How to find potential over a line charge

[baltimorebest]

Homework Statement

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)

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.

 

[vela]

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

 

[baltimorebest]

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

 

[baltimorebest]

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

 

[vela]

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.

 

[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

 

[vela]

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.

 

[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?

 

[vela]

Yup!

 

[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.

 

[vela]

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.

 

[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.

 

[vela]

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.

 

[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.

 

[vela]

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

 

[baltimorebest]

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

 

[vela]

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

 

[baltimorebest]

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

 

[vela]

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.

 

[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?

 

[vela]

There's no need to write in any units. Checking the units is just a quick way to see if you may have made a mistake. Any valid formula will always work out unit-wise.

 

[baltimorebest]

Ok so I hope what I posted above is correct for the first part.

This is what I have so far for the 2nd part, and I got stuck at the bottom.

 

[vela]

Looks good so far. You just need to plug the limits in. What happens to each term in the log as z goes to infinity?

 

[baltimorebest]

well the term without the square root goes to 0. The other term is the one I am confused on. Does it go to 1?

 

[vela]

No. When z>>L, you can neglect the L² term in the radical. Does that help?

 

[baltimorebest]

Well that's what I thought, but even then, doesn't it become sqrt(z^2)/z?

 

[vela]

Yes, it does. What is the square root of z² equal to?

 

[baltimorebest]

well it becomes z. And that is over z isn't it? That's why I thought it would go to 1.

 

[vela]

Oh, yeah, it does. For some reason, I thought you asked if it went to 0. So the argument of the log goes to 1; log 1 = 0, so at infinity, the potential is 0.

 

[baltimorebest]

Ok so here is my answer to the second part. It doesn't look equal to my answer for the first part, but maybe it's just because the form is so different?

 

[vela]

Yup, that's right. If you rationalize the denominator in your first result, you can turn it into your second result.

 

[baltimorebest]

Oh ok well I will check that later and take your word on it for tonight. Thank you so much for your help on this problem.

 

[baltimorebest]

I tried to plug in values to see if my two answers were equivalent, and it wasn't working out. Maybe it's just because I am tired, but I just thought I would let you know. Thanks again.