Point Charge on Semicircular Ring

[Tater]

Homework Statement

A point charge Q is located at point P(0,-4,0) while a 10nc charge is uniformly distributed along a semicircular ring as shown in the figure. find the value of Q such that E(0,0,0)=0

Homework Equations

Q=ρ_Ldl
dl = ρd∅ (because ρ and z are constant)
E=$\frac{Qρ}{4piεr^2}$

The Attempt at a Solution

Okay so first I'd have to find Q using the formula above. I'd set my limits from 0 to pi.

Now looking at the formula for point charge, the only missing factor is R. I don't know how to express this. I know that the equation of a hemisphere is very similar to that of a circle. I came up with y=$\sqrt{4-x^2}$. But now that I look at it, if I tried to convert that to polar using the relationships x=rcos$\vartheta$ and y=rsin$\vartheta$, it really seems wrong.

What's the best procedure in doing this? What should I say to myself every time I see a ring? Because to be honest, I really get confused when I see an odd figure labeled in terms of (x,y,z) even though it's much easier in another system.

Also, after finding R, I simply just stick to the formula and I will obtain my solution, right?

Thank you for any help :)

 

[runnergirl]

First off you want to treat each case separately. By symmetry of the problem we know that the E-field will only be in the y-direction. Since the charge is distributed uniformly along the arc, you can deal with it as a line charge density which is what you did:
$dQ_1=\rho_l(\rho d\phi)$ now integrating that out you can get the line charge density in terms of $Q_1$ (which you'll want in the end).

So setting up the integral for the E-field:
Since your r is constant (r=2) you'll only be integrating over the angular change.
$dE_y=\frac{\rho_l}{2\pi\epsilon_or}sin\phi d\phi$

Integrating that out should give you the E-field in the y-direction due to the hemisphere.

The point charge is just the E-field due to a point charge which is:
$E = \frac{Q}{4\pi\epsilon_oR^2}$ where Q is the charge you're solving for and R is the distance to the charge.

The total E-field:
$E_t = E_{ring}-E_{charge}$ and just solve for Q.

Hope this helps.

 

[runnergirl]

Also, a nice site I found for a hemisphere line charge if you want more of a visualization of what is going on:
http://www.phys.uri.edu/~gerhard/PHY204/tsl329.pdf