How Does Electric Force Affect a Charged Sphere Between Parallel Plates?

[arsene2conde]

A small sphere of mass 1.0 x 10^-6 kg carries a total charfe of 2.0 x 10^-8 C. The sphere hangs from a silk thread between two large parallel conducting plates. The excess charge on each plate is equal in magnitude, but opposite in sign. The thread makes an angle of 30 with the positive plate. Consider the sphere to be a point particle and the electric field between the plates to be uniform.

Homework Equations

1)One of the question was to draw a FBD. I drew the sphere with the force of the tension pointing up towards one of the plate...in this problem, it is the positive plate. Then the force from the negative plate pointing directly to the right and the force from the positive plate also pushing the sphere to the right.

2)How many more protons than electrons are on the sphere?
I used q=Ne
since q and e are given, I get N = 1.25 x 10^11 protons.

3)Calculate the magnitude of the electric force on the sphere.
This is where I am stock. I have sumed up all the force on the x and y axis. Then squared both forces and took the square root. I have on the x-axis (2.0 x 10^-8)sin30 and on the y-axis I have -(1 x 10^-6)(9.8) + (2.0 x 10^-8)cos30.
I am not sure if I am doing this right at all.

4)Finally it is asking for the magnitude of the electric field between the plates.
I order to move on, I believe I need answer from question 3.
I believe I would need to use E = sigma/e0

I would appreciate if someone could help me out on this

Thank you

The Attempt at a Solution

 

[kuruman]

1)One of the question was to draw a FBD. I drew the sphere with the force of the tension pointing up towards one of the plate...in this problem, it is the positive plate. Then the force from the negative plate pointing directly to the right and the force from the positive plate also pushing the sphere to the right.

The tension points along the thread and away from the sphere. Is that what you meant? Also, what about the force of gravity?

 

[arsene2conde]

The tension points along the thread and away from the sphere. Is that what you meant? Also, what about the force of gravity?

yea that' s what I meant and the force of the gravity points directly downward, forgot to mention it.

 

[kuruman]

This is where I am stock. I have sumed up all the force on the x and y axis. Then squared both forces and took the square root. I have on the x-axis (2.0 x 10^-8)sin30 and on the y-axis I have -(1 x 10^-6)(9.8) + (2.0 x 10^-8)cos30.
I am not sure if I am doing this right at all.

It is not right. You need to draw a free body diagram and say that sum of all the forces in the horizontal direction is zero and the sum of all the vertical forces is zero. This gives two equations and two unknowns. The unknowns are the tension T and the electrical force qE.