There are two hollow cylinders with same lengths "l" as shown in the figure below. The smaller inner cylinder is negatively charged. The outer one is now induced to become positively charged. I am trying to derive a general equation for Electric field that would give field everywhere around these cylinders- outside them, between them and at the center.

enter image description here

So I started like this:

$$ \textrm{Electric field at point r,} \\ E_{r} = E_{big} + E_{small}\\ E_{big} = \frac{kQ}{r^2}\\ Q = \sigma (2\pi bl)\\ \therefore E_{big}=\frac{\sigma (2\pi bl)}{r^2} \\ \textrm{Similarly for the smaller inner cylinder}\\ E_{small} = - \frac{\sigma (2\pi al)}{r^2}\\ \therefore E_r = \frac{2\pi l\sigma}{r^2}(b-a) $$

This is all I got. Is this correct. If yes, how should I use it for calculating the field inside, outside and in between the cylinders? Will these equations change if the outer cylinder is grounded?

Second attempt (only for inner cylinder and assuming L is much much larger than r):

$$ \textrm{Surface charge density of inner cylinder,}\\ \sigma = \frac{Q_{enclosed}}{Area}\\ \sigma = \frac{Q_{e}}{2\pi aL}\\\\ \therefore Q_e = 2\sigma \pi aL \\ \textrm{where L is the length of the cylinder} \\ \phi = \int_{0}^{L} E.dA_{Gauss} = \frac{Q_e}{\epsilon_0}\\ E\int_{0}^{L}dA_G = E (2\pi r L) = \frac{2\sigma\pi aL}{\epsilon_0} \\ \therefore E=\frac{\sigma a}{r\epsilon_0} $$

I am not sure if I should be doing a closed integral rather than from 0 to L. It won't effect the result either way, will it?

  • $\begingroup$ Second attempt look good assuming $E=E_{small}$ and $r>a$ and all the assumptions you mentioned in the second attempt. $\endgroup$
    – Timaeus
    Commented Jan 18, 2015 at 0:53
  • $\begingroup$ Well, your second attempt is reasonable. Yet I'd like to point out, if you don't mind, that there is still something to improve for the sake of rigorousness of your solution. I'll add it to my answer because it has to include some attachment. $\endgroup$
    – Vim
    Commented Jan 18, 2015 at 7:15

2 Answers 2


No, that's not correct (except as a far-field approximation - see the note at the bottom). You're barking down the wrong tree. Your second equation $$E_{big} = \frac{kQ}{r^2}$$

is valid only for a point charge! You're approximating your cylinders as though they were single point charges at the same point in space, which is of course going to lose most of the complexity of your situation.

There are two ways to go - you can either integrate the electric field contribution from every differential element of charge on both cylinders. That's gonna get ugly.

The best way to go is to use Gauss's law with a cylindrical gaussian surface. This relates the flux through the surface to the charge contained inside the surface. If you pick a convenient symmetrical surface, you can deduce the electric field. This, unfortunately, is only possible if you're willing to assume that $l >> r$ (that is, the cylinders are very long compared to how far away from them we are).

Also, note that you dropped the $k$ accidentally - your last equation should read

$$E_r = \frac{k 2\pi l\sigma}{r^2}(b-a)$$

Interesting side note - your final equation will be valid for $r >> l$, since when you are very far from the cylinders, they are indistinguishable from two superimposed point charges. So you have in fact found what the electric field approaches as you get very far away from your two cylinders. So, in a way, your equation is just as correct as the one you'd get from Gauss's law, just for a different area of space.

  • $\begingroup$ So, if we were to imagine that these two hollow cylinders were indeed infinitely long, can we apply Gauss' Law? $\endgroup$ Commented Jan 17, 2015 at 23:14
  • $\begingroup$ Yes - Gauss's law will only give you an exact analytical solution in the case of infinitely long cylinders, but it's a good approximation as long as $l>>r$. $\endgroup$
    – Brionius
    Commented Jan 18, 2015 at 0:30

First, we should clarify that it is not the outer cylinder, but the inner surface of the outer cylinder that is positively charged. Since at the beginning the outer cylinder is neutral, to maintain the conservation of charge, its outer surface must be negatively charged to make the sum remain zero.
Another important thing, all of your calculation should be based on one assumption that $l>>b$ which is the prerequiste for the application of Gauss's Theorem. If this doesn't hold, I'm afraid it would be extremely hard (actually not possible) to accomplish your calculation. (Well, the reason is that without this assumption there would be no symmetry for $\vec{E}$, say, it would not be radial, and then even applying Gauss's Theorem would come to nothing.)
As for the case where the outer cylinder is grounded, I think it is best to explain by the property of field lines. Note that any field line begins from positive charge or infinity, and ends up in negative charge or infinity. Now that the outer cylinder is grounded, it is plain to see that the outer surface of the outer cylinder will no longer be charged (say, completely neutralized). If not, then there would be field lines end up in the (negative) charge on the surface, which have no alternative but to start from infinity, and thus causing a voltage rise from the outer cylinder to infinity. But since the outer cylinder is grounded, we should be aware that $V_{ground} = V_{infinity}$, Absurdum! Therefore the outer surface of the outer cylinder must be neutral. And, what about the inner surface of the outer cylinder? The answer is everything remains: the charge distribution does not change! To explain why, consider the fact we just deduced that there is no field lines outside the outer cylinder, in other words, $\vec{E}$ is constantly 0. By applying Gauss Theorem using a proper Gaussian surface, it is easy (well, I think you are completely able to do it yourself :-) ) to further deduce that the inner surface of the outer cylinder must be charged with the equal amount of the charge of the inner cylinder. So the inner surface of the outer cylinder is still charged with $+Q$. And according to the uniqueness of electrostatic field (or more simply, the symmetry), the distribution doesn't change either.

About your second attempt:
1).Although your result is correct, I have some doubt whether your selection of the Gaussian surface is proper. You should be aware about why we must first assume that $L>>r$: for symmetry, well, and to avoid the so-called "fringe effect" which will break the radial symmetry of the field ,thus making $\vec{E}$ not uniform on the Gaussian surface and unable to derive from Gauss's Theorem. So, actually what we are working on is indeed $\vec{E}$ at the points that are "exempt" from "fringe effect". When you select your Gaussian surface, you should not let it extend into the "fringe area", say, area not so far away from the two ends of the inner and outer cylinders. So I think the Gaussian surface should be a "short cylinder" that is far from the two ends, only in this way can we technically derive $\vec{E}$. And for an example of such a selection, see below(for visibility the outer cylinder is not shown) :enter image description here

2).Technically , yes, you should be doing a closed integral $\oint \vec{E} \centerdot d\vec{A}$. But you only need to consider the integral on the side surface of the Gaussian cylinder because for the two ends it is zero. But it is important to know that you are actually supposed to do a closed integral whenever applying Gauss's Theorem.

  • $\begingroup$ I have made another attempt. Could you take a look at it? $\endgroup$ Commented Jan 17, 2015 at 23:42

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