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?


1 Answer 1


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

  • $\begingroup$ 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

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