Homework Help: How to find potential over a line charge

1. Sep 22, 2010

baltimorebest

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

You are given a charge distribution that is a uniform line charge (λ) extending from –L to L along the x-axis. Calculate the potential along the z axis. Use the given value for the electric field to calculate the line integral and verify part a.

The given value for the electric field is...

k(2*Lamda*L)/(Zsqrt(z^2+L^2)

3. The attempt at a solution

My initial thought process was to use the equation V=k*integral[(lamda/r)dl] where r is equal to sqrt(z^2+l^2). I attempted to solve this integral using tables and got...

V=k*lamda*[ln{2sqrt(z^2+l^2)} +2l] evaluated from l=L to l=-L.

I need help proceeding from here, especially with the second part of the problem.

2. Sep 22, 2010

vela

Staff Emeritus
The result of your integration seems to be wrong. Try using the substitution l=z tan θ.

3. Sep 22, 2010

baltimorebest

Ok when I did it this way I got the integral of sec(theta) dtheta

4. Sep 22, 2010

baltimorebest

I feel like that can't be right....

5. Sep 22, 2010

vela

Staff Emeritus
That's right. You have to have more confidence in yourself!

If you're familiar with the hyperbolic trig functions, the substitution l=z sinh θ also works. It's a bit simpler, but either way works.

6. Sep 22, 2010

baltimorebest

Hmm ok so that leaves me with a value of..

k*Lamda* ln{sec(theta) + tan(theta)

I need to find the values of theta. I believe tan(theta) is equal to l/z from our substitution. Not too sure about sec(theta) though

7. Sep 22, 2010

vela

Staff Emeritus
Draw a right triangle with l for the length of the opposite side and z for the length of the adjacent side so that tan(theta)=l/z. You should be able to figure out what sec(theta) is equal to.

8. Sep 22, 2010

baltimorebest

Ok I got sec to be sqrt(z^2 + l^2)/z.

so would that make my final integral

k*Lamda* ln{sqrt(z^2 + l^2)/z + l/z) and evaluate l from L to -L?

9. Sep 22, 2010

vela

Staff Emeritus
Yup!

10. Sep 22, 2010

baltimorebest

Ok cool. So I attached my final answer for the first part of this problem.

For the second part, I know you take the line integral of the given value of the electric field dotted with dl, but I am not sure how to carry that out.

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11. Sep 22, 2010

vela

Staff Emeritus
You can combine the logs using the property log(a/b)=log a - log b. Check your algebra. You made a few minor errors, and I think you may have accidentally changed l/z into l/2 as well.

For the second part, I suggest you integrate along from z axis from z=∞ to z=z.

12. Sep 22, 2010

baltimorebest

Ok here is the updated version. I am not sure where my algebra was wrong, but I did fix the other things.

I also forgot to mention for the 2nd part that I already set up the problem so that I integrate along the z axis from z=∞ to z=z.

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13. Sep 22, 2010

vela

Staff Emeritus
Look at the units of what's under the radical. The units of the first term is length squared, but the second term is unitless. That can't possibly be correct.

Your set-up for the second part looks fine.

14. Sep 22, 2010

baltimorebest

Oh I had an extra 'z' term. But it appears that I still have under the radical...

z^2 + L^2/z

those units don't match up either :/ I have to be overlooking something simple.

And for the 2nd part, I am trying to find a trig substitution that will allow me to solve that integral.

15. Sep 22, 2010

vela

Staff Emeritus
Whenever you have something like sqrt(a2+x2) in the integrand, it suggests a trig substitution of the form x=a tan θ.

16. Sep 22, 2010

baltimorebest

Ok I will work on that now. What about the first part of my statement?

17. Sep 22, 2010

vela

Staff Emeritus
Before combining the logs, try putting what's inside each log over the common denominator z, and then combine the logs.

18. Sep 22, 2010

baltimorebest

I am still confused. This is what I did. I understand everything you are telling me, but my units aren't checking.

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19. Sep 22, 2010

vela

Staff Emeritus
You're starting with the wrong expression. Your expression back in post 8 is correct.

Try checking the units on each step. When it stops working, you know you did something wrong.

20. Sep 22, 2010

baltimorebest

Ahh I think I got it now. So that is the answer for the first part of the question. Should I write the units, or are they understood based on the variables?

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