1
$\begingroup$

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

Improve this question
$\endgroup$
1
  • $\begingroup$ As others have said, it's infinite, there is no edge. Everywhere, the field points radially outwards $\hat \rho$ $\endgroup$
    – jensen paull
    Commented Mar 4, 2022 at 14:31

2 Answers 2

Reset to default
1
$\begingroup$

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 $\theta=\pi/2$, $cos(\theta)=0$).

Improve this answer
$\endgroup$
5
  • $\begingroup$ 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
    Commented Oct 2, 2015 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
    Commented Oct 2, 2015 at 7:39
  • 2
    $\begingroup$ 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
    Commented Oct 2, 2015 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
    Commented Oct 4, 2015 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. $\endgroup$
    – BioPhysicist
    Commented Aug 14, 2018 at 17:36
0
$\begingroup$

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.

Improve this answer
$\endgroup$

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.