Solving Charged Circular Arc Problem: Q, R, \Delta E_x

[Punchlinegirl]

A uniformly charged circular arc AB is of radius R covers a quarter of a circle and is located in the second quadrant. The total charge on the arc is Q > 0. This problem has 4 parts, I got the first 2.
1. The direction of the electric field E due to the charge distribution at the origin is in quadrant 4.
2. Determine $$\Delta E_x$$, the x-component of the electric field vector at the origin O due to the charge element $$\Delta q$$ locate at an angle $$\theta$$ subtended by an angular interval $$\theta$$.
$$\Delta E_x = kQ/R^2 * 2\Delta \theta / \pi * cos \theta$$
3. Find E_x, the electric field at the origin due to the full arc length for the case where Q= 2.3 $$\mu C$$ and R= 0.37 m. Answer in units of N/C.

I have no idea how to find the value for theta. Can someone tell me what I should do?

 

[quantumdude]

You're not looking for *a* value of theta, you're looking for a range of values. That's because you have to integrate the expression that you found in part 2.

So between what two angles is the second quadrant bounded?

 

[Punchlinegirl]

The arc is bounded between $$\pi$$ and $$\pi /2$$.
so when I go to integrate it, would it just be the integral just be 2 $$\Delta \theta cos \theta$$? Since kQ/r^2 $$\pi$$ is all constant?

 

[lightgrav]

Your latex isn't loading for me...
cos(theta) is NOT constant, but all the other terms are.
you have to integrate cos(theta) from 0 to pi/2 .

 

[Punchlinegirl]

So if I integrate I get
$$KQ/r^2 2 \Delta/ \pi sin \theta$$ from pi/2 to 0.
I'm assuming Delta pi would just be pi/2-0= pi/2.
So plugging in gives me,
2.3 x 10^-6 k/(.37)^2 * 2(pi/2)/pi * sin pi/2
Which = 151205 N/C
This isn't right.. I think my delta theta is messed up.
Also, why wouldn't the bounds be pi and pi/2 since it's located in the 2nd quadrant?

 

[lightgrav]

What? $$\Delta \theta = d \theta$$ , which is your integration variable!

We can't tell where you're measuring theta from , without a diagram.
The result is the same, either way.

 

[xdtran3x]

anyone?