# How do I apply Gauss's law to coaxial conducting cylinders?

$$\vert \vec E\vert =\frac{\lambda}{2\pi \varepsilon_0 r}$$ So I know this is the magnitude of the electric field of a line of charge using a cylindrical Gaussian surface. But, now let's say I have two coaxial metal conducting cylinders, one with the inner cylinder radius $a$ that is positively charged, and the outer cylinder with a larger radius $b$ which we can just say is negatively charged. My goal is to find the Electric field at a distance $a > r > b$.

Now I know that charge density $\lambda$ = (Charge)/(Length of wire) in the line of charge formula, which is how the length cancels out from the denominator. But when I introduce the inner cylinder with radius $a$, how do I adjust this formula to account for the fact that I now have an inner conducting cylinder with some radius? Does the length portion of the charge density have to change to something that represents the area of the inner cylinder?

Part of me thinks the formula might be the same because, if the electric field just shoots radially outward from the inner cylinder then the radius doesn't matter. Is that the case? Or do I need to account for the size of this inner cylinder?

• It would help you if you don't start out solving the problem with the answer in mind. Use Gauss' law afresh, without worrying about the cancellation. If you have a cylinder, take a gaussian cylinder of length L around it. You can easily find the charge inside. The electric field is still going to be radial (what else could it be allowed to be?) and the dot product becomes normal multiplication. Then the usual stuff. Just remember that you always take the enclosed charge only. Sep 29, 2020 at 9:12

I think the figure below shows something like the geometry you have in mind: this is a cross-sectional view of an infinitely long cylinder, with inner solid cylinder of radius $a$ coaxial with a hollow cylinder of inner radius $b$.

The key point to observe is that a Gaussian cylinder of radius $a<r<b$ will only enclose the charge of the inner solid cylinder. Hence, a long as $a<r<b$, the $\vec E$ will be that of the inner cylinder alone. When $r$ goes beyond $b$ and encloses some or all of the charge of the outer hollow cylinder, the geometry will not change but the net charge enclosed will be reduced so the field will be reduced accordingly. If the outer hollow cylinder has the same charge per unit length as the solid inner one, then the net charge enclosed for $r>c$ will be $0$ and the field will thus be $0$ outside the arrangement.

[Figure credit: modified from Young and Freedman's University Physics]

The answer should be the same for the cylindrical capacitor with a inner radius $a$ and outer radius $b$.

For all $a < r < b$, the answer of the electric field for the cylinder will be equal to

$$E = \frac{\lambda}{2 \pi \epsilon_0 r}$$

Where $\lambda$ is the charge per length.

• So is the equation the same because I can draw the same Gaussian surface around the inner cylinder as I would a line of charge? Dec 10, 2017 at 19:59
• Yes. Similar to how hollow sphere field is zero inside of sphere, hollow cylinder field is zero inside of cylinder. Field will only depend on internal charges ie in this case solid cylinder of radius a. @studyingforphysicsrightnow Dec 13, 2019 at 8:43

Yes, those are the wonders of symmetry.

• Gauss' law only needs "charge inside the surface". It doesn't matter how that charge is distributed: both a line and a cylinder produce the same flux, no matter how the charge is distributed.
• But, the distribution is important if you're lookign for the electric field. NEver forget that Gauss law talks about the electric flux. The flux does not vary wether there is a wire or a cylinder. However, if you want to extract the electric field from the flux, you need the distribution to be symmetric. In this case, it is correctly symmetric, so that the electric field has the same value along the whole surface.

That's why, a cylinder behaves as if all chage were in the core line. Same with spheres: all charge can be considered to be at the center. This does not apply if the distribution is not uniform.