Charge is uniformly distributed with charge density $ρ$ inside a very long cylinder of radius $R$.

Find the potential difference between the surface and the axis of the cylinder.

Express your answer in terms of the variables $ρ$, $R$, and appropriate constants.


I am struggling with determining which Gaussian surface to use. If I use a cylinder, then the cylinder would have an infinite area, right? How can I deal with that? If I use a sphere (since I am trying to find the potential difference between only two points, one on the surface and one on the axis), what will be the charge inside the sphere?

If I use a sphere as my Gaussian surface, I get:

$$\int \overrightarrow{E}.d\overrightarrow{A}=\frac{Q }{\epsilon _{0}}$$

$$\Delta V = -\int_{i}^{f}\overrightarrow{E}.d\overrightarrow{s}$$

$$E = \frac{\rho }{4\pi R^{2}\epsilon _{0}}$$

$$\Delta V = \frac{\rho }{4\pi R^{2}\epsilon _{0}} \int_{0}^{R}dR=\frac{\rho }{4\pi R\epsilon _{0}}$$

But this is wrong.


4 Answers 4


All the above answers are correct, although none gives you an answer as to WHY you should NOT use a sphere and none addresses your "infinite gaussian surface" problem and you seem to be a bit confused on how things actually work (you have to understand how things work before you delve into the mathematics part).

So, if you use a sphere, then the integral in Gauss' law will not be easy because most of the electric field vectors will not be perpendicular (and thus you can get $E$ out of the integral). You are simply not using the symmetries of the system.

You must use your intuition to guess where are the electric field vectors are pointing (here outwards cylindrically because you have an infinite cylinder), so you must use a cylinder. Think about the symmetries of the sources to figure out the symmetries of the fields.

Now, about your infinite gaussian surface problem: You only have to create a gaussian surface of finite size, because Gauss' law gives you the NET (total) electric field -created from all its surroundings- on one side of the equation (in the integral) but you can find it by only the enclosed charge (other side of the equation). So, you must not have an infinite gaussian surface.


Actually, using a cylinder for your Gaussian surface is your best approach. The fact the area is infinite should not matter, if you expression the infinite length of the cylinder as a variable, say $l$. Noting that the Gaussian surface area, $A = 2\pi Rl$, and that $Q = \rho l$, the $l$ term should eventually cancel out in your working out.

  • $\begingroup$ Thank you very much for answering. If I do that, I get $$\Delta V = \frac{\rho }{2\pi R\epsilon_{0}}\int_{0}^{R}dR=\frac{\rho }{2\pi \epsilon_{0}}$$ Is this right? It doesn't look right to me since it is independent of R. $\endgroup$
    – user57015
    Commented Aug 31, 2014 at 11:51
  • $\begingroup$ I'm not 100%, but shouldn't the $R$ be inside the integral? $\endgroup$
    – Involute
    Commented Aug 31, 2014 at 11:57
  • $\begingroup$ So now I get $$\frac{\rho }{2\pi \epsilon_{0}}ln(R)$$ for the answer. Is there a way I can be sure this is correct? $\endgroup$
    – user57015
    Commented Aug 31, 2014 at 12:07

Like Eternal Code said, using a cylinder inside the original problem cylinder is the right approach. If you use Gauss' Law, you should find that the electric field inside the infinitely long, uniformly charged cylinder is

$$E=\frac{ρr}{(2ε_0)} $$

Now, to calculate the potential difference between the surface and axis of the cylinder,

$${\Delta V}=-\int_0^R \frac{ρr}{(2ε_0)}dr$$

This gives the potential difference between the surface and axis of the cylinder as being

$${\Delta V}=\frac{-ρ(R^2)}{4ε_0}$$

  • $\begingroup$ Could you please explain to me how you got $$E = \frac{\rho r}{2\epsilon_{0}}$$ $\endgroup$
    – user57015
    Commented Aug 31, 2014 at 12:35
  • $\begingroup$ Would it make sense to combine your two answers into a single post? $\endgroup$
    – Floris
    Commented Apr 26, 2016 at 15:55

By Gauss' Law,

$E\cdot A=\frac {q}{\epsilon_0}$ (assuming that the Electric Field is constant at every $dA$ and that it is always parallel to $dA$, which it is in this case)

Let us define the charge contained in the original problem cylinder as being $Q$ whereas the charge in the smaller Guassian cylinder as being $q$.

Therefore, the charge in the smaller Guassian cylinder is dependent on the ratio between the volumes of the two cylinders due to the uniform charge distribution:


This simplifies to


Also we know that $Q=ρV=ρπ(R^2)L$

Substituting this in we get

$$E\cdot A=\frac{ρπ(R^2)L(r^2)}{ε_0(R^2)}$$


Cancelling out from top and bottom gives us the answer



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