Potential difference between two infinite cylinders

1. Jul 16, 2014

rshalloo

1. The problem statement, all variables and given/known data
Find the potential difference between two oppositely charged, infinite cylinders of radii R whose axes lie at $$y =+\frac{d}{2}$$ and $$y = -\frac{d}{2}$$ They have surface charge densities of magnitude $$\sigma$$

2. Relevant equations
The family favourite - Gauss' Law:
$$\int \textbf{E.da} = \frac{q_{encl}}{\epsilon_{0}}$$
and of course how could we forget
$$V = -\int \textbf{E.dl}$$

3. The attempt at a solution
Using gauss' law and a cylindrical gaussian surface I obtained an electric field for a single infinite cylinder located at the origin to be:
$$\textbf{E} =\frac{\sigma R}{\epsilon_{0} r}\hat{\textbf{r}}$$

I now use this to find the E-field on the line connecting the centers of the cylinders only. Obviously I am also confining my region further to exclude the radii of the cylinders. So my range of applicability is $$[-\frac{d}{2}+R,+\frac{d}{2}-R]$$

So placing the positive cylinder in the lower half of the plane I get:
$$\textbf{E_{+}} = \frac{\sigma R}{\epsilon_{0} (y+d/2)}\hat{\textbf{y}}$$
and placing the negative sphere in the upper half of the plane I get:
$$\textbf{E_{-}} = \frac{\sigma R}{\epsilon_{0} (d/2-y)}\hat{\textbf{y}}$$

Using superposition the total field is simply the sum of these.

I then use the formula for the potential listed above and use the integration limits:
$$[-\frac{d}{2}+R,s]$$
that is finding the potential difference assuming zero potential at the lower sphere. s is my variable height
I get the following answer but it doesn't match the solution.
$$V = -\frac{\sigma R}{\epsilon}ln\left (\frac{(s+d/2)(d-R)}{(d/2-s)(R)}\right )$$

Is my reasoning correct or have I made a fundamental error somewhere?

2. Jul 16, 2014

BruceW

It looks pretty close to the final answer. why don't you put in the value for 's' that is needed?

3. Jul 16, 2014

rude man

Just a quick look at it & it seems:
1. your E fields are OK.
2. What's with the "s" for the upper limit of integration? Put in the right value as poster #2 suggests.

Then I got an expression in inverse hyperbolic tangents but as I said it's a quick look. Might simplify. Will try to investigate and keep track of your thread.

Last edited: Jul 16, 2014
4. Jul 17, 2014

haruspex

I'm not sure what the question is asking. Is that the exact wording? You seem to have interpreted it as
Find the potential function between ... ​
whereas other posters so far have read it as
Find the potential difference between the axes of ...​

5. Jul 17, 2014

rude man

Seems clear enough. It's the absolute value of the integral of the E field set up by the two charged cylinders from one surface to the other. The integral is (conveniently) evaluated over a distance of d - 2R, not from y = -d/2 to y = +d/2.

@OP, probably the inverse hyperbolic tangents can be removed by fooling with tanh-1(x) = 1/2 ln{(1+x)/(1-x)}.

6. Jul 17, 2014

rshalloo

I feel kind of silly now, you're completely right, I don't know why I was trying to get a general formula for the potential at a point s rather than the difference between the cylinders. Too close to the problem I suppose!

(also thanks haruspex and rude man for seeing this too!)

So evaluating at
$$s = \frac{d}{2} - R$$
I get:
$$V = -\frac{2\sigma R}{\epsilon}ln\left (\frac{d-R}{R}\right )$$

Would you all agree?

7. Jul 17, 2014

rude man

Looks about right, but I'd have to translate fro my inverse hyperbolic tangents to the log expression exactly to be sure. Too late for tonite!

8. Jul 17, 2014

rshalloo

Too early for some of us across the globe! Thanks again. Your word is enough, I'm sure it makes sense physically because as d->2R the minimum it can be, the log tends to ln(1) which is 0. ie when the cylinders are touching there is zero potential between them? I know its just one case, but a nice little sanity check!

9. Jul 17, 2014

BruceW

hehe. Yes, I get the same answer. p.s. this is the potential difference going from the positive cylinder to the negative cylinder. (i.e. it will be a negative number, if you check a few examples). So if you want to give the absolute value of potential difference, then you can just remove that minus sign at the front.

Yeah, that's a nice check. It's interesting that even in the limit of the cylinders almost touching but not actually touching, the potential difference still tends to zero. I guess the electric field is still a finite value. But the distance between the cylinders tends to zero, so the potential difference does too.

10. Jul 17, 2014

rshalloo

What do you mean here, surely we would expect the Potential difference to tend to zero (but still remain finite) as the distance between the cylinders tended to zero? Obviously this has limits, this theory is not valid for instance on an atomic scale as we have assumed a continuous charge distribution. But in the realm of our theory would this not be expected?

11. Jul 17, 2014

BruceW

ah, yes you're right. There was never anything to worry about, because we used a charge distribution that is finite at every point. I guess I was feeling cautious because this problem is similar to the problem of two lines of charge, in which case the electric field blows up to infinity.

12. Jul 17, 2014

rude man

Good. BTW I got the inverse hyperbolic tangents by integrating both E fields in one integral. The rest incl. yourself apparently integrated one at a time and added, gettng the log function instead.
At any rate, that's what Wolfram Alpha gave me for doing the joint E-fields integration.