Net electric field of a charged arc

[Jrlinton]

Homework Statement

So this was a problem worked in class by the professor in class.
Find the net electric field at the origin due to the arcs

Homework Equations

L=2πr/4
λ=q/L
E=kQ/r²

The Attempt at a Solution

So the professor gave the answer using the fromula
E_net=λ₁(2sin45°)/(4πε0r₁)+λ₂(2sin45°)/(4πε0r₂)+λ₃(2sin45°)/(4πε0r₃)

I can understand the formula except for where the 2sin45° comes from. I mean the arc is a quarter circle of 90° in the second quadrant but I am unsure where this term comes from and how it would change if the parameters of the problem were to change.

 

[TSny]

See if this diagram helps

 

[Jrlinton]

So if U am understanding this correctly then it is the added sine values of the two angles created when drawing the vector for the field?

 

[TSny]

So if U am understanding this correctly then it is the added sine values of the two angles created when drawing the vector for the field?

I'm not following what you are saying here. In order to see why there is a factor of 2sin(45^o) in the answer, you need to go through the derivation. This means setting up and evaluating the integral for the net electric field. Are you having trouble setting up the integral?

 

[Jrlinton]

Yes, I am having trouble coming up with the integral.

 

[TSny]

Yes, I am having trouble coming up with the integral.

OK. Show us your attempt at setting up the integral and we can go from there. It helps to first consider the direction of the net electric field.