Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm somewhat confused as to why the electric field outside a spherical shell is $\frac{Q}{4\pi r^{2}\epsilon_{0}}$

Going through the work:

$$ Q = 4 \pi r^{2} \sigma ; \frac{4\pi r^{2}\sigma}{ε_{0}} = \int \int_{s} \vec{E} \cdot d \vec{A} $$

Which, doing the integral leaves = $$\frac{4\pi r^{2}\sigma}{ε_{0}}= 4\pi R^{2}E$$ which means that $$E = \frac{r^{2} \sigma}{ε_{0}R^{2}}$$

Is this correct, or am I making some silly mistake?

share|cite|improve this question

Yes, that is right. Now $4\pi r^2\sigma=Q\implies r^2\sigma=\frac{Q}{4\pi}$, hence $E=\frac{Q}{4\pi\epsilon_0 R^2}$. Of course, in your notation you have switched $r$ for $R$, so that is why you may be confused.

share|cite|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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