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Need guidance on finding charge.

  • Thread starter cidr1337
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  • #1
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Homework Statement


System with 3 charged particles red, yellow, and blue. The blue and red have a positive charge, while the yellow has a negative charge. The blue particle lies at (0,0). The red charge lies at (d1,0). The Yellow charge lies at (d2cos(θ), -d2sin(θ)). The Fon blue is (0,-F); F>0. q of yellow is 2q. What is the charge on the red particle in terms of q, d1, d2, and theta.


Homework Equations


Coulomb's law. F=(K*q1*q2)/d^2


The Attempt at a Solution


http://i249.photobucket.com/albums/gg221/cidr1337/394254_297418573698401_1519950841_n.jpg
edited for kushan's question.
 
Last edited:

Answers and Replies

  • #2
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Is the answer asking in terms of q , qr( charge of red) , theta , d1 , d2 .
I have an intuition that q is charge of red particle . Can you check you question ?
 
  • #3
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Is the answer asking in terms of q , qr( charge of red) , theta , d1 , d2 .
I have an intuition that q is charge of red particle . Can you check you question ?
in terms of q, theta, d1, and d2.
 
  • #4
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So what is the charge of red particle ?
 
  • #5
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yes.
 
  • #6
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You know the resultant of force is along -Fj . And you also know that force between blue and red would be along -i , and the force between blue and yellow makes and angle theta with positive x axis .
Solve these vector equations .
 
  • #7
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Not sure i understand... sorry.
My new attempt i set 0 = Fredonblue+Fyellowonblue
0=(K*qr*qb)/d1^2+ (k*qy*qb)/(d2cos(theta))^2
-(K*qy*qb)/(d2cos(theta))^2=(K*qr*qb)/d1^2 qb and K cancels out and setting qy=2q yields
|qr|= -(2q(d1)^2)/(d2cos(theta))^2
does this make any sense or am i grabbing at straws?
 
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  • #8
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shameless self bump.
 
  • #9
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I don't think that is true since it states that F>0.

But I think that you have the right idea that F is just the sum of individual forces and then you just solve for qr in terms as stated.
 
  • #10
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Can you help me figure out the force that yellow exerts on blue? I think I'm having a problem with the actual equation. In the x direction? I'm saying it is Fyonb= (K*|2q|*|qb|)/(d2)^2cos(theta) but it says the answer doesn't rely on K or qb
 
  • #11
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[itex] \vec{F_{blue}} = \vec{F_{blue red}} + \vec{F_{blue yellow}} [/itex]

I read the problem again and I think you're going to have to solve a system of equations to get it in the required terms.
I'd start off by writing each of the forces acting on the other and then the net force for blue that you are given. Then see if you can set any of them equal to eliminate variables.

I haven't started working out the problem yet so I can't make any promises but it seems like a reasonable approach.
 
  • #12
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Thanks for trying to help me guys.
 
  • #13
TSny
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cidr1337,

You are given that the net force on the blue charge has zero x-component. Do you see that the only way for that to happen is: The x-component of the force on the blue charge due to the yellow charge must cancel the x-component of the force on the blue charge due to the red charge?

Can you construct expressions for these x-components of the forces?

Can you use these expressions to set up an equation that expresses that the net x-component of the force on the blue charge is zero?
 
  • #14
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cidr1337,

You are given that the net force on the blue charge has zero x-component. Do you see that the only way for that to happen is: The x-component of the force on the blue charge due to the yellow charge must cancel the x-component of the force on the blue charge due to the red charge?

Can you construct expressions for these x-components of the forces?

Can you use these expressions to set up an equation that expresses that the net x-component of the force on the blue charge is zero?
*That is my problem. I can't set this up. I tried 0=Fyonb+Fronb but my y on be expression is wrong. I don't know how to express this. Apparently it doesn't depend on K or qb, I said F yonb = (K*|2q|*|qb|)/(d^2)*cos(theta).
I'm probably going to feel stupid when I figure this out. :)
 
  • #15
TSny
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I said F yonb = (K*|2q|*|qb|)/(d^2)*cos(theta).
Yes :smile:, that's the correct expression for the x-component of the force on the blue charge due to the yellow charge (where d = d2, right?)

Now write out an expression for the x-component of the force on the blue charge due to the red charge (including the correct sign).
 
  • #16
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Yes :smile:, that's the correct expression for the x-component of the force on the blue charge due to the yellow charge (where d = d2, right?)

Now write out an expression for the x-component of the force on the blue charge due to the red charge (including the correct sign).
My problem is I went into the hints and the question asked me, "Find the x component of the force that the yellow sphere exerts on the blue sphere.
Express your answer in terms of q, d2and theta. I input Fyonb and it tells me the answer doesn't depend on K or qb. It's telling me it's not right :(.
 
  • #17
TSny
Homework Helper
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My problem is I went into the hints and the question asked me, "Find the x component of the force that the yellow sphere exerts on the blue sphere.
Express your answer in terms of q, d2and theta. I input Fyonb and it tells me the answer doesn't depend on K or qb. It's telling me it's not right :(.
Your expression for the x component of Fyonb is correct. It does depend on k and qb, as well as on q, d2, and theta.

You're ready to continue. So, go on and find an expression for the x component of Fronb.
 
  • #18
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Ok so the x component is (K*(2q^2)*cos(theta))/(d2^2). idk how they got qb=q if i could have gotten that I would have got the problem.
 
  • #19
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render.gif

Can you guys help me figure out why qblue=q
 
  • #20
TSny
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Ok so the x component is (K*(2q^2)*cos(theta))/(d2^2). idk how they got qb=q if i could have gotten that I would have got the problem.
Where did you get this expression? Who is "they"? The expression is correct if qb = q, but your original statement of the problem did not specify a value for qb. Now, it actually doesn't matter what the value of qb is. You will see that if you set up the equation that states that the sum of the x-components of the two forces acting on the blue charge is 0, then you won't need to know the blue charge in order to find the red charge.
 
  • #21
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Where did you get this expression? Who is "they"? The expression is correct if qb = q, but your original statement of the problem did not specify a value for qb. Now, it actually doesn't matter what the value of qb is. You will see that if you set up the equation that states that the sum of the x-components of the two forces acting on the blue charge is 0, then you won't need to know the blue charge in order to find the red charge.
That's what im saying it's a program called mastering physics, that my university uses. and you're right about qb=q not mattering for the over all equation this program has given me a headache. I don't have my book i had been solving for the angle while solving for the force this whole time. I didn't know we had to solve for force and then solve for the components. Thanks for your help guys it is appreciated.
 

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