Charged ball hanging in equilibrium between two parallel plate capacitors.

[bitterbilly]

1. An experiment is run to determine the validity of numerous electrostatic principles. A small ball of mass 6.50 x 10^-3 kg with a charge of ±1.50 μC is suspended from a non-conductive wire and hangs between two parallel plate capacitors. An angle of 30.0° is measured between the wire and the vertical while the ball is in equilibrium.
a. Determine the sign of the charge on each of the plates and the small ball.
b. Draw a free body diagram for the small ball.
c. Determine the magnitude of the electric field produced by the parallel plate capacitor.
d. Solve for the potential difference between the two plates that are separated by a distance of 1.00 x 10^-4m.
e. A charge of 10.0 C is stored on each plat. Determine the surface area of each plate.
f. Calculate the amount of flux that passes through one plate.
g. Discuss the change in the angle of θ if the same experiment were immersed in non-conductive oil. (Assume the dielectric constant for the oil is greater than that of air.)


I am having a lot of trouble trying to figure out what to do.



So far, for (a) I have the signs as + ---------> -
+ ---------> - etc.
For (b) I have weight going down, the Electric Force, F_E going right, and Tension going North/West at 30°.

For (c) I set Fy = W, and so Fy = 6.37 x 10^-2N. Then using trig I found that Fx = 3.68 x 10^-2N, which is then equal to the Force.

This is where I'm stuck, I don't really know how to find the value for the E field. Any help would be much appreciated.

 

[technician]

I agree with your Fy and Fx. Fx is the horizontal electric force on the 1.5μC charge.
The equation for the force on a charge, q, in an electric field, E is F = q x E.
You should be able to find the Electric field strength E.
Field strength is measured in Volts per metre so if you know the separation of the plates you should be able to find the voltage across them.
BUT... part (d) in your question states that the plates are 1 x 10^-4 m apart. This is only 1/10 of a mm... are you sure about this?