manually inserting different values for $a$ and $b$ in Mathematica, it seems that a good approximation is:

$$V\approx\frac{\sigma}{\pi\epsilon_0}\frac{b^2}{2a}$$

Therefore, setting $\sigma ab = CV$ gives you

$$C=2\pi\epsilon_0\frac{a^2}{b}$$

This means that the charge density $\sigma$ is completely determined by $a$, $b$, and $V$:

$$\sigma=\frac{CV}{ab}=2\pi\epsilon_0\frac{a}{b^2}V$$

• Thank you for the thoughtful answer! So my charge density and thus electric field is more a result of my particular power supply characteristics? The fact that I had it set to say 100V does not factor in? – Andrew Apr 2 '18 at 6:02
• Well, that might not be true. Let me think for a moment. I'll delete this answer until I have a definitive answer. – probably_someone Apr 2 '18 at 6:02
• @Andrew Edited to include the estimation of $\sigma$. – probably_someone Apr 2 '18 at 6:33
• Thank you, this model roughly lines up with what is found after modeling this in Comsol. – Andrew Apr 3 '18 at 18:00