Find charge due to two particles

In summary, the conversation discusses the calculation of the charge q_red on a red sphere located at (d1,0) with a positive blue sphere at the origin and a negative yellow sphere located at (d2cos(theta), -d2cos(theta)). The net force F is in the direction of -y. The equations for Fx on blue due to the yellow and red spheres are set equal to each other to solve for q_red, resulting in the formula q_red= 2qd1^2/[(d2cos(theta))^2]. The individual is seeking confirmation on the correctness of their equations and formula.
  • #1
StephenDoty
265
0
Suppose that the magnitude of the charge on the yellow sphere is determined to be 2q and at (d2cos(theta), -d2cos(theta). Calculate the charge q_red on the red sphere at (d1,0) with a positive blue sphere,q, at the origin and whose net force,F, is in the direction of -y. The yellow sphere is negative and the red sphere is positive. ( See picture bellow)
Express your answer in terms of q, d1,d2, and theta.

Well the Fx on blue due to the yellow sphere= k(2q)(q)/(d2cos(theta))^2
and the Fx on blue due to the red sphere = - k(q_red)(q)/(d1)^2

Now what? I am unsure as to what to do now. Any help would be appreciated.

Thank you.
Stephen

attachment.php?attachmentid=15199&d=1219858491.jpg
 
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  • #2
Would you set them equal to each other?
 
  • #3
I would? because of the fact that there is no net force along the x axis. Hence all x components of the force must cancel out to equal 0. Do you not get the right answer by doing that?
 
  • #4
I set them equal to each other and solved for q_red and got:

q_red= 2qd1^2/(d2^2)(cos(theta)^2)

Did I set up the original equations correctly? And did I find q_red correctly in terms of q, d1,d2, and theta?

Thank you. I really appreciate the help.

Stephen
 
  • #5
I have to turn this in soon so any help would be appreciated.

thank you.

Stephen
 
  • #6
I don't see anything wrong with it. Does anyone else see anything wrong with it?
 
  • #7
bump
Does everyone agree with nova-ex? Are my equations correct?
 
  • #8
Not to be a bother but I have to turn this in tomorrow. So if you can please tell me whether or not I set the problem up correctly and got the correct formula for q_red: q_red= 2qd1^2/[(d2cos(theta))^2]?

Thanks I would really appreciate it.

Stephen
 
  • #9
Any opinions?
 

1. How do you calculate the charge due to two particles?

The charge due to two particles can be calculated using the Coulomb's law, which states that the force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

2. What is the unit of charge used to calculate the charge due to two particles?

The unit of charge used to calculate the charge due to two particles is Coulomb (C). It is defined as the amount of charge that passes through a wire in one second when a current of one ampere is flowing through it.

3. Can the charge due to two particles be positive and negative at the same time?

No, the charge due to two particles cannot be positive and negative at the same time. Each particle has a specific charge, either positive or negative, and the total charge due to two particles is the algebraic sum of their individual charges.

4. How does the distance between two particles affect the charge due to them?

The charge due to two particles is inversely proportional to the square of the distance between them. This means that as the distance between the particles increases, the force between them decreases, and vice versa.

5. Is the charge due to two particles affected by the medium between them?

Yes, the charge due to two particles can be affected by the medium between them. If the medium is a conductor, it can alter the charge distribution between the particles, resulting in a different charge due to them. However, if the medium is an insulator, the charge due to the particles remains the same.

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