# Electric potential on long insulating cylinder

• Seraph404
In summary, the voltmeter would read 8.5E-6 C when connected between the surface of the cylinder and .04 m above the surface.
Seraph404

## 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.

"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 $$\int$$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:
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$$\pi$$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?

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:

## 1. What is an insulating cylinder?

An insulating cylinder is a type of cylindrical object that does not allow electric current to pass through it. This means that it is a non-conductive material that can act as a barrier for electricity.

## 2. How is electric potential defined on a long insulating cylinder?

Electric potential on a long insulating cylinder refers to the amount of electrical energy that is stored in the cylinder due to the presence of an electric field. It is a measure of the potential difference between two points on the surface of the cylinder.

## 3. How does the electric potential vary on a long insulating cylinder?

The electric potential on a long insulating cylinder varies depending on the distance from the center of the cylinder. As the distance increases, the electric potential decreases.

## 4. What factors affect the electric potential on a long insulating cylinder?

The electric potential on a long insulating cylinder is affected by the electric field strength, the distance from the center of the cylinder, and the dielectric constant of the material the cylinder is made of.

## 5. How is the electric potential calculated on a long insulating cylinder?

The electric potential on a long insulating cylinder can be calculated using the equation V = kQ/r, where V is the electric potential, k is the Coulomb's constant, Q is the charge on the cylinder, and r is the distance from the center of the cylinder.

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