1
$\begingroup$

Imagine two coaxial cylinders, one that is a volume with radius $Ra$ (and charge per unit lenght $-\lambda$) and another one that is just a surface with radius $Rc$ (and charge per unit lenght $+\lambda$), with $Rc > Ra$. What is the magnitude of the electric field everywhere in this distribution?

So the answer I have is this: $$ E(r) = \begin{cases} 0, & \text{, if $r < Ra$} \\ \frac{1}{2\pi \mathcal{E}_0}\frac{\lambda}{r}, & \text{, if $Ra<r<Rc$} \\ 0, & \text{, if Rc < r} \end{cases} $$

My questions regard all the three positions. Why is the Electric field zero in the $r<Ra$ and the $Rc<r$ regions? And in the middle region, shouldn't the electric field be the sum of the fields created by the two charges?

$\endgroup$

1 Answer 1

0
$\begingroup$

This can be answered easily on the theorem of Gauss law.. In the three cases consider a imaginary cylinder at first radius

Notice that charge enclosed in first case is 0 second case is λ*l and third case is again 0

Thus E.2πr*l=charge enclosed/epsilon Putting values give desired result

$\endgroup$
3
  • $\begingroup$ Yes, but why are the charges enclosed in the first and third cases zero? $\endgroup$
    – BSD
    Commented Apr 29, 2017 at 18:30
  • $\begingroup$ Charges reside on the surface and in first case it encloses no charged surface $\endgroup$ Commented Apr 29, 2017 at 22:39
  • $\begingroup$ In third case final charge adds up to 0 due to induction $\endgroup$ Commented Apr 29, 2017 at 22:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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