# Potential of a finite rod

## 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.

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

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rude man
Homework Helper
Gold Member
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?

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.

BvU
Science Advisor
Homework Helper
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

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

rude man
Homework Helper
Gold Member
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!