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.

 

[paranormal]

I would like to provide a response to this content. Firstly, it is important to note that the electric force acting on the sphere is a result of the electric field between the two parallel conducting plates. The excess charge on each plate creates an electric field, and the charged sphere experiences a force due to its presence in this field.

To address the first question, drawing a free body diagram (FBD) is a good approach to visualize the forces acting on the sphere. As mentioned, the tension force from the silk thread will be acting upwards, while the electric forces from the positive and negative plates will be acting to the right and left respectively. It is important to note that the electric forces will be equal in magnitude and opposite in direction, resulting in a net force of zero in the horizontal direction.

In order to answer the question about the number of protons and electrons on the sphere, we can use the equation q=Ne, where q is the charge on the sphere, N is the number of charges (protons or electrons) and e is the elementary charge. Since both the charge and the number of electrons are given, we can calculate the number of protons on the sphere to be 1.25 x 10^11.

Moving on to the calculation of the magnitude of the electric force on the sphere, we can use the equation F=qE, where F is the force, q is the charge on the sphere and E is the electric field. Since the electric field between the plates is uniform, we can use the equation E=V/d, where V is the potential difference between the plates and d is the distance between them. In this case, the potential difference can be calculated as V=Ed, where E is the electric field and d is the distance between the plates. Therefore, the magnitude of the electric force on the sphere can be calculated as F=qV/d.

Finally, to calculate the magnitude of the electric field between the plates, we can use the equation E=V/d, where V is the potential difference between the plates and d is the distance between them. As mentioned earlier, the potential difference can be calculated as V=Ed, where E is the electric field and d is the distance between the plates. Therefore, the magnitude of the electric field between the plates can be calculated as E=V/d=Ed/d=q/d.

In conclusion, the electric force on the charged sphere can be calculated using the equation F