Potential of charged cylinders

  • #1
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Homework Statement


An infinite solid cylinder of radius A and uniform charge distribution ρ is surrounded by a thin cylindrical envelope of radius B and linear charge distribution λ. The two cylinders are co-axial.
Find the potential V(r) as a function of r from r=0 to r=∞.

Homework Equations


V(r)-V(r0)=-∫r0rE⋅dl

The Attempt at a Solution


Quick notes:
  1. Do not worry about the constants in front of the integrals; my questions do not concern them, so I will omit them.
  2. Ignore the V(A) I have in my solutions, it's another unimportant constant.
  3. I am trying to find the potential function, not the potential difference between specific points.
I have already found the electric fields for the different regions; E(0<r<A), which is inside the solid cylinder, E(A<r<B), which is in the gap between the solid cylinder and the cylindrical surface and E(r>B), outside the whole system. The E in the first region is proportional to r, and the other two are proportional to r-1. I also have no problem finding V(0<r<A), it's just the two other domains that give me trouble.

What I am having trouble with is that I don't know how to structure my integral in the regions A<r<B and r>B such that V goes to zero as r increases. Currently, my solutions increase without bound. I do not know how to mathematically "choose" V(∞)=0 in this situation. My first thought was to integrate from r=A to r=r in the first region and r=B to r=r in the second, but this is what I got:


ΔV(A<r<B)=-∫Ar E(A<r<B)⋅dl

V(A<r<B)-V(A)=-∫Ar(1/r)dr

V(A<r<B)=V(A)+ln(A/r)

==> V(r>B)=V(B)+ln(B/r)=V(A)+ln(A/B)+ln(B/r)


Which doesn't work because the functions in both regions increase without bound (or at least the function in the first region would increase without bound had its domain not been restricted). But if I try to choose r=∞ as my reference point, I get


V(A<r<B)=V(A)+ln(∞/r)

==> V(r>B)=V(B)+ln(∞/r)=V(A)+ln(∞/B)+ln(∞/r)


Which also doesn't work because ∞ is in the actual expression and the constant ln(∞/b) makes no logical sense.

What am I doing wrong?

Again, I am trying to find the potential function for these regions, so structuring an integral from r=A to r=B does not help me because that gives the potential difference.
 

Answers and Replies

  • #2
Delta2
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1) You should choose the same reference point for all 3 cases and i believe that should be the point at r=0,z=0.
2) You only care for the case where r>B for the formula to have V(r)=0 as r goes to infinity. In the two other cases u already have a bound for r which doesnt allow it to go to infinity.
 
  • #3
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1) You should choose the same reference point for all 3 cases and i believe that should be the point at r=0,z=0.
2) You only care for the case where r>B for the formula to have V(r)=0 as r goes to infinity. In the two other cases u already have a bound for r which doesnt allow it to go to infinity.
Ok, but if my reference point is zero in every one of the cases, how do I keep the potential from going to infinity as r goes to infinity?
 
  • #4
Delta2
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Can you give me the expressions u have for electric field for 0<r<A, A<r<B and B<r<infinite?
 
  • #5
rude man
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Ok, but if my reference point is zero in every one of the cases, how do I keep the potential from going to infinity as r goes to infinity?
You don't.
You cannot have V(∞) = 0.
You can define V(r=0) = 0 however.
You'd have one expression for the potential difference between r=0 and r=a; another for a<r<b, and a third for r > b. At any point r the potential would be the sum of the previous region's max. potential plus the difference in potential between that boundary and r. For example, the potential in region 2 (betw. the cylinder and the shell) would be the potential at r = a + the potential difference between r=a and r, a < r < b.
 
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  • #6
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Can you give me the expressions u have for electric field for 0<r<A, A<r<B and B<r<infinite?
E(0<r<A)=ρ/(2ε0)r

E(A<r<B)=ρ/(2ε0)A2r-1

E(r>B)=(λ+πρA2)/(2πε0)r-1
 
  • #7
Delta2
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Ok so the potential for r>B will be [itex]V(r)=\int_{0}^{r}Edl=\int_{0}^{A}E(0<r<A)dl+\int_{A}^{B}E(A<r<B)dl+\int_{B}^{r}E(r>B)dl[/itex]

Ok i see as rudeman said you cant have V(infinite)=0, this is probably because of the infinite dimension along the z-axis of the charge distribution.
 
  • #8
rude man
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I think you know the right formulas but you're a bit sloppy in writing them.
Example: in region 1 it's E(0<r<A)=ρr/(2ε0). If you don't use enough brackets to remove all ambiguity then multiplication is assumed before division in general.
Fix the other two also. They all suggest you know what you're doing.
 
  • #9
Delta2
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Give him a break he is not the only one in this world having trouble with Latex and in general with formating equations in text :).
 
  • #10
collinsmark
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You don't.
You cannot have V(∞) = 0.
You can define V(r=0) = 0 however.
.
Actually the convention is the other way around, me thinks. Electrical potential is of course relative, but if the reference point is not specified, the convention is that V(∞) = 0. Of course that's just a convention; it's all relative. That said, the problem statement did not specify a point in space where V=0, so this is at least a little ambiguous. But since it wasn't specified, then V(∞) = 0 isn't a bad assumption in my eyes. :smile:

On a different note, this is one of my favorite post/threads regarding [itex] \LaTeX [/itex] from micromass (yea! @micromass !)
https://www.physicsforums.com/threads/physics-forums-faq-and-howto.617567/#post-3977517
 
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  • #11
collinsmark
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Anyway, @Carmanman, break up the integratons so that you integrate over [itex] r = \infty [/itex] to [itex] r = B [/itex], then once more to [itex] r = B [/itex] to [itex] r = A[/itex].

So what can you say about the electric field in the region between [itex] r = A[/itex] to [itex] r = 0[/itex]? If the electric field is zero in that region (or is it?) then what does that say about the change in potential in that region? :wink:
 
  • #12
rude man
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Actually the convention is the other way around, me thinks. Electrical potential is of course relative, but if the reference point is not specified, the convention is that V(∞) = 0. Of course that's just a convention; it's all relative. That said, the problem statement did not specify a point in space where V=0, so this is at least a little ambiguous. But since it wasn't specified, then V(∞) = 0 isn't a bad assumption in my eyes. :smile:
Tell you what, when you have nothing better to do, try to set V(∞) = 0 in this problem and compute the potential at any finite distance r from the axis.
 
  • #13
rude man
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Give him a break he is not the only one in this world having trouble with Latex and in general with formating equations in text :).
Well, delta, you look very intimidating :smile: but I must insist.
Besides, the OP did not use Latex in his post (#6). Not that I fault him for that. I never use it myself, have better things to do than learn a new language at my advanced age.
 
  • #14
collinsmark
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Tell you what, when you have nothing better to do, try to set V(∞) = 0 in this problem and compute the potential at any finite distance r from the axis.
You are correct. I should have paid closer attention to the problem statement. When working with charged, infinitely long lines, cylinders or planes it makes more sense to define [itex] V(r=0) = 0 [/itex].

The [itex] V ( r \rightarrow \infty) = 0 [/itex] convention makes more sense for point charges (and works well for any finite charge distribution for that matter), but that doesn't apply here.

Sorry for the confusion. :oops:
 
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  • #15
rude man
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When working with charged, infinitely long lines, cylinders or planes it makes more sense to define [itex] V(r=0) = 0 [/itex].
You can't have V(0) = 0 for a line charge situated at the origin either. It was OK to do so in this case since the field inside the cylinder went as r but for a line charge it goes as 1/r.
 
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  • #16
collinsmark
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You can't have V(0) = 0 for a line charge situated at the origin either. It was OK to do so in this case since the field inside the cylinder went as r but for a line charge it goes as 1/r.
Ah, you're right again!

In the case of the uniformly charged, infinitely long [one dimensional] line, you can't define the reference potential at either [itex] r = 0 [/itex] or [itex] r \rightarrow \infty [/itex]. Neither will make sense. The best you can do is compare the potential between two, finite yet non-zero distances from the line!

However yes, as you have already mentioned for this problem (the problem in the OP), it works to set [itex] V(r=0) = 0[/itex] in this case.
 
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  • #17
rude man
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Ah, you're right again!

In the case of the uniformly charged, infinitely long [one dimensional] line, you can't define the reference potential at either [itex] r = 0 [/itex] or [itex] r \rightarrow \infty [/itex]. Neither will make sense. The best you can do is compare the potential between two, finite yet non-zero distances from the line!

However yes, as you have already mentioned for this problem (the problem in the OP), it works to set [itex] V(r=0) = 0[/itex] in this case.
Exactly!
 
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  • #18
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Also, I tried copy and pasting a latex written version of what I was trying to say and it was blank...
 
  • #19
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I'll try again. This is what I've got for the voltages of each section.
 
  • #20
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Nope, it's not working. It doesn't like the </> signs in my code. I'll just type it out.


V(0<r<A) - V(0) = - ∫0rE(0<r<A)dr , V(0)≡0
V(0<r<A) = - ∫0rE(0<r<A)dr

V(A<r<B) - V(0) = - ∫0rE(A<r<B)dr = - ∫0AE(0<r<A)dr - ∫ArE(A<r<B)dr
V(A<r<B) = V(0<r<A)|r=A - ∫ArE(A<r<B)dr

V(r>B) - V(0) = - ∫0rE(r>B)dr = -∫0AE(0<r<A)dr - ∫ABE(A<r<B)dr - ∫BrE(r>B)dr
V(r>B) = V(A<r<B)|r=B - ∫BrE(r>B)dr


Does this look right? Keep in mind that for the last one, V(A<r<B)|r=B will include the constant V(0<r<A)|r=A, since that constant is part of V(A<r<B).
If you want, I can post what these actually equal once the integrals have been evaluated based on my derived expressions for the electric fields stated earlier.
 
  • #21
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Wait, how do I put Latex code into my post? I click on the "Latex" button in the bottom left corner of the text box and paste my code but it doesn't appear in the post when I post it. I'm confused.
 
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  • #22
rude man
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Nope, it's not working. It doesn't like the </> signs in my code. I'll just type it out.


V(0<r<A) - V(0) = - ∫0rE(0<r<A)dr , V(0)≡0
V(0<r<A) = - ∫0rE(0<r<A)dr

V(A<r<B) - V(0) = - ∫0rE(A<r<B)dr = - ∫0AE(0<r<A)dr - ∫ArE(A<r<B)dr
V(A<r<B) = V(0<r<A)|r=A - ∫ArE(A<r<B)dr

V(r>B) - V(0) = - ∫0rE(r>B)dr = -∫0AE(0<r<A)dr - ∫ABE(A<r<B)dr - ∫BrE(r>B)dr
V(r>B) = V(A<r<B)|r=B - ∫BrE(r>B)dr


Does this look right? Keep in mind that for the last one, V(A<r<B)|r=B will include the constant V(0<r<A)|r=A, since that constant is part of V(A<r<B).
If you want, I can post what these actually equal once the integrals have been evaluated based on my derived expressions for the electric fields stated earlier.
OK except for the ones in red.
A pure mathematician would have you use a dummy variable, e.g V(0<r<A) = -∫0rE(r')dr' but engineers play fast & loose with math anyway so I wouldn't deduct any points :smile:
Look around the site, there is at least one very good Latex post in there somewhere.
 
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  • #23
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OK except for the ones in red.
A pure mathematician would have you use a dummy variable, e.g V(0<r<A) = -∫0rE(r')dr' but engineers play fast & loose with math anyway so I wouldn't deduct any points :smile:
Look around the site, there is at least one very good Latex post in there somewhere.
I'm a math & physics major so I'll use that notation from now on :)
 
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  • #24
rude man
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I'm a math & physics major so I'll use that notation from now on :)
Good for you! It's the right way. And it becomes obvious when after integrating you apply the limits. What sense does it make to say I'm setting x = x? But setting x' = x does.
 

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