# Finding the potential between two coaxial cylinders

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1. Oct 17, 2016

### Jaco Leo

1. The problem statement, all variables and given/known data.

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. Relevant 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!

2. Oct 18, 2016

### 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.

3. Oct 18, 2016

### Jaco Leo

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!

4. Oct 18, 2016

### 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 ?

5. Oct 18, 2016

### Jaco Leo

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 cant 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?

6. Oct 18, 2016

### 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 ?

7. Oct 18, 2016

### vela

Staff Emeritus
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.

8. Oct 19, 2016

### Jaco Leo

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 cant figure that out, and like Vela said, the ln(b) term is really throwing me off.

9. Oct 19, 2016

### 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 ....

10. Oct 19, 2016

### Jaco Leo

Here's the actual answer, this might help you. But that ln(b) term still doesnt make sense to me.