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This tutorial explains how to calculate the potential at a point outside a long insulating cylinder of charge, using a simplified approach and without the need for complex integration. It also explores the concept of potential energy in cases where the charge distribution is unknown.
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Example 5: calculate the potential at a point outside a very long insulating cylinder of radius R and positive uniform linear charge density . Reason #1: a mathematical pain. What do I pick for my charge element dq? Little cubes? Ugh. Circular rings that I have to integrate from 0 to R and from - to along the axis? Long cylindrical shells that I have to integrate from 0 to R? There must be an easier way. I would prefer to not start with and integrate. Why? Reason #2: I suspect that we would eventually find (read your text) that V= at any finite distance from the cylinder. Not very useful. Reason #3: we have previously derived a simple expression for outside a cylinder of charge. This can easily be integrated to find V. To be worked at the blackboard in lecture…
r r=a >0 E dr r=R R Start with and to calculate
r r=a >0 E dr r=R R
If we let a be an arbitrary distance r, then If we take V=0 at r=R, then
Things to note: If we tried to use V=0 at r= then (V is infinite at any finite r). That’s why we can’t start with V is zero at the surface of the cylinder and decreases as you go further out. This makes sense! V decreases as you move away from positive charges.
Things to note: For >0 and r>R, Vr – VR <0. Our text’s convention is Vab = Va – Vb. This is explained on page 759. Thus VrR = Vr – VR is the potential difference between points r and R and for r>R, VrR < 0. In Physics 1135, Vba = Va – Vb. I like the Physics 1135 notation because it clearly shows where you start and end. But Vab has mathematical advantages which we will see in Chapter 24.
See your text for other examples of potentials calculated from charge distributions, as well as an alternate discussion of the electric field between charged parallel plates. Remember: worked examples in the text are “testable.” Make sure you know what Vab means, and how it relates to V. Vif = Vf – Vi so Vif = -Vif
Special Dispensation For tomorrow’s homework only: you may use the equation for the electric field of a long straight wire without first proving it: Of course, this is relevant only if a homework problem requires you to know the electric field of a long straight wire. You can also use this equation for the electric field outside a long cylinder that carries charge.
Homework Hint! Problems like 23.32 and 23.33: you must derive an expression for the potential outside a long conducting cylinder. See example 23.10. V is not zero at infinity in this case. Use If 23.32 and 23.33 are not assigned, don’t be disappointed. We can still get you on this in problems in chapter 24!
Homework Hints! In energy problems involving potentials, you may know the potential but not details of the charge distribution that produced it (or the charge distribution may be complex). In that case, you don’t want to attempt to calculate potential energy using . Instead, use If the electric field is zero everywhere in some region, what can you say about the potential in that region? Why?
