- #1
rbrayana123
- 44
- 0
Homework Statement
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?
Homework Equations
U = kq/r
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?