- #1

CharlieCW

- 56

- 5

## Homework Statement

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##.

## Homework 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

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

## 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).