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I'm confused with the Gauss' Law to calculate the electric field for two coaxial cylindrical conductors of finite length.

I know that we can use the Gauss' Law to calculate the electric field for two coaxial cylindrical conductors of INFINITE length, but I don't see why we can use the Gauss' Law for the one of finite length.

Ok here is the problem.

enter image description here Two coaxial cylindrical conductors are shown in perspective and cross-section above. The inner cylinder has radius a = 2 cm, length L = 10 m and carries a total charge of Qinner = + 8 nC (1 nC = 10-9 C). The outer cylinder has an inner radius b = 6 cm, outer radius c = 7 cm, length L = 10 m and carries a total charge of Qouter = - 16 nC (1 nC = 10-9 C). What is Ex, the x-component of the electric field at point P which is located at the midpoint of the length of the cylinders at a distance r = 4 cm from the origin and makes an angle of 30o with the x-axis?

The problem I have is that, I thought, in order to use the Gauss' law, the electric field has to be constant through out a gaussian surface, but I don't see why we can use a cylindrical shell of radius $4cm$ and length $10m$ to calculate the electric field at $P$ since I think the elctric field is not the same if I move from $0m$ to $10m$ on z-axis. Help me.

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You are correct that you can't really use a gaussian of length 10 meters. Instead, consider using a very short gaussian cylinder near your area of interest. In this limit, you should still be able to approximate the electric field as "nice and symmetric" with respect to the area vectors $d\vec{A}$.

By the way, the same thing is done for finite sheets of charge. If you use a small enough gaussian surface that is located near the center of the sheet, the procedure and results become the same as that due to an infinite sheet. The only difference is the limited region of applicability.

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  • $\begingroup$ What do you mean by "The only difference is the limited region of applicability"? $\endgroup$ – User Jun 21 '14 at 16:07
  • $\begingroup$ By that confusing phrase I mean that the result for the electric field you obtain is only valid for regions near where you initially placed the gaussian surface. (Why would it be valid anywhere else?) And since you can really only place the gaussian surface near the "center" to obtain a result for the electric field (since you need nice symmetry to extract $E$ from the integral), it also means that your result for the electric field is only valid near the "center". That is, it has a limited spatial region of applicability. $\endgroup$ – BMS Jun 21 '14 at 16:10
  • $\begingroup$ Oh, so, you mean that the Gauss' Law can be used for the finite sheets of charge or finite line of charge as long as the point we are interested is located at the above of the center of the finite sheet or the center of the length of the finite line with a small gaussian surface? $\endgroup$ – User Jun 21 '14 at 16:15
  • $\begingroup$ Yes, though I'm not sure about the word "above" in your description. Either way I wouldn't recommend memorizing that bit of information. My opinion is that it's better to realize that to use Gauss' law to solve for $E$, you need to be able to take $E$ out of the integral, which means that it's constant over the surface(s). $\endgroup$ – BMS Jun 21 '14 at 16:18

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