Potential difference between two infinite cylinders

AI Thread Summary
The discussion revolves around calculating the potential difference between two oppositely charged infinite cylinders using Gauss' Law. The initial attempt led to an electric field expression, but the potential difference calculation did not match the expected solution. Participants clarified that the integration limits should reflect the distance between the cylinders, specifically evaluating at the correct height. After adjustments, the potential difference was expressed as V = -2σR/ε ln((d-R)/R), with agreement on its correctness. The conversation also touched on the physical implications of the result, particularly regarding the behavior of potential as the distance between cylinders approaches zero.
rshalloo
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Homework Statement


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 $$\sigma$$


Homework Equations


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


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?
Thanks in advance for your help!
 
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It looks pretty close to the final answer. why don't you put in the value for 's' that is needed?
 
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:
rshalloo said:
Find the potential difference between two oppositely charged, infinite cylinders of radii R whose axes lie at $$y =+\frac{d}{2}$$ and $$y = -\frac{d}{2}$$
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 ...​
 
haruspex said:
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 ...​

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

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?
 
rshalloo said:
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?

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!
 
rude man said:
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!

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!
 
rshalloo said:
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?
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.

rshalloo said:
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!
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
BruceW said:
It's interesting that even in the limit of the cylinders almost touching but not actually touching, the potential difference still tends to zero

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
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
rshalloo said:
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!

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.
 
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