# Potential above a charged shell

• Melawrghk
In summary, the conversation discussed how to find the potential of a point located 4.4cm above the surface of a long cylindrical shell with a radius of 6.90 cm and a linear charge density of 8.40 microC/m. The conversation involved using Gaussian surface to find the electric field created by the shell and then integrating to find the potential. The correct method was determined to be using d as the distance to the center of the cylinder and integrating from 0 to d with the resulting equation V=Q/(2pi*d*epsilon_0)ln((d-r)/r).
Melawrghk

## Homework Statement

A very long insulating cylindrical shell of radius 6.90 cm carries charge of linear density 8.40 microC/m spread uniformly over its outer surface. Find the potential of a point located 4.4cm above the surface of the shell

## The Attempt at a Solution

I used Gaussian surface to find the electric field created by the shell outside (Q = linear charge density):
QL/$$\epsilon_{0}$$ = E*2*pi*d*L,
where L is length, d is distance to point from the axis of the cylinder.
The Ls cancel out, rearranging, I got:
E=$$\frac{Q}{2pi*d*\epsilon_{0}}$$

I then integrated that to get the potential:
V=$$\frac{Q}{2pi*\epsilon_{0}} \int {(1/d)dd}$$
The integral's upper limit is d-r, and lower limit is r.
I got the equation below as a result:
V=$$\frac{Q}{2pi*d*\epsilon_{0}} ln(\frac{d-r}{r})$$

Substituting in the numbers I got V=-6.8*10^4, but that's wrong. Is it my method? Should I slice the cylinder into rings and do V from each one, then integrate the whole thing?

Last edited:
Melawrghk said:
I then integrated that to get the potential:
V=$$\frac{Q}{2pi*d*\epsilon_{0}} \int {(1/d)dd}$$
What's that extra 1/d doing out front of the integral?

The integral's upper limit is d-r, and lower limit is r.

Why is that?

gabbagabbahey said:
What's that extra 1/d doing out front of the integral?
Why is that?

First one: typo, sorry :)

Second one: now that I think about, I'm not sure. I was thinking of using distance from the cylinder's axis and then r would be the smallest so that you're still on the outside (well, surface) and d-r would be anything beyond that. Should those limits be d and 0 instead?..

EDIT: I got it! Instead of using d as distance to center of the cylinder, I used d as distance to surface and for gaussian surface I had:
E=Q/(epsilon*2*pi*(d+r))
And integrated taht with upper limit being 'd' and lower being 0. And it worked! :D Thanks so much.

Last edited:

## 1. What is the potential above a charged shell?

The potential above a charged shell is the amount of energy needed to move a unit charge from infinity to a point above the shell.

## 2. How is the potential above a charged shell calculated?

The potential above a charged shell can be calculated using the equation V = kQ/r, where V is the potential, k is the Coulomb constant, Q is the charge of the shell, and r is the distance from the center of the shell to the point above it.

## 3. Does the potential above a charged shell depend on the charge distribution?

Yes, the potential above a charged shell does depend on the charge distribution. The potential is directly proportional to the total charge of the shell, so a shell with a higher charge will have a higher potential above it.

## 4. How does the potential above a charged shell change with distance?

The potential above a charged shell decreases with distance. This is because the electric field created by the shell weakens as you move further away from it, resulting in a lower potential.

## 5. Can the potential above a charged shell be negative?

Yes, the potential above a charged shell can be negative. This would occur if the charge of the shell is negative, which would result in an attractive force between the shell and a positively charged test charge above it.

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