Potential

**heathrowjohnny** · April 7th 2008, 01:31 AM

Find the potential a distance $s$ from an infinitely long straight wire that carries a uniform line charge $\lambda$ .

So $\bold{E} = \frac{1}{4 \pi \epsilon_{0}} \frac{2 \lambda}{s} \bold{\hat{s}}$ .

So you can't set the reference point at $\infty$ because the charge extends out to $\infty$ .

So is it $V(s) = -\int_{a}^{s} \left(\frac{1}{4 \pi \epsilon_{0}} \frac{2 \lambda}{s} \right) ds = -\frac{1}{4 \pi \epsilon_{0}} 2 \lambda \ln \left(\frac{s}{a} \right)$ ?

**mr fantastic** · April 7th 2008, 06:07 AM

Originally Posted by heathrowjohnny

Find the potential a distance $s$ from an infinitely long straight wire that carries a uniform line charge $\lambda$ .

So $\bold{E} = \frac{1}{4 \pi \epsilon_{0}} \frac{2 \lambda}{s} \bold{\hat{s}}$ . Mr F says: Correct (and there's an obvious trivial simplification). I hope you used Gauss's Law!

So you can't set the reference point at $\infty$ because the charge extends out to $\infty$ .

So is it $V(s) = -\int_{a}^{s} \left(\frac{1}{4 \pi \epsilon_{0}} \frac{2 \lambda}{s} \right) ds = -\frac{1}{4 \pi \epsilon_{0}} 2 \lambda \ln \left(\frac{s}{a} \right)$ ?

$V(s) = -\int_{+\infty}^{s} \left(\frac{1}{4 \pi \epsilon_{0}} \frac{2 \lambda}{s'} \right) ds'$ .

And that's just the way it is!

There's not really any trouble because in practice the important physical quantity is the electric potential difference between two points:

$V(b) - V(a) = - \int_{a}^{b} \left(\frac{1}{4 \pi \epsilon_{0}} \frac{2 \lambda}{s} \right) ds = .....$

where I've assumed that a and b are along the same radial line from the wire.

**topsquark** · April 7th 2008, 07:36 AM

Originally Posted by heathrowjohnny

Find the potential a distance $s$ from an infinitely long straight wire that carries a uniform line charge $\lambda$ .

So $\bold{E} = \frac{1}{4 \pi \epsilon_{0}} \frac{2 \lambda}{s} \bold{\hat{s}}$ .

So you can't set the reference point at $\infty$ because the charge extends out to $\infty$ .

So is it $V(s) = -\int_{a}^{s} \left(\frac{1}{4 \pi \epsilon_{0}} \frac{2 \lambda}{s} \right) ds = -\frac{1}{4 \pi \epsilon_{0}} 2 \lambda \ln \left(\frac{s}{a} \right)$ ?

As Mr. Fantastic says, it's all about the potential difference. However I thought a quick additional word might be useful.

The form of your potential is
$V(s) = -\frac{1}{4 \pi \epsilon_{0}} 2 \lambda \ln (s) + C$

Typically we would like to set either the potential at s = 0 to be 0 or the potential at infinity to be 0. We can do neither in this case.

The whole trouble is that the line charge extends to infinity, so the constant we need to subtract to make this potential realistic is also infinite. This is not a very good Mathematical procedure, so we'd like something better.

The solution to this dilemma has exactly the same form that you wrote:
$V(s) = -\frac{1}{4 \pi \epsilon_{0}} 2 \lambda \ln \left(\frac{s}{a} \right)$

The whole point of this solution is the physical interpretation of a: The coordinate s = a is the point where we are defining the potential to be 0.

I hope this helps clear up some of the confusion.

-Dan

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