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mrtubby
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Homework Statement
There is a charged wire aligned/centered along the Y 2L in length. A test charge is placed D distance from the wire on the X axis. The wire has a total charge of Q. calculate the integral to figure out how much force the test charge experiences.
a semi decent pic of the layout
Homework Equations
Columbs law + some calc
The Attempt at a Solution
My question isn't about getting the answer, I have it (too bad I didn't have the problem right when I looked at it), but I don't understand some of it and am hoping for some clarification.
[tex]F_x = \int_{-L}^{L} \frac{kQq}{2Lr^2}cos(\theta)dy[/tex]
This is all well and good, the integral is over the length of the wire, the rest of it is basically Columbs law with Q/2L * dy as the charge density * tiny section of wire for integration. It took me a while (and this is the part i am hung up on) but the cos theta is in there because we need to deal with the fact that all the little tiny chunks of wire are each at a different angle to the test particle so different ratios of force are felt in the X and Y directions.
now for what I don't understand when this integral is worked you get this
[tex]F_x = \frac{kQq}{2L}\int_{-L}^{L} \frac{D}{\sqrt{D^2 + y^2}^{3/2}}dy[/tex]
Inside the integral the denomanator is obviously pythagorians theorum to find the distance between wire chunk and particle. How does [itex]cos{\theta}[/itex] convert to just plain D (the distance between the wire and particle at the x axis)?
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