Why is dE(y) = dEcos(theta) in Electric Field Calculations?

[Lavace]

Homework Statement

http://www.ph.qmul.ac.uk/~phy210/Ec2.pdf
Question. 3)

Alright I'm just going over things again so I'm ready for year 2 (this is year 1 question).

I already know the method of answering (take a small part dE of the line, form an question of it's electric field on P, intergrate over the length L).

http://www.ph.qmul.ac.uk/~phy210/ECAns2.PDF
There's the answers, I don't understand why, when he also stated that all x-direction forces will cancel due to symmetry, and then says dE(y) = dEcos(theta) <-- isn't that the x-component of the field? I was intending to use sin(theta) until I saw he done that.

Is this because when intergrated we end up with sin(theta)? Why do we use this process?

Thanks!

 

[AUMathTutor]

He uses cos(theta) because he places his theta such that the leg adjacent to theta is the one perpendicular to the line of charge. This is the direction the force will be in, since the vertical components will cancel out.

If you choose the other angle for theta (not the right angle, mind you) then you would in fact use sin(theta). Basically, whether you use sin or cos depends on which of the two angles you're calling theta. The important part is that what matters is that you use the right legs of the triangle...

 

[javadazimi]

That's right .