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Potential of a finite rod

  • Thread starter camira
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  • #1
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Homework Statement


Hi all,
I need to find the potential of a positively charged rod with charge Q. Assuming to the right as positive,the center of the rod is on the origin, and it extends to -L/2 in the negative x direction, and L/2 in the positive x direction.
There is a point at distance x, located to the right of the rod, on the x axis.
I have to express the answer in terms of the variables Q, L, x, and appropriate constants.


Homework Equations


V= ∫K*dQ/R
λ=Q/L

The Attempt at a Solution


I know this question has been asked here before, but I wasn't able to come to a conclusion about it myself based on these posts...so here we go:
I started by choosing an arbitrary point on the rod, called dQ, that is a distance dx away from the point.
λ=dQ/dx
dQ=λ*dx
I have my equation
V=∫K*dQ/R
sub in for dQ
V=∫K*λ*dx/R
I know I need to integrate over the length of the rod. So i decided to use the bounds as 0 to L/2, and then multiply the integral by 2.
This is where I am stuck. I don't know what the value for R is...I am leaning towards x+L..but I am not sure. Also I seem to have gotten rid of my Q variable..so im not sure I should have subbed in λ*dx for it...
Any suggestions would be greatly appreciated!
 

Answers and Replies

  • #2
rude man
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I started by choosing an arbitrary point on the rod, called dQ, that is a distance dx away from the point.
x is a finite distance away from the end of the rod. What you say here makes no sense to me.
Consider various elements of charge dQ = λ dx along the rod. The element of dQ closest to x will be at what distance from x? The element of charge dQ farthest from x will be at what distance from x? So what should be your limits of integration?
 
  • #3
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x is a finite distance away from the end of the rod. What you say here makes no sense to me.
Consider various elements of charge dQ = λ dx along the rod. The element of dQ closest to x will be at what distance from x? The element of charge dQ farthest from x will be at what distance from x? So what should be your limits of integration?
The closest element of dQ to x will be...L/2 + x...I'm not sure about the furthest...I am thinking x+L?
I am still not sure about the radius R though, because the small section of charge being summed can be at any distance on the rod away from the set point.
 
  • #4
BvU
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If you already have an x axis, it's better to have point at position x, not at distance x (that's confusing). Make a drawing to find out what is meant with the variables involved. Also you want to clearly distinguish between the integration variable (I see a dx) and the ('fixed' *) x coordinate of the point where you want to calculate the potential.

*) 'during' the integration

So i decided to use the bounds as 0 to L/2, and then multiply the integral by 2.
Not wise. The integrand isn't the same for the other interval
 
  • #5
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Thanks everyone for the replies
After contemplating this further, I think I didn't do a very good job of conceptualizing the situation
So starting from the beginning:
Let's call the distance the point is away(so from the origin to the point) x.
Rod is the same -L/2 to L/2
The distance that my charge, dQ, is at, from the origin, is D.
λ=dQ/dD=Q/L, so λdD=(Q/L*dD)
I know that the distance from the point to my dQ is x-D, so R=x-D
I want to integrate from -L/2 to L/2
so:
K∫λdD/R
Kλ∫dD/(x-D)
Integrate:
Kλ [-ln (x-D)] from -L/2 to L/2
Kλ[-ln (x- L/2) + ln(x+L/2)]
using log rules:
Kλ[ln((x+L/2)/(x-L/2))
to express in terms of Q, just sub in for λ=Q/L
I think this is more accurate
 
  • #6
rude man
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Kλ[ln((x+L/2)/(x-L/2))
to express in terms of Q, just sub in for λ=Q/L
I think this is more accurate
This is correct. Good!
 

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