# The Potential of a Cylinder

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• 0:03 What is Potential?
• 1:00 Gauss' Law for a Cylinder
• 3:21 Example Problem
• 4:10 Lesson Summary
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Lesson Transcript
Instructor: David Wood

David has taught Honors Physics, AP Physics, IB Physics and general science courses. He has a Masters in Education, and a Bachelors in Physics.

After watching this video, you will be able to explain what electric potential is and solve problems using an expression for the electric potential of a cylinder. In AP Physics C, you should be able to derive the expression. A short quiz will follow.

## What Is Potential?

Electric potential is the amount of electric potential energy that each unit charge would have at a particular point in space. It's measured in Joules per coulomb or volts. Two points in space have different electric potentials due to their different positions inside a field.

Since electric potential is quite hard to picture, one way to make it easier to imagine is to think in terms of something we experience every day: gravity. The gravitational potential is how much gravitational potential energy each kilogram of mass (instead of charge) would have at a particular location.

If you raise a ball to a height, h, and drop it, it moves because it had gravitational potential energy at that height. The amount of energy a 1 kilogram mass would have is the potential at that position. And if you drew field lines, they would point from the raised ball towards the ground - potential decreases as you follow the field lines. Electric potential works in exactly the same way.

## Gauss' Law for a Cylinder

In another lesson, we introduced Gauss' law. Gauss' law is used to come up with equations for the electric field created by a charged object. This could be any shape; it could be a charged sphere or, for today's lesson, a charged cylinder. Once we have an expression for electric field, we can use it to come up with an expression for electric potential.

Gauss' law says that the electric field times the area of a surface is equal to the charge enclosed by that surface divided by the permittivity of free space, which is always equal to 8.85 * 10^-12. The next step is to plug in an equation for the surface area of a particular object we're looking at.

Today, we're looking at a cylinder. But to make it easier, we're going to make it a long cylinder. By long, we mean that it's long enough that the edges of the cylinder are far away and don't affect the electric field result. For a long cylinder, the surface area is equal to the circumference of the cylinder (which is 2pir) multiplied by the length of the cylinder. Plug that into Gauss' law and rearrange to make E the subject, and we get this equation: E equals Q over 2pi epsilon-zero rL.

Last of all, we need to get rid of the length because with it being a long cylinder, we don't want length to be involved. To do that, we define the charge per unit length, lambda, where lambda equals Q divided by L. Our equation already has Q divided by L in it, so we can replace that with lambda instead. When you do that, you get this final equation, where lambda is the charge density per unit length of the cylinder, epsilon-zero is a constant that is always equal to 8.85 * 10^-12, and r is the distance you are from the center of the cylinder.

But we don't want an equation for electric field, we want an equation for the electric potential. To get this, we would have to do some calculus. The integral of the electric field with respect to the radius will give us the potential. So, we have to integrate the equation we have so far with respect to r.

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