# Homework Help: Electric potential due to long wire

1. Jun 1, 2007

### vs5813

1. The problem statement, all variables and given/known data
Using Gauss' law, or otherwise, find an expression for electric potential at a distance r from an infinitely long straight line of charge of length h with linear charge density lambda per unit length.

2. Relevant equations

q=lambda*h

3. The attempt at a solution
The first part is easy, using gauss' law to find electric field..the result i get is

E=lambda/(2*pi*r*epsilon_0)

..which should be correct. But then i'm not sure how to find potential in this case. I tried using the equation:

so

V = integral[E*dl]

V= -lambda/(2*pi*epsilon_0) integral_from_infinity_to_r[(1/r) dr]

V = -lambda/(2*pi*epsilon_0) [ln(r)]

where the ln(r) is evaluated from infinity to r...and im not sure that looks right, or where to go from there.. I don't know what the solution is supposed to be and I couldn't find any explanations when i googled it..anyone have any idea??? :uhh: thankyou sooo much!!!!

2. Jun 1, 2007

### Dox

Hello.

In fact that's the solution. That's life!

3. Jun 2, 2007

### vs5813

heh thankyou!!

4. Jan 25, 2010

### Cintdrix

how do you evaluate ln at infinity and r??

5. Feb 10, 2010

### andurill13

pick an arbitrary point a, and integrate with respect to that point. you're correct that you can't integrate to an infinite potential at the wire.

6. Apr 19, 2010

### damnedcat

the only way i can think of is that u have to assume the boundary condition that V=0 at infinity. but then that seems forced because when evaluating the similar case for a sphere u get a 1/r term and when u plug in infinity there it goes to zero (see griffiths 3rd ed example 2.6). can anyone reconcile these cases? i don't see any explanation above addressing this.

7. Sep 22, 2010

### Tvdmeer

Hi,

You cannot use infinity, because the problem states that there is an infinite length of wire. This tells us that the charge is infinite. Using a variable like "a" is your best bet for solving the problem.