# Uniform charge across a rod, Professor couldn't answer

1. Sep 21, 2008

### Hydroshock

Alright, well I wanted to see what help I could get here, my professor had assigned this problem and when someone in class asked a question, he started doing it on the board and after using ~15min of class time decided there was an issue with the problem. The problem got progressively harder (which leads me to believe he must have made a mistake) and he ended up deciding we don't have to do it, but i'm interested in getting it done anyway.

1. The problem statement, all variables and given/known data
"A rod of length L has a total charge Q uniformly distributed alone its length. The rod lies along the x axis with its center at the origin. (a) What is the electric potential as a function of position along the x axis for x > L/2? (b) Show that for x >> L/2, your result reduces to that due to a point charge Q.

2. Relevant equations

V = int|kdq/r
dq = λdx
λ = Q/L
r = x0 - x

3. The attempt at a solution

Well what the problem was when we got to (b). Which after setting up the equation we ended up with

$$k\lambda\int_{-L/2}^{L/2}\frac {dx} {x_0-x})$$

where after substitution we get

$$k\lambda\int_{-L/2}^{L/2}ln(x_0-x)$$

but since there's units (meters) for x, it can't be run through a transcendental function, then I got lost here, but with more substitution we ended up with

$$V = k\lambdaln\frac {x_0-1/2} {x_0+1/2}$$

then when the problem got going, it expanded and expanded and didn't get any simplification going, which is odd for a textbook question. I guess it's more of needing a check with the math work done in the problem?

Last edited: Sep 21, 2008
2. Sep 21, 2008

### CaptainMarvel

Hi!

What you've done looks pretty good, but after subbing in the limits from the integral on this line:

$$\frac { \lambda } { 4 \pi \epsilon_o} ( ln (x_o + L/2) - ln (x_o - L/2) )$$

instead of combining the logs (using $$ln(A) - ln (B) = ln (A/B)$$) like you've done on the next line, why not factorise out an x_o in each log? Then you can pull the x_o out of the log by using the log rule ln (AB) = ln A + ln B. Something like this:

$$\frac { \lambda } { 4 \pi \epsilon_o} ( ln (x_o) - ln (1 + \frac{L}{2x_o}) - ln(x_o) - ln (x_o - \frac{L}{2x_o}) )$$

Notice how the $$ln(x_o)$$ cancels.

Here's the trick now though. Try expanding each of the logs to O^2 in terms of a taylor series. If you've done it right, you'll hopefully get some canceling and be left with a an equation which you can replace the charge density and length by the total charge. You're final equation should look like this:

$$\frac { Q } { 4 \pi \epsilon_o x_o}$$

Hope this helps :-)

3. Sep 21, 2008

### gabbagabbahey

For starters I'd use $$x'$$ to represent the position of the source charge, and $$x$$ to represent the field point along the x-axis. Then Coulomb's Law giives:

$$V(x)=k \lambda \int_{-L/2}^{L/2} \frac{dx'}{|x-x'|} = -k \lambda ln \left( |x-x'| \right) |_{-L/2}^{L/2} = -k \lambda ln \left( \frac{|x-L/2|}{|x+L/2|} \right) = k \lambda ln \left( \frac{|x+L/2|}{|x-L/2|} \right)$$

For $$x> L/2, \quad |x-L/2|=x-L/2, \quad |x+L/2|=x+L/2$$

So, your solution for (a) is:

$$V(x)=k \lambda ln \left( \frac{x+L/2}{x-L/2} \right)$$

Now for (b), expand the log in a Mercator Series:

$$V(x)=k \lambda ln \left( \frac{x+L/2}{x-L/2} \right) =k \lambda ln \left( \frac{1+\frac{L}{2x}}{1-\frac{L}{2x}} \right) = k \lambda ln \left( 1+\frac{L}{2x} \right) - k \lambda ln \left( 1-\frac{L}{2x} \right) \approx k \lambda \left( \frac{L}{2x} - \frac{L^2}{8x^2} \right) - k \lambda \left( - \frac{L}{2x} - \frac{L^2}{8x^2} \right)=k \frac{ \lambda L}{x} = k \frac{Q}{x}$$

Last edited: Sep 21, 2008
4. Sep 22, 2008

### Hydroshock

That works gabbagabbahey :) Thanks for the help to you two.