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I'm working on a problem with an (infinitely) long cylinder with a charge density, and I'm trying to find the electric field. Using the charge density, I found the enclosed charge of a proportional, enclosed Gaussian surface, so

$$Q_enc=\frac{2\pi \ell s^3 k}{3}$$

As such, we know for the Gaussian surface

$$\int{E\cdot \mathrm da}=\frac{Q_{enc}}{e_0}$$

So, you can simply solve this if you know the area, and I know the solution is

$$E = \frac{k s^2 {\hat{\mathbf{s}}}}{3 \epsilon_0}$$

My question is, why do the charges on either end of the cylinder contribute nothing to E? My book tells me this is a result of E being perpendicular to $\mathrm da$, but I don't understand exactly why it that eliminates its contribution. I think it has something to do with the reliance of Gauss' Law on the divergence of E, but I don't entirely understand this in the physical sense. Hopefully the diagram helps, but let me know.

Gaussian surface inside long cylinder

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If $E$ is perpendicular to the surface, then the dot-product in the integration $E \cdot da$ is just zero, because the dot-product is defined as such (remember the $cos(\theta)$ in the definition; when $\theta=\pi/2$, $cos(\theta)=0$).

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  • $\begingroup$ Right, but what's the physical reason for this? Why can't the surface charge of the cylinder have E on the outside edges? $\endgroup$ – A4Treok Oct 2 '15 at 4:03
  • $\begingroup$ Wow, I blanked hard. Physical answer is that the charges on either side of the cylinder's lengthwise section will cancel out their respective (x,y) coordinate forces. All force is in z direction because of symmetry. $\endgroup$ – A4Treok Oct 2 '15 at 7:39
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    $\begingroup$ As I see it, it has more to do with the fact that the electric field has no component along the axis of the cylinder; the endcircles of the cylinder in the illustration is only the Gaussian box (which we can choose to be any form) - the cylinder itself extends to infinity and therefore doesn't have such endcircles. $\endgroup$ – Bobson Dugnutt Oct 2 '15 at 9:33
  • $\begingroup$ Yeah, that's a good point as well. Better explanation of why the ends contribute nothing. Thanks for the help! $\endgroup$ – A4Treok Oct 4 '15 at 5:54
  • $\begingroup$ This needs to be changed to say the dot product is $0$ if $E$ is parallel to the surface. Not perpendicular. This is because the area vector is perpendicular to the surface. $\endgroup$ – Aaron Stevens Aug 14 '18 at 17:36
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Since the cylinder is INFINITE at any point in space there is as much charge to the left as there is to the right. Therefore any E to the left cancels any E to the right. Hence no contribution from the sides.

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