I am trying to generalize the electric field strength equation as a function of radius from the anode for Gas Electron Multiplier (GEM). I think you can think this as a parallel plate capacitor (Mansfield p. 457). The potential difference between two points is (Mansfield p. 447, ch 15) etc in the parallel plate capacitor:

\begin{equation} V := \int_{A}^{B} \mathscr{E} \, d \mathbf l = \mathscr{E} \int_{A}^{B} d \mathbf l = \mathscr{E} r \end{equation} and hence $E = \frac{V}{r}$.

#1 proposal.

\begin{equation} \mathscr{E}(r) = \frac{V} {\omega \cdot r} \end{equation} where $\mathscr{E}$ is the electric field strength, $r$ radius, $V$ applied voltage and $\omega$ characteristic morphological factor, which is for instance in case of cylindricer $\omega = \ln (a/b)$ (Knoll et al) where $a$ and $b$ are constants. The linear relationship of $r$ cannot hold for all morphologies so I started to suspect if the equation can hold even for GEM. I think I should present by an integral form but I could not find one.

#2 form (p. 451).

\begin{equation} \mathscr{E}(r) = \frac{Q}{4 \pi \epsilon r^{2}} \hat{r} \end{equation}

and integrating it gives

\begin{equation} V(r) =- \int \mathbf E d \mathbf l = - \int_{a}^{r} \frac{Q}{4\pi \epsilon r^{2}} \hat{r} d r = - \int_{a}^{r} \frac{Q}{4 \pi \epsilon r^{2}} |\hat{r}| |d r| = - \int_{a}^{r} \frac{Q}{4\pi \epsilon} \, \frac{dr}{r^{2}} = - \frac{Q}{4\pi \epsilon} \left[ - \frac{1}{r} \right]^{r}_{a} \end{equation}

implying $V(r) = \frac{Q}{4 \pi \epsilon r} ln(r/a)$.

Some formulee

The Gauss's law is in the differential form: \begin{equation} \nabla \cdot \mathscr{E} = \frac{\rho} {\epsilon_{0}}. \end{equation}


  • Understanding Physics, M. Mansfield


  • So you need two integrals for inside and outside. I think the inside is the hole in GEM foil, while the outside is the integral over the square GEM container. However, not sure. How can you present those inside and outside integrals?
  • How can you present the electric field strength as a function of radius for a GEM?
  • 1
    $\begingroup$ It differs depending on where you are, but just use Gauss' law to integrate over the charge distribution with a different integral for inside and outside. $\endgroup$ – honeste_vivere Nov 3 '15 at 23:24
  • $\begingroup$ @honeste_vivere How can you present those inside and outside integrals? $\endgroup$ – Léo Léopold Hertz 준영 Nov 3 '15 at 23:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.