This is a problem from my homework, but I'm lost conceptually on how to do it. I'm not looking for someone to do the work for me, I'm looking for some guidance on how I can set it up/understand it

The Charge density of an electron's cloud in a Hydrogen atom is

$$\rho = - \frac{e}{\pi a^3} \exp(-2 r / a) \, .$$

Where $a$ is the Bohr radius, $r$ is the distance to the proton and the proton has a charge of $+e$. Investigate $E$ at small $r \ll a$ and large $r \gg a$ distances.

In addition to how the question is worded, I also know that we are supposed to use the delta function in our solution, though I'm a tad confused on how to do that.

Thinking about it logically for a moment though, it appears to me that $r \ll a$ should result in a positive electric field, while $r \gg a$ should result in a net $E=0$. Is this assumption correct?

  • 1
    $\begingroup$ Hello, and welcome to Physics Stack Exchange. Please note that this site supports mathjax so that you can write nice equations in your questions. I edited the question to use it. Please click the "edit" button to view the changed source and see how it works. Also you're more likely to get good answers if you ask specific questions. This is particularly important on homework-like questions because people vote to close those unless they ask a clear conceptual question. This question is pretty vague and will probably get closed. Can you make it more specific? $\endgroup$
    – DanielSank
    Sep 17, 2015 at 21:33

1 Answer 1


The most simple way of solving this is using Gauss' law and spherical symmetry. Then $$\int_{\delta \Omega} \mathbf{E}\; \text{d}\mathbf{F} = \int_{ \Omega} \nabla \mathbf{E} \; \text{d}V = \int_{ \Omega} \frac{\rho}{\varepsilon_0} \; \text{d}V $$ In the case of spherical symmetry this can be simplified to $$\int_{ \Omega} \mathbf{E}\; \text{d}\mathbf{F} = 4\pi r^2 E_0(r) = 4\pi \int \limits_0^r r'^2\frac{\rho(r')}{\varepsilon_0} \; \text{d}r' $$


Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.