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## Infinite Line Charges

 Quote by exiztone Patrick, or anyone else who might be able to help. After many attempts I got the following: $$[\sinh^{-1} ({\frac{x}{2}})]^{+\infty}_{-\infty}$$ But I can't go on from there. I have to get rid of the infinite somehow, but I can't see how. Can you point me in the right direction? Thank you so much!
Hi.. sorry I replied so late. Got a phone call yesterday night and by the time I was done you had logged out. Also, it was a bit more tricky than I initially thought (for the reason I mention below) so I had to give it some thought.

The website http://integrals.wolfram.com/index.jsp is useful to do integrals.

This is a tiny bit more tricky that I was expecting because I had in mind the calculation of the *electric field* at a point, which involves a simpler integral which gives an arctan.

But we still can do it. going back to the form in terms of x and y (that will make things a bit simpler), we get
$$\int { dx \over {\sqrt{ x^2 + y^2} } } = ln(|x + {\sqrt{ x^2 + y^2}}|)$$

which I think you gave earlier in the thread! (sorry if I went in a different direction! It's because I still had in mind the result for the calculation of the electric field which involves an arctan). Now, this integral still diverges at infinity!!

The trick is that we will calculate the difference of potential between the two points before evaluating at x= plus or minus inifnity!

So the difference of potential between y = 4 and y =2 meters is (I assume that we want the former minus the second one) given by

$${ \lambda \over 4 \pi \epsilon_0} ln {| x + {\sqrt{x^2+16}} |\over| x + {\sqrt{x^2 + 4}}|}$$

Now we need to evaluate this between x= + infinity and x=-infinity. As x goes to infinity, the argument of the ln goes to 1 so we get ln(1)=0. The case minus infinity is more interesting. Notice that as x goes to minus infinity, we get

$$x + {\sqrt{x^2 + C}} \rightarrow x + |x| {\sqrt{1 + {C \over x^2}}} \approx x +|x| + {C \over 2 |x|}$$
which, as x goes to minus infinity, goes to
$${C \over 2 |x| }$$

Therefore, as x goes to minus infinity, the expression
$$ln {| x + {\sqrt{x^2+16}}| \over| x + {\sqrt{x^2 + 4}}|}$$
becomes
$$ln ( {16 \over 2 |x|} {2 |x| \over 4}) = ln({16 \over 4})$$

Finally plugging this back in the equation for the potential at 4 minus the potential at 2 meters, we get
$$- { \lambda \over 4 \pi \epsilon_0} ln({ 16 \over 4}) = - 2{ \lambda \over 4 \pi \epsilon_0} ln({ 4 \over 2})$$

The sign is correct (I took the potential at 4m minus the potential at 2 meters and you expect the potential at 4 meters to be smaller than at 2 meters).

Notice that you can see the general rule right away: for the difference of potential between a point at r_1 and at r_2, we get

$$V(r_2) - V(r_1) = - 2{ \lambda \over 4 \pi \epsilon_0} ln({ r_2 \over r_1})$$
(the sign is correct..if r_2 > r_1, the difference of potential is negative).

Notice that we can only calculate the difference of potential between two points. Often people will choose to set V=0 at r=1 meter , in which case the equation becomes

$$V(r) = - 2{ \lambda \over 4 \pi \epsilon_0} ln(r )$$
with the choice of ground at r=1 meter!

You may ask: why did we get infinities in the first place? It is because the formula kq/r for a point charge assumes a ground (surgace of zero potential) at r=infinity. This choice is *not* possible for an infinite line of charge.

Phew...that was quite a post

I hope it makes sense. Don't hesitate to ask questions if anything is not completely clear.

Patrick
 Patrick, thanks a million! This is a huge help. I know the general formula now, but more importantly I understand how that was derived. In the exam, I'll probably use this formula but explain how I got it (this way, I don't have to do that horrible integration). You've been a huge help, thanks a million!