Electric potential on long insulating cylinder

AI Thread Summary
A long insulating cylindrical shell with a radius of 0.06 m and a linear charge density of 8.5E-6 C is analyzed for electric potential difference. The approach involves using Gauss's Law to find the electric field, which simplifies due to the cylinder's effective infinite length, allowing height to cancel out in calculations. The potential difference can be calculated using the integral of the electric field, with the formula V = ∫ E ds, where E is constant across the surface. The discussion emphasizes that the height of the cylinder is not necessary for the calculations, as it cancels out. The final expression for the electric field and potential difference is confirmed to be correct after recalibrating the calculator.
Seraph404
Messages
63
Reaction score
0

Homework Statement



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?

Homework Equations



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


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.
 
Physics news on Phys.org
"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.
 
Is \intdl 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:
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
 
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.
 
So...

S = 2\pirh
ES = \lambda*h/\epsilon0

E(r) = \lambda*h/(\epsilon02\pirh)

h cancles...

E(r) = \lambda/(\epsilon02\pir)

And now...

v = \int E(r) * dr

Is that right?
 
looks good to me!
 
Nevermind. I think I just need to clear the memory out the memory in my calculator. Thanks a lot!
 
Last edited:
Back
Top