How to Calculate Electric Potential Along the Axis of a Charged Tube?

[gabbagabbahey]

I tried to do that integral by hand, and I used a substitution of u=Rtan(theta).
I ended up with an integral of sec(theta) d(theta) which integrated to ln[sec(Theta)+tan(theta)] which transformed back into ln{sqrt(R^2 + u^2)/z + u/z)

Don't you mean \ln\left(\frac{\sqrt{R^2+u^2}+u}{R}\right)? If so, keep in mind that \ln\left(\frac{f(u)}{R}\right)=\ln(f(u))-\ln(R)[/itex], and \ln(R) is just some constant that can be absorbed into the integration constant (same thing with the factor of 2 in the result I posted earlier)

 

[gabbagabbahey]

Ohh yes of course. I see that now. Ok so I think I am done with the endcaps.

Are you sure? Changing the denominator of the integrand means changing the way you integrate...

 

[maherelharake]

Curved part:
Oh yes that's what I meant. Sorry. So then I just plug in the bounds and it should give me the correct potential?

Endcaps: Ohh of course. Now I can use substitution and set u=s'^2 +(z-z')2

 

[maherelharake]

Here is my work for the endcaps. I believe that both endcaps should give the same result after substitution

 

[maherelharake]

Ok I have attached (what I hope is) my final draft for this problem. One attachment is for the end caps, and one is for the curved part. I am going to sleep, and these are due tomorrow morning. If you have any more pointers, I will read them before I go to campus. Thank you for your help.