Electrostatics, Even Distrobution on wire dumb question

[mrtubby]

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.

$$F_x = \int_{-L}^{L} \frac{kQq}{2Lr^2}cos(\theta)dy$$

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
$$F_x = \frac{kQq}{2L}\int_{-L}^{L} \frac{D}{\sqrt{D^2 + y^2}^{3/2}}dy$$

Inside the integral the denomanator is obviously pythagorians theorum to find the distance between wire chunk and particle. How does $cos{\theta}$ convert to just plain D (the distance between the wire and particle at the x axis)?

 

[gabbagabbahey]

Hi mrtubby, welcome to PF!

now for what I don't understand when this integral is worked you get this
$$F_x = \frac{kQq}{2L}\int_{-L}^{L} \frac{D}{\sqrt{D^2 + y^2}^{3/2}}dy$$

Inside the integral the denomanator is obviously pythagorians theorum to find the distance between wire chunk and particle. How does $cos{\theta}$ convert to just plain D (the distance between the wire and particle at the x axis)?[/color]

It doesn't!

Remember your unit circle...$\cos\theta$ equals adjacent/hypoteneuse. And take note of the fact that the denominator in your integrand is actually [tex](D^2+y^2)^{3/2}[/itex] and Pythagoras Law tells you $r^2=D^2+y^2$.[/tex]

 

[mrtubby]

Fantastic, I had come up with some kind of flawed reasoning as to why i was looking at $(D^2+y^2)^{3/2}$ that basically boils down to some weak algebra skills. Now the math makes sense and I can continue on with my day.

Thanks Bunches!