Surface charge of infinite conducting cylinder

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Homework Help Overview

The discussion revolves around the electrostatics of an infinite conducting cylinder with a line charge positioned outside of it. The original poster attempts to find the potential and surface charge induced on the cylinder due to the line charge using the method of images and integration of the electric field.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of the method of images and the application of Gauss's Law to find the electric field and potential. There are questions about the differentiation process and the interpretation of variables r and r' in the context of the problem.

Discussion Status

Participants are actively engaging with the problem, clarifying the relationships between variables and discussing the implications of their calculations. Some guidance has been provided regarding the differentiation process and the use of cylindrical coordinates, indicating a productive direction in the discussion.

Contextual Notes

There is an ongoing exploration of the definitions and properties of grounded versus neutral conductors, as well as the implications of the line charge's position relative to the cylinder.

dcrisci
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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\frac{\partial V}{\partial r}

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!
 
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dcrisci said:
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})

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.
 
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:
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.
 
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?
 
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).
 
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?
 
Yes.
 
TSny said:
Yes.

Thank you so much for your help!
 

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