Finding charge based on net force and other charges

  • #1

Homework Statement



Blue charge is at origin with positive q charge
Red charge at point (d1,0) with unknown positive charge q_red
Yellow charge at point (d2cos(theta),-d2sin(theta)) with negative 2q charge

The net electric force on the blue sphere has a magnitude F and is directed in the - y direction.

Suppose that the magnitude of the charge on the yellow sphere is determined to be 2q. Calculate the charge q_red on the red sphere. Express your answer in terms of q, d1, d2, and theta.

Homework Equations



electric force F = kq_1q_2/r^2 where k = 9*10^9, q_1 and q_2 represent point charges, and r is distance between point charges

The Attempt at a Solution



F = [(k*q_yellow*q_blue)/d2^2 ] + [(k*q_red*q_blue)/d1^2]
F = [(k*(-2q)*q)/d2^2] + [(k*q_red*q)/d1^2]
d1^2[F - ((k*(-2q)*q)/d2^2)] / kq = q_red

i think i'm on the right track but i did not use theta which i need to, where did i go wrong
 

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Answers and Replies

  • #2
does this involve the coordinate location of the yellow charge because it contains theta which i think i need in my final answer, but the distance is stated as d2 so why should i need it?
 
  • #3
LowlyPion
Homework Helper
3,090
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Homework Statement



Blue charge is at origin with positive q charge
Red charge at point (d1,0) with unknown positive charge q_red
Yellow charge at point (d2cos(theta),-d2sin(theta)) with negative 2q charge

The net electric force on the blue sphere has a magnitude F and is directed in the - y direction.

Suppose that the magnitude of the charge on the yellow sphere is determined to be 2q. Calculate the charge q_red on the red sphere. Express your answer in terms of q, d1, d2, and theta.

Homework Equations



electric force F = kq_1q_2/r^2 where k = 9*10^9, q_1 and q_2 represent point charges, and r is distance between point charges

The Attempt at a Solution



F = [(k*q_yellow*q_blue)/d2^2 ] + [(k*q_red*q_blue)/d1^2]
F = [(k*(-2q)*q)/d2^2] + [(k*q_red*q)/d1^2]
d1^2[F - ((k*(-2q)*q)/d2^2)] / kq = q_red

i think i'm on the right track but i did not use theta which i need to, where did i go wrong
I can't see your picture as yet, but I would suggest that you separate the forces into their x,y components and then add them. They tell you the result vector is acting in the -Y direction only, so x components must add to 0. Force is a vector and adding the magnitudes if they are not acting along the same line is not the way to do it.
 
  • #4
how about this, i put the net forces into components:

fnet_x = [k(2q)(q)/(d2cos(theta))^2] + [k(q_red)(q)/(d1^2)]
fnet_y = [k(2q)(q)/(d2sin(theta))^2] + 0

F = sqrt((fnet_x)^2 + (fnet_y)^2)
[sqrt[F^2 - (fnet_x)^2](d2sin(theta))]/(2q)(k) = q_red

is that correct now?

how do i factor in the -y net force direction?
 
Last edited:
  • #5
LowlyPion
Homework Helper
3,090
4
how about this, i put the net forces into components:

fnet_x = [k(2q)(q)/(d2cos(theta))^2] + [k(q_red)(q)/(d1^2)]
fnet_y = [k(2q)(q)/(d2sin(theta))^2] + 0

F = sqrt((fnet_x)^2 + (fnet_y)^2)
[sqrt[F^2 - (fnet_x)^2](d2sin(theta))]/(2q)(k) = q_red

is that correct now?

how do i factor in the -y net force direction?
Not quite.

First simply identify the force between the Blue/Red and the Blue/Yellow.
These are the forces that you must treat as vectors.

Hence
F(b/r) = kqb*qr/(d1)2*x-hat + 0*y-hat
F(b/y) = kqb*qy/(d2)2*Cosθ*x-hat + kqb*qy/(d2)2*Sinθ*y-hat

Now when you add F(b/r) and F(b/y) you add the components.
But you also know that the x-components (x-hat terms) must add to 0
And you also know that the charge on Yellow is -2*q and the charge on Blue is +1*q. The qr is the one that is unknown. Figure it must be a positive charge since Red is positive Yellow negative and otherwise they could never add to 0.
 
  • #6
so this is what i got, since we know the x components must sum to zero soo...

[(q_red)(+q)k]/d1^2 + [(+q)(-2q)k]/d2cos(theta)^2 = 0
so q_red = [-k(+q)(-2q)(d1^2)]/[(d2cos(theta)^2)(+q)(k)]

is this right?
 
  • #7
LowlyPion
Homework Helper
3,090
4
so this is what i got, since we know the x components must sum to zero soo...

[(q_red)(+q)k]/d1^2 + [(+q)(-2q)k]/d2cos(theta)^2 = 0
so q_red = [-k(+q)(-2q)(d1^2)]/[(d2cos(theta)^2)(+q)(k)]

is this right?
No. I don't think you read the equations I gave you carefully.

[tex] \vec{F_{BR}} = \frac{k*Q_B*Q_R}{d_1^2} *\hat{x} + 0*\hat{y}[/tex]

[tex] \vec{F_{BY}} = \frac{k*Q_B*Q_Y}{d_2^2}*Cos\theta* \hat{x} + \frac{k*Q_B*Q_Y}{d_2^2}*Sin\theta *\hat{y} [/tex]
 
Last edited:
  • #8
okay i see now, aside from the issue with the sine and cosine, was my approach for solving for q_red correct?

here it is with corrections

[(q_red)(+q)k]/d1^2 + [(+q)(-2q)(k)(cos(theta))]/d2^2 = 0

q_red = -[k(+q)(-2q)(d1^2)(cos(theta))]/[(d2^2)(+q)(k)]

q_red = -[(-2q)(d1^2)(cos(theta))]/[(d2^2)] = (2q)(d1^2)(cos(theta))]/[(d2^2)

how about now?
 
  • #9
LowlyPion
Homework Helper
3,090
4
okay i see now, aside from the issue with the sine and cosine, was my approach for solving for q_red correct?

here it is with corrections

[(q_red)(+q)k]/d1^2 + [(+q)(-2q)(k)(cos(theta))]/d2^2 = 0

q_red = -[k(+q)(-2q)(d1^2)(cos(theta))]/[(d2^2)(+q)(k)]

q_red = -[(-2q)(d1^2)(cos(theta))]/[(d2^2)] = (2q)(d1^2)(cos(theta))]/[(d2^2)

how about now?
That looks more like it.
 
  • #10
thanks so much, it was right!!!
 

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