# What is the true electric potential at the terminals of a battery?

I recently studied Laplace's equation and numerical methods for solving it, and so I'm trying to simulate an electron gun.

(assuming cylindrical symmetry on the Wehnelt, anode, and beam.) The setup is simple. The filament releases electrons, the Wehnelt cylinder is the cathode and helps focus electrons, and the electrons accelerate towards the anode. This was enough for me to set everything up and solve for the electric potential, with potential=0 as distance goes to infinity. There's a problem though: There are an infinite number of ways to set the potential of the cathode/anode that leave the voltage difference the same.

For example, with a voltage of 1V between the plates...

Cathode set to -1V:

Cathode set to 0V:

(the visualization is a slice through the potential. It is cylindrically symmetric. Space has been compactified so that the midpoint is at the origin but the edges represent infinity, according to the equation $x_{\text{world}}=x/(1-x^2)$, where $-1<x<1$ are the coordinates in the image and both the radial/axial directions behave the same.)

The gradients of the potentials are different, so these two potentials represent physically different situations. I realized that I don't know which one is more correct/reasonable. I understand that potentials are relative, but for example if a lab frame is taken to have a potential of 0... which picture is the true picture if I brought in a 1v battery? Are the terminals at $-0.5$V and $0.5$V? Why?

## 1 Answer

Interesting compactification, although I admittedly have no idea whether there are any mathematical pitfalls that one must avoid when doing such a geometric manipulation. However, assuming the apparatus is reasonably well-separated from nearby objects and that the apparatus isn't floating at a significantly different voltage than the surroundings, the choice of the potential of the surroundings shouldn't have much impact in real life.

Typically in simulations representing objects in free space, a boundary condition for the air "walls" is $\nabla\cdot\mathbf{D}=0$. With this condition you don't have to worry about specifying the potential of the surroundings, and the potential within the finite domain is uniquely specified, and the field is uniquely specified as well. Unless you have reason to believe the surrounding environment is floating at a significantly different potential than the lens (which it shouldn't, since your circuit diagram indicates the device is grounded), $\nabla\cdot\mathbf{D}=0$ should be pretty accurate.

Also, I'm willing to bet there's a lot of old books written on this subject and, more importantly, actual experimental results and techniques from people who built and trouble-shooted them previously. Try checking your local library, they might have some helpful hints!

• I just covered $\mathbf{D}$. In a vacuum we have $\mathbf{D}=\mathbf{E}$ right? I thought that $\nabla \cdot \mathbf{E}$ is precisely Laplace's equation. The equation to be satisfied for the whole volume (save for boundaries), NOT the boundary conditions. – user12029 Nov 29 '13 at 2:56
• Is one answer that the top or bottom picture is correct depending on which one is grounded? But then my question becomes: what if there is no ground wire? [in a vacuum] – user12029 Nov 29 '13 at 2:57
• Whoops, sorry! I meant to write $\mathbf{n}\cdot\mathbf{D}=0$, not $\nabla\cdot\mathbf{D}=0$. I'll edit my response to fix that. You're correct about $\mathbf{D}=\epsilon_0\mathbf{E}$ in free space, and that $\epsilon_0$ is irrelevant since you get Laplace's equation anyways for the volume. – DumpsterDoofus Nov 29 '13 at 4:18