# Potential Difference of Hollow Circular Cylinder

1. Dec 22, 2012

### rbrayana123

1. The problem statement, all variables and given/known data
A hollow circular cylinder of radius a and length b, with open ends, has a total charge Q uniformly distributed over its surface. What is the difference in potential between a point on the axis at one end and the midpoint of the axis?

2. Relevant equations

U = kq/r

3. The attempt at a solution

dq = σdS

I decide to integrate along the length of the cylinder so:
dS = 2∏a dx
σ = Q/2∏ab

Therefore, dq = Q/b dx

Now, for the r, it's equal to √(a^2 + x^2) with x ranging from 0 to b for the endpoint and 0 to b/2 for the midpoint. (However, I double it)

So my integral for the endpoint:

kQ/b∫dx/√(a^2 + x^2) from 0 to b is kQ/b * [ln(sqrt(a^2 + b^2) + b) - ln(a)]

My integral for the midpoint is:

2kQ/b∫dx/√(a^2 + x^2) from 0 to b/2 is 2kQ/b * [ln(sqrt(a^2 + b^2/4) + b/2) - ln(a)]

Now I actually have a couple of questions regarding the integration. Is it correct? I used trig sub to integrate secx to get ln (secx + tanx) and substituted back in. Wolfram Alpha appears to agree: http://www.wolframalpha.com/input/?i=integral+of+dx/sqrt(a^2+++x^2)

Next, for the second integral, why does integrating from -b/2 to b/2 get me a different answer?

Lastly, the potential difference seems messy. Is there a more elegant way to calculate out the potentials for both the endpoint and midpoint or is this appropriate?

2. Dec 22, 2012

### haruspex

Looks right.
Are you sure it's different?
ln(sqrt(a^2 + b^2/4) - b/2) = 2 ln(a) - ln(sqrt(a^2 + b^2/4) + b/2)
I'm not aware of any easier way. It's not as though the final answer is paticularly simple and elegant.

3. Dec 22, 2012

### rbrayana123

I could either use a symmetry argument and integrate from 0 to b/2 but multiply by two OR integrate from -b/2 to b/2. It appears different when comparing those cases mostly because of the "+ x" term in the ln for the integral.

EDIT: I'm a little confused about the above equality you gave 0_0
EDIT2: Ahh, using ln identites and multiplying by the conjugate gets me that equality.

Thanks for the help sir. =)

Last edited: Dec 22, 2012
4. Dec 22, 2012

### haruspex

You didn't post what you got for the other half integral, so I presumed it was
kQ/b * [ln(a) - ln(sqrt(a2 + b2) - b) ], i.e. swap the sign of b and swap the signs on the two terms. My identity was to show this is actually the same value.