Electric potential of cylinder

1. Feb 17, 2008

743344

1. The problem statement, all variables and given/known data

A hollow cylindrical shell of length L and radius R has charge Q uniformly distributed along its length.

What is the electric potential at the center of the cylinder?

2. Relevant equations

n = Q/A
V = kQ/r

3. The attempt at a solution

1) the cylinder has uniform surface charge densitiy n = Q/A;
where A = 2piR^2 + 2piRL = area of cylinder
therefore: n = Q/(2piR^2 + 2piRL)

2) now i am stuck.......because for the two circle parts of the cylinder (2 disks) i know i can find the potential by dividing the disk into rings and then using integration..........but i don't know how to incorporate the length of the cylinder in to the problem before intergation........i mean i know know how to start dividing the cylinder in to small pieces and then using the surface charge desity and then intigrating to find the equation?

please give me as much info about this quesition as posible.....since the assignment is due tomorrow and this is the only problem i have left to do

thank you

2. Feb 17, 2008

pam

As I read the problem, I think the ends are open, so the charge is only on the curved surface. Use the potential on the axis of a ring and integrate to get your answer.

3. Feb 17, 2008

marlon

The ends don't play any role because the electrical field lines will be tangetial with respect to these surfaces !

What matters is this :
a) is the charge distributed along the surface ? If yes : Gauss's law teaches us that E=0 inside the cylinder

b) is the charge distributed along the volume of the cylinder ? If yes, E will be non zero.

To calculate E and then the potential, one can apply the law of Gauss ! You will need the electrical flux through a second cylinder, inside the given cylinder + the total charge inside the second cylinder!

marlon

marlon

Last edited: Feb 17, 2008
4. Feb 18, 2008

pam

Since the problem refers to "charge Q uniformly distributed along its length", it implies that the ends are open.
Gauss's law does not imply that E=0 or anything else.
Gauss's law cannot be used to find the potential at the center, because the configuration does not have enough symmetry with a finite length cylinder. Use the method in post #2.

5. Feb 24, 2008

marlon

Not enough symmetry ? It's a cylinder with a charge that is uniformely distributed along the surface ??? What extra symmetry do you need ?

Seems to me you have some conceptual problems with the Law of Gauss...

marlon

Last edited: Feb 24, 2008
6. Feb 24, 2008

pam

It is of finite length, so E is not constant along its length.
One of us does.

7. Feb 24, 2008

Dr Transport

The problem says "At the center of the cylinder.".....

8. Feb 25, 2008

pam

Is that relevant to the use of Gauss's law?

9. Feb 25, 2008

marlon

yes what matters is whether you are inside or outside the cylinder

marlon

10. Feb 25, 2008

Ulysees

You might have to

1. divide the cylinder into circular rings
2. pair them up about the centre (which cancels any component of the E-field in the direction of the cylinder)
3. work out the E-field component in the direction perpendicular to the cylinder by integrating over one of the two rings and doubling.
4. integrate this E-field component from the centre of the cylinder to infinity, in a direction perpendicular to the cylnder to get the potential (by convention the potential is set to zero at infinity)

But have you been taught integration?