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deedsy
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Homework Statement
A charge Q is uniformly distributed along a straight line of length a. Calculate V by integration, choosing the origin of the coordinates as the center of the charge.
Next, expand this value of V up to terms in 1/r^2
Homework Equations
[itex] V = \frac{\lambda}{4 \pi \epsilon_0 r} [/itex]
The Attempt at a Solution
I let the line of charge sit along the z-axis, with its center at z=0, in the z-x plane
I know the charge density [itex]\lambda=\frac{Q}{a} [/itex]
[itex] V = \frac{Q}{4\pi a \epsilon_0} \int_{-a/2}^{a/2} \frac{1}{\sqrt{z^2 + x^2}} \, dz[/itex]
Solving the integral, this gives me the solution [itex] V = \frac{Q}{4\pi a \epsilon_0} \ln \frac{\sqrt{x^2 + \frac{a^2}{4}}+ \frac{a}{2}}{\sqrt{x^2 + \frac{a^2}{4}}- \frac{a}{2}}[/itex]
However, this is not the right answer. My book says the answer is [itex] V = \frac{Q}{4\pi a \epsilon_0} \ln \frac{\sqrt{r^2 - az + \frac{a^2}{4}}+ \frac{a}{2} -z}{\sqrt{r^2 + az + \frac{a^2}{4}}- \frac{a}{2} -z} [/itex]So, now I realize I solved for the potential only straight above the center point of the line, on the x-axis... So, I'm guessing my book wants a solution for the potential anywhere, and that is why their answer looks more complicated, with more variables...
So, I'm having trouble generalizing my V formula to work for any point around the line of charge. I know I need to change [itex] \frac{1}{\sqrt{z^2 + x^2}} [/itex], but I'm not sure how exactly to change it..
Also, I'm not sure what r and z correspond to in the book's solution - I'm not even sure how they set up their axis (no picture)