# Electric potential on long insulating cylinder

1. Feb 15, 2009

### Seraph404

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

A long, insulating cylindrical shell of radius .06 m carries a linear charge density of 8.5E-6 C spread uniformly over its outer surface. What would a voltmeter read if it were connected between the surface of the cylinder and .04 m above the surface?

2. Relevant equations

potential difference for a distribution of charge v = $$\int$$ dq/r

3. The attempt at a solution

I forget how to do a problem like this. I looked at some sources online, and for this, I first thought I could use gauss's law to find electric field, and then integrate E*dl... but I don't know what the height of my cylinder is in order to calculate it's surface area for Gauss's Law.

If possible, I would just like a few hints before I try to attempt the problem again. Maybe I'm forgetting about something.

2. Feb 15, 2009

### Delphi51

"long" means effectively infinitely long so you don't have to worry about the ends of your Gaussian cylinder around the real cylinder.

Express your charge with an h for height of the cylinder in it, express your Gaussian area also with an h in it and the h's should cancel out when you find the electric field.

3. Feb 15, 2009

### Seraph404

Is $$\int$$dl the distance from the surface of the insulating cylinder to the point at which I'm measuring potential? I'm guessing not because I'm totally not getting the right answer.

Last edited: Feb 15, 2009
4. Feb 15, 2009

### Gear300

Instead of using V = $$\int$$ dq/r use V = $$\int$$ E-ds in which - is dot product. Using Gauss's law, find an expression for the electric field as one goes further from the cylinder. You do not need the height for the cylinder; you'll find that it will cancel out (as Delphi51 said). Use the expression for E in the integral and note that the radial displacement in your expression is along the same path as ds, thus the integral could reduce to V = $$\int$$ E ds

5. Feb 15, 2009

### Delphi51

I don't think you need to integrate because E is the same everywhere on the surface of the cylinder. Integral of E*ds is just ES, where S is the surface area of the cylinder (ignore the ends). Gauss' law says ES = Q/epsilon. Put in expressions for S and Q involving h and r and do some canceling to get a nice expression for E as a function of r. Of course you will have to integrate to get V.

6. Feb 15, 2009

### Seraph404

So...

S = 2$$\pi$$rh
ES = $$\lambda$$*h/$$\epsilon$$0

E(r) = $$\lambda$$*h/($$\epsilon$$02$$\pi$$rh)

h cancles....

E(r) = $$\lambda$$/($$\epsilon$$02$$\pi$$r)

And now...

v = $$\int$$ E(r) * dr

Is that right?

7. Feb 15, 2009

### Delphi51

looks good to me!

8. Feb 15, 2009

### Seraph404

Nevermind. I think I just need to clear the memory out the memory in my calculator. Thanks a lot!

Last edited: Feb 15, 2009