Potential for an infinite line charge

1. Oct 21, 2012

blerb795

1. The problem statement, all variables and given/known data
For the single line charge, derive an expression for Electric Potential.

2. Relevant equations
V(r)=-$\int$E$\bullet$dr
E for infinite line = $\frac{\lambda}{2\pi r\epsilon}$
3. The attempt at a solution
The integration is straightforward enough—my question is as to what the limits of the integration should be for a conceptual problem such as this. 0 to r makes the most sense to me but I am entirely stumped.
Not paying attention to the limits, I just got $\frac{\lambda ln(r)}{2\pi \epsilon}$ but I am again not sure of the sign.

2. Oct 21, 2012

Simon Bridge

$\vec{E}$ in the integrand is a vector. So is $d\vec{r}$.
Don't forget to include the coordinate system. Which direction is $d\vec{r}$?

Curious:
Why not start with the known solution for a point charge and use the superposition principle?

3. Oct 21, 2012

frogjg2003

You could do the indefinite integral. This will give you the potential plus a constant. Physically, the constant doesn't matter, so you can set it to any value you want, usually 0.

4. Oct 21, 2012

blerb795

I'm not sure if that was a rhetorical question to get me thinking, but dr⃗ is intended to be radially outward from the line charge.

I'm not sure how I would use the superposition principal here for a point charge...

frogjg, so in that case with the indefinite integral, my answer should still be negative?

5. Oct 21, 2012

frogjg2003

Think about it graphically. E=dV/dr is always positive, so V should be sloping up. Is +ln(r) or -ln(r) sloping up?

6. Oct 21, 2012

Simon Bridge

If you put the line on the z axis, a short length of the line, dz, at position z, has charge dq ... you know how to find the potential due to dq at some particular point a distance r away from the line. Adding up all the contributions (superposition principle) amounts to integrating along the line.

There are any number of worked examples online for this.

But you may be happier with the integral you've got. frogjg2003 can help you with that ;)

7. Oct 22, 2012

TSny

For an infinite line of charge there's a difficulty in integrating over the line if you use kdq/r as the potential of a charge element dq = λdz. The integral will not converge. That's because kdq/r assumes you're taking V = 0 at infinity. But, for an infinite line of charge, taking V = 0 at infinity will make V infinite at any finite distance from the line.

8. Oct 22, 2012

Simon Bridge

Yeah - one needs be more careful than that... though it is also why OP is having trouble picking limits to the integration. Excuse: I don't want to clutter the thread further with this side-chat.

I want to watch frogjg do this :)

9. Oct 22, 2012

frogjg2003

Now I feel stupid, I completely missed the minus sign when I was talking about the slope of the curve.
The correct way of thinking:
E=-dV/dr is positive, so V must have a negative slope. Since all the other constants are positive, that means there's only one source of negatives, the only place for there to be a negative is -ln(r).

As for what the constant at the end of the indefinite integral should be, it doesn't matter. The physics is in the E field, which doesn't see the constant. This usually means the constant is just assumed zero. Other times, the problem gives natural points to consider where to put the zero, and therefore what value to use for that constant.