# Surface charge of infinite conducting cylinder

1. Oct 17, 2014

### dcrisci

1. The problem statement, all variables and given/known data
There is an infinite conducting cylinder positioned at the axis with radius R. An infinite line charge (+λ) is placed distance d from the axis and d>R. I was supposed to 1. Find the potential and then 2. find the surface charge on the cylinder.

2. Relevant equations

V = -∫ E⋅dr
δ = -ε0$\frac{\partial V}{\partial r}$

3. The attempt at a solution
I used method of images and placed a wire of -λ inside the cylinder on the same axis as the original line charge. I calculated the electric field using Gauss's Law, and then integrated it with respect to r (saying that the potential does not depend on θ in this case). I found the electric field of both line charges, integrated them to get the potential from each line charge, then added them together and simplified them to get:

V = $\frac{λ}{2πε}\ln\frac{r'}{r}$

then was supposed to find the surface charge on the cylinder that is induced by the line charge. Using the equation above I was stuck with:

δ = $\frac{-λ}{2π}\frac{\partial }{\partial r}ln(\frac{r'}{r})$

Upon differentiating the ln function I am left with -1/r, and the surface charge becomes

δ = $\frac{λ}{2πr}$

which is positive. However due to the positive line charge positioned at distance d from the axis of the cylinder, I would have thought the induced surface charged would have to be negative? I looked up the answer to the potential I found for the area outside the cylinder and its correct, so where am I going wrong here?

Edit: also I am curious as to what the difference between a conductor that is grounded and neutral. I looked up a definition for them and it gave the definition in terms of electrical circuits and the voltages at ground and neutral wires, however I was looking for more of a general definition that can be applied to these problems. Such as do both have V = 0 on the surface? Both have E = 0 inside the conductor? and any other properties between the two. Thank you!

2. Oct 17, 2014

### TSny

Make sure you are clear on the meaning of r and r'. Are you sure you want to take the derivative with respect to r to find the electric field? The field at the surface of the cylinder is related to the normal derivative of V at the surface.

3. Oct 17, 2014

### dcrisci

So in this case, r is the from the line charge +λ to the point on the surface of the cylinder and r' is from the -λ line charge to the same point on the surface of the cylinder. So I guess I was wrong as the r vector is actually pointing in the wrong direction, so I would differentiate with respect to r'?

Last edited: Oct 17, 2014
4. Oct 17, 2014

### TSny

No. In general, the normal direction to the surface of the cylinder is not in either the r or r' direction. If you choose an axis running through the center of the cylinder and let s, say, denote distance from the axis, then the normal derivative would be in the direction of increasing s.

5. Oct 17, 2014

### dcrisci

So I would have to put r and r' in terms of their components in cylindrical coordinates and differentiate with respect to s only?

6. Oct 17, 2014

### TSny

Yes. If you choose cylindrical coordinates (s, θ, z), then you should be able to express r and r' in terms of s and θ. So, V can be expressed in terms of s and θ (as well as R and d).

7. Oct 17, 2014

### dcrisci

Okay so I think Im getting there. I have found that

r' = sqrt( s^2 + b^2 - 2Rbcosθ) where b is the distance from the origin to the -λ line charge
r = sqrt( s^2 + d^2 - 2Rdcosθ) where d is the distance from the origin to the +λ charge

Differentiate r'/r with these functions with respect to s?

8. Oct 17, 2014

### TSny

Yes.

9. Oct 17, 2014

### dcrisci

Thank you so much for your help!