I was trying to calculate the electric field on any point of the $z$-axis of a layer with the following properties:

  1. It as a thickness of $a$
  2. It is placed along the XY plane
  3. It is infinite in extension
  4. It has a uniform density of charge $\rho_o$

So I decided to do the following: let's imagine that the entire surface is made out of concentric pipe segments centered in $(0,0,0)$:


(Here you can see why I'm not a renowned artist). Each segment has a thickness of $dr$ and therefore a volume of $dV=(2\pi r) a dr$. It also has a charge $dq=\rho_o dV$.

For any point $(x,y,z)$ outside of the layer itself the electric field must point upwards and it must be the sum of the fields from all the infinitesimal rings under it. The further a right is, the bigger the angle $\theta$ between $dE$ and the $z$-axis. So we have:

$$E=\int_S dE = \int_S \frac{1}{4\pi \epsilon_o} \frac{\rho_o dV}{r^2+z^2}\cos{\theta} = \int_S \frac{1}{4\pi \epsilon_o} \frac{\rho_o dV}{r^2+z^2}\frac{z}{\sqrt{r^2+z^2}}$$

Taking the constants out, integrating and simplifying we get:

$$E= \frac{\rho_o a}{2\epsilon_o}$$

Now, I don't think what I got is right even if I'm not able to see where I failed. The field apparently doesn't depend on how close or far away we are from the layer, and that seems wrong. So I either made a math mistake or my premises were wrong. Where did I get it wrong and how?

PS: I realize I could express the field in vectorial form, but it's the process of deriving the equation above that I'm interested in.


Result isn't very intuitive, but absolutely correct. Infinite uniformly charged list gives uniform electric field. It's a well known task, you can always check your results by googling something like "electric field infinite plate".

  • $\begingroup$ It truly is counterintuitive! I hadn't thought about it like that, but you're right. It puzzles me to think that the field would have the same intensity both an inch and a thousand kilometers away from the plane, but I checked and apparently that's the way it is. Thanks for your answer :) $\endgroup$ – mar Apr 12 '17 at 18:05

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