Finding the potential between two coaxial cylinders

[Jaco Leo]

Homework Statement

.

Trying to find the potential between a variable capacitor that is made up of two coaxial cylinders of radii a and b, with (b-a) << a, when inner cylinder displaced by a distance y along axis.

2. Homework Equations

E = λ / 2piε₀r
V = λ/2piε₀ * ln(b/a) when there is no displacement

3. The Attempt at a Solution

I already calculated the potential when there is no displacement by ∫E dr with respect to a and b. But Honestly I don't even know where to start for finding the potential when the inner cylinder is displaced a distance y along the axis. Any help would be appreciated, thanks!

[/B]

 

[BvU]

You forgot to tell us what $\lambda$ is. But I strongly suspect that there's the key to your answer. From the $(b-a)<<a$ you may assume there only is a field (and therefore a nearby charge ) between the cylinders in the area where they are close to each other.

 

[Jaco Leo]

You forgot to tell us what $\lambda$ is. But I strongly suspect that there's the key to your answer. From the $(b-a)<<a$ you may assume there only is a field (and therefore a nearby charge ) between the cylinders in the area where they are close to each other.

λ is the line charge, and I actually already know the answer to this question, it's (λ*L/2piε0y) * ln(b) + (λ*L/2piε0(L-y)) * ln(b/a)), where L is the length of the cylinder and y is the displacement of the inner cylinder along the axis. I'm just not sure how they got to this answer. I'm not sure if you have to integrate the E-field again to arrive at this or if you can just do it intuitively, any help would be appreciated, thanks!

 

[BvU]

Charges are expressed in Coulombs. My guess is that $\lambda$ is the linear charge density, expressed in Coulombs per meter. That way your answer comes out in V instead of in meter x V.

You say you did part a) already. When you compare that with the given answer to part b, do you recognize anything ?

 

[Jaco Leo]

Charges are expressed in Coulombs. My guess is that $\lambda$ is the linear charge density, expressed in Coulombs per meter. That way your answer comes out in V instead of in meter x V.

You say you did part a) already. When you compare that with the given answer to part b, do you recognize anything ?

So it looks like to find the voltage when the inner cylinder moves a distance y, it takes into account two different electric fields to be integrated? because you've got two terms in the answer. Also there's a y in the denominator in the first term and a (L-y) in the denominator in the 2nd term. These must mean the displacement in someway but I honestly can't wrap my head around it atm. Can you explain to me how you arrive at finding the voltage once the inner cylinder moves a distance y along the axis?

 

[BvU]

If the inner cylinder is displaced over a distance of y, the charge is inclined to stick to the area that is opposite the outer cylinder (where there is an opposite charge tugging at it). The total charge stays the same, so the charge density ( did I guess right in post # 4? ) goes from $\lambda$ to $\lambda {L\over L-y}$. Now do you recognize one of the terms ?

 

[vela]

λ is the line charge, and I actually already know the answer to this question, it's (λ*L/2piε0y) * ln(b) + (λ*L/2piε0(L-y)) * ln(b/a)), where L is the length of the cylinder and y is the displacement of the inner cylinder along the axis.

That expression doesn't look right to me because of the ln(b) term. The argument of the logarithm should be able to be written in a unitless form, but I don't see how you can do that with the expression you wrote.

 

[Jaco Leo]

If the inner cylinder is displaced over a distance of y, the charge is inclined to stick to the area that is opposite the outer cylinder (where there is an opposite charge tugging at it). The total charge stays the same, so the charge density ( did I guess right in post # 4? ) goes from $\lambda$ to $\lambda {L\over L-y}$. Now do you recognize one of the terms ?

Ok yeah that makes sense, so the (λ*L/2piε0(L-y)) * ln(b/a)) term is describing the displacement of the inner cylinders potential. But I'm still very confused about the second term (λ*L/2piε0y) * ln(b). Is this describing the potential left within the system? Honestly can't figure that out, and like Vela said, the ln(b) term is really throwing me off.

 

[BvU]

Same here. At first I thought it had something to do with a change in the potential at b, but the more I look at it, the more I think that term shouldn't be there at all ...

 

[Jaco Leo]

Same here. At first I thought it had something to do with a change in the potential at b, but the more I look at it, the more I think that term shouldn't be there at all ...

Here's the actual answer, this might help you. But that ln(b) term still doesn't make sense to me.