Transcript Slide 1
Example 5:
insulating cylinder of radius R and positive uniform linear charge density .
calculate the potential at a point outside a
very long r
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.
Reason #2: I suspect that we would eventually find (read your text) that V= at any finite distance from the cylinder. Not very useful.
E outside a cylinder of charge. This can easily be integrated to find V.
To be worked at the blackboard in lecture…
E r r=a dr r=R >0 R
E 2 λ πε
0
r
i f
V
R a
.
E r r=a dr r=R
V
a
V
R
V
aR
V
R a
R a
R a
R a
λ 2πε r
0
dr λ 2πε
0 R a
dr r λ 2πε
0
R a a R >0 R
V
a
V
R
λ 2πε
0
λ 2πε
0
a ln R λ 2πε
0
ln R a
If we let a be an arbitrary distance r, then
V
r
V
R
λ 2πε
0
ln R .
r
If we take V=0 at r=R, then
λ 2πε
0
ln R .
r
Things to note:
λ 2πε
0
ln R .
r
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.
V
r
V
R
λ 2πε
0
ln R r
If we tried to use V=0 at r= then
V
r
V
λ 2πε
0
ln r
(V is infinite at any finite r).
That’s why we can’t start with
dV k dq .
r
Things to note:
V
r
V
R
λ 2πε
0
ln R r
For >0 and r>R, V r – V R <0.
Our text’s convention is V 84. Thus V rR = V r – V R r and R and for r>R, V rR ab = V a is the potential difference between points < 0. – V b . This is explained on page In Physics 23, V b a = V a – V b . I like the Physics 23 notation because it clearly shows where you start and end. But V ab has mathematical advantages which we will see in Chapter 4.
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 V ab V.
V i f = V f – V i means, and how it relates to so V i f = -V if
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:
E
line
2
0
r .
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 3.32 and 3.33: you
must
derive an expression for the potential outside a long conducting cylinder. See example 3.10. V is not zero at infinity in this case. Use i f If 3.32 and 3.33 are not assigned, don’t be disappointed. We can still get you on this in problems in chapter 4!
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
U k q q U q V .
r
12 If the electric field is zero everywhere in some region, what can you say about the potential in that region? Why?