Surface charge of infinite conducting cylinder

In summary, the conversation discusses the process of finding the potential and surface charge on an infinite conducting cylinder with an infinite line charge placed a distance d from the axis. By utilizing the method of images and Gauss's Law, the electric field is calculated and the potential is found to be V = (λ/2πε)ln(r'/r). However, there is confusion on how to find the surface charge on the cylinder, as the direction of the normal vector is not clear. It is suggested to express r and r' in terms of cylindrical coordinates (s, θ, z) and differentiate with respect to s in order to find the normal derivative of V.
  • #1
dcrisci
45
0

Homework Statement


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.

Homework Equations



V = -∫ E⋅dr
δ = -ε0[itex] \frac{\partial V}{\partial r}[/itex]

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 = [itex] \frac{λ}{2πε}\ln\frac{r'}{r}[/itex]

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 = [itex] \frac{-λ}{2π}\frac{\partial }{\partial r}ln(\frac{r'}{r}) [/itex]

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

δ = [itex] \frac{λ}{2πr} [/itex]

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!
 
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  • #2
dcrisci said:
V = [itex] \frac{λ}{2πε}\ln\frac{r'}{r}[/itex]

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 = [itex] \frac{-λ}{2π}\frac{\partial }{\partial r}ln(\frac{r'}{r}) [/itex]

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
TSny said:
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.

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:
  • #4
dcrisci said:
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'?
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
TSny said:
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.

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
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
TSny said:
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).

Okay so I think I am 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
Yes.
 
  • #9
TSny said:
Yes.

Thank you so much for your help!
 

1. What is the surface charge of an infinite conducting cylinder?

The surface charge of an infinite conducting cylinder is zero, as the charge distributes evenly throughout the surface of the cylinder due to the conductive material.

2. How does the surface charge of an infinite conducting cylinder affect its electric field?

The surface charge of an infinite conducting cylinder creates an electric field that is perpendicular to the surface and does not vary with distance from the cylinder. This results in a constant electric field around the cylinder.

3. Can the surface charge of an infinite conducting cylinder be changed?

No, the surface charge of an infinite conducting cylinder cannot be changed as it is determined by the material of the cylinder and the charge distribution will always be even due to the conductivity of the material.

4. What is the relationship between the surface charge density and the electric field on an infinite conducting cylinder?

The surface charge density and the electric field on an infinite conducting cylinder are directly proportional. This means that as the surface charge density increases, the electric field also increases.

5. How does the shape of an infinite conducting cylinder affect its surface charge?

The shape of an infinite conducting cylinder does not affect its surface charge, as long as it remains an infinite cylinder. Any changes in shape may alter the distribution of charge and therefore affect the surface charge.

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