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Potential inside a cylindrical shell with a line charge

  • Thread starter CharlieCW
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1. The problem statement, all variables and given/known data

Find the electric potential of an infinitely long cylinder shell of radius ##R## whose walls are grounded, when in its interior a line charge, parallel to the cylinder, is placed at ##r=a## (with ##a<R##) and that has a lineal charge density ##\lambda##.

2. Relevant equations

Potential of a line charge:

$$\phi(r)=A-\frac{\lambda}{2\pi \epsilon_0}ln(r)=\frac{\lambda}{4\pi \epsilon_0}ln(\frac{B^2}{r^2})$$

Poisson equation

$$\nabla^2 \phi=\rho / \epsilon_0$$

3. The attempt at a solution

My first idea was to start with Poisson's equation in cylindrical coordinates and try to solve for the Dirichlet boundary conditions ##\phi(r=R)=0## and ##\phi(r\rightarrow \infty## for all ##\theta\in[0,2\pi]##. However, this leads to a group of three differential equations which are very hard to solve analytically.

Therefore, I tried using the method of images by placing instead a charge line outside of the cylinder at some distance ##r=b##. Therefore, the potential is the sum of the original line charge and the image line charge:

$$\phi(r)=\frac{\lambda}{4\pi \epsilon_0}ln(\frac{B^2}{r_1^2})+\frac{\lambda'}{4\pi \epsilon_0}ln(\frac{B^2}{r_2^2})$$

Changing to cylindrical coordinates:

$$\phi(\rho,\phi)=\frac{\lambda}{4\pi \epsilon_0}ln(\frac{B^2}{\rho^2+a^2-2\rho acos\phi})+\frac{\lambda'}{4\pi \epsilon_0}ln(\frac{B^2}{\rho^2+b^2-2\rho bcos\phi})$$

Applying the boundary condition ##\phi(r\rightarrow \infty)=0## we can find that ##\lambda'=-\lambda##, which makes sense since the total charge should be zero.

Applying the boundary condition ##\phi(r=R)##, we get:

$$\phi(R,\phi)=\frac{\lambda}{4\pi \epsilon_0}ln(\frac{B^2}{R^2+a^2-2\rho acos\phi})-\frac{\lambda}{4\pi \epsilon_0}ln(\frac{B^2}{R^2+b^2-2\rho bcos\phi})=0$$

Where we can easily see that it's true only if ##b=a##, which means that the virtual charge line is exactly where the original charge line used to be, so it doesn't make sense.

Therefore, I'm not sure if it's even correct to apply the method of images here, moreover since the original line charge is inside the cylinder and not outside (like this solution https://www.ebah.com.br/content/ABAAAhCJgAJ/jackson-solutions-jackson-2-11-homework-solution).
 

BvU

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can you explain ##r_1## and ##r_2## in cylindrical ?
 

PeroK

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$$\phi(r)=\frac{\lambda}{4\pi \epsilon_0}ln(\frac{B^2}{r_1^2})+\frac{\lambda'}{4\pi \epsilon_0}ln(\frac{B^2}{r_2^2})$$

Changing to cylindrical coordinates:

$$\phi(\rho,\phi)=\frac{\lambda}{4\pi \epsilon_0}ln(\frac{B^2}{\rho^2+a^2-2\rho acos\phi})+\frac{\lambda'}{4\pi \epsilon_0}ln(\frac{B^2}{\rho^2+b^2-2\rho bcos\phi})$$
The method of images should work. The trick (and it is tricky) is to get a cylinder of equipotential about the first line. You can do this with another equal and opposite line change at a certain distance.

Hint: first place just the two wires an equal distance from the x-axis, say, and find the equation of the equipotential surfaces. Maybe you have already done that problem previously?
 
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can you explain ##r_1## and ##r_2## in cylindrical ?
Sure. The ##r_1## and ##r_2## are the distances of the line charges to the observation point. Using the law of cosinus, we can express this distance in terms of the radial distance ##\rho## in cylindrical coordinates as ##r_1^2=\rho^2+a^2-2\rho a cos\phi##. I don't have the drawing, but imagine a triangle with side ##\rho## and ##a## (I also checked in the link that this is correct).

The method of images should work. The trick (and it is tricky) is to get a cylinder of equipotential about the first line. You can do this with another equal and opposite line change at a certain distance.

Hint: first place just the two wires an equal distance from the x-axis, say, and find the equation of the equipotential surfaces. Maybe you have already done that problem previously?
I checked the potential for an infinite straight wire, which can be written like the one I wrote (depending on how I choose my constant) or like this:

$$\phi(r)=\frac{\lambda}{2\pi \epsilon_0}ln(\frac{2A}{r})$$

I'll try with this one to see if it makes thing easier. Thank you by the way, I was unsure about using the method of images because I had only solved problems where the charges were at the center or outside of the conductor shell. I solved before the case for punctual charges, so I'll work the problem in analogy for the line charges and post the solution in a while.
 

BvU

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The method of images should work. The trick (and it is tricky) is to get a cylinder of equipotential about the first line. You can do this with another equal and opposite line change at a certain distance.

Hint: first place just the two wires an equal distance from the x-axis, say, and find the equation of the equipotential surfaces. Maybe you have already done that problem previously?
I worked out a similar case where the cylinder is not grounded but the line charge makes the cylinder an equipotential surface with respect to it. I do this by applying the method of images again and applying the boundary conditions ##\phi(r\rightarrow \infty)=0## and ##\phi(r=R)=V_0##.

From the first boundary condition the procedure is the same as my original post and I get ##\lambda'=-\lambda##. From the second boundary condition I get instead:

$$\phi(R,\phi)=\frac{\lambda}{4\pi \epsilon_0}ln(\frac{B^2}{R^2+a^2-2\rho acos\phi})-\frac{\lambda}{4\pi \epsilon_0}ln(\frac{B^2}{R^2+b^2-2\rho bcos\phi})=V_0$$

Which after some algebra gives me ##b=\frac{R^2}{a}##. Knowing this it's just a matter of substituting in the previous equations to find the potential inside and outside of the cylinder (outside it's zero since the image line charge doesn't really exist and the field is contained inside the cylinder like a Faraday's cage).

However, when I ground the cylinder I effectively make it's potential equal to zero and I arrive to the issue of my original post (that is, I predict the image line charge is directly over the original line charge).

Nevertheless, the fact that I found that the distance ##b## is independent of ##V_0## makes me think that the solution is also valid even if the cylinder is grounded. Am I correct?
 

PeroK

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I worked out a similar case where the cylinder is not grounded but the line charge makes the cylinder an equipotential surface with respect to it. I do this by applying the method of images again and applying the boundary conditions ##\phi(r\rightarrow \infty)=0## and ##\phi(r=R)=V_0##.

From the first boundary condition the procedure is the same as my original post and I get ##\lambda'=-\lambda##. From the second boundary condition I get instead:

$$\phi(R,\phi)=\frac{\lambda}{4\pi \epsilon_0}ln(\frac{B^2}{R^2+a^2-2\rho acos\phi})-\frac{\lambda}{4\pi \epsilon_0}ln(\frac{B^2}{R^2+b^2-2\rho bcos\phi})=V_0$$

Which after some algebra gives me ##b=\frac{R^2}{a}##. Knowing this it's just a matter of substituting in the previous equations to find the potential inside and outside of the cylinder (outside it's zero since the image line charge doesn't really exist and the field is contained inside the cylinder like a Faraday's cage).

However, when I ground the cylinder I effectively make it's potential equal to zero and I arrive to the issue of my original post (that is, I predict the image line charge is directly over the original line charge).
Nevertheless, the fact that I found that the distance ##b## is independent of ##V_0## makes me think that the solution is also valid even if the cylinder is grounded. Am I correct?
##b=\frac{R^2}{a}## is correct. Yes, ##b## determines the radius of the equipotential.

That gives you an equipotential on the cylinder, which must be ##0## if the cylinder is grounded. All you need to do then is set the potential equal to ##0##. I would stick with Cartesian coordinates:

##\phi = \frac{\lambda}{4\pi \epsilon_0}\ln(\frac{(x-b)^2 + y^2}{(x-a)^2 + y^2}) + V##

You can then find the constant ##V## that sets the potential to the conventional ##0## for a grounded conductor.

Note that ##b=a## is another solution for zero potential, but that one kills all the internal charge and isn't valid.
 
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