Finding Equilibrium Angle in Uniform Electric Field

[z_sharp]

Hey all, I'm not sure where to start on this question. Any ideas would be spectacular.

A tiny conducting ball of mass 1.00 g and charge 20.0 times $$10^{-6}$$ is hung from a non-conducting, massless thread. The electric field of magnitude $$10^3N/C$$ existing in this region is uniform and horizontal. Find the maximum angle of deviation between the thread and the vertical for the equilibrium, position of the ball.

So far this is what I have
For equilibrium sum of all forces must equal 0
Therefor...
$$\begin{equation*}<br /> \begin{split}<br /> \ F_{{net}_x}=0 \\<br /> 0=F_{field}-{T_x} \\<br /> 0= ? - T\sin\theta<br /> \end{split}<br /> \end{equation*}$$

$$\begin{equation*}<br /> \begin{split}<br /> \ F_{{net}_y}=0 \\<br /> 0=F_g-{T_y} \\<br /> 0=mg-T\cos\theta<br /> \end{split}<br /> \end{equation*}$$

I'm not to sure what I do for the force of the electric field and how I encorperate that into the equation.

Thanks Everyone

 

[clive]

Your equilibrium conditions are OK:

$$F_{field}-Tsin \theta=0$$
$$mg - T cos \theta=0$$

Now you must take into account that $$F_{field}=qE$$ (electric charge times electric field magnitude). You obtain then (by eliminating T):

$$tg \theta = \frac{qE}{mg}$$.

 

[z_sharp]

Thanks for your help, I was able to get the problem with the assistance you provided.