Where on the x-axis will a third charge have no net force

In summary, the net force on a (-2) micro Coulomb charge would be zero when it is positioned at d = 6 on the x-axis. The equation F = k*(q1q2)/d^2 should only be used to calculate the magnitude of the force, and the negative sign should not be included in the calculation for the (-4) micro Coulomb charge.
  • #1
bbraves7
15
0

Homework Statement



Where on the x-axis would the net force on a (-2) micro Coulomb charge be zero?

*picture attached*

Homework Equations



F = k*(q1q2)/d^2

The Attempt at a Solution



I think the answer is d= 6, making the position on the x-axis = 7. However, when solving my equation, I have to take the square root of a negative number, which obviously has no real answer. I couldn't seem to logic my way around it, so what am I missing? Other than expressing the answer in terms of i, how can I get around this??
 

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  • #2
9 μC = 9×10‒6 C , Similarly ‒4 μC = ‒4×10‒6 C and ‒2 μC = ‒2×10‒6 C.

You have written them as 9‒6 C, ‒4‒6 C, and ‒2‒6 C .

(Maybe you are simply using a non-standard notation.)
 
  • #3
You need F1 + F2 = 0.

So that F1 = ‒ F2

That should take care of your signs.
 
  • #4
SammyS said:
9 μC = 9×10‒6 C , Similarly ‒4 μC = ‒4×10‒6 C and ‒2 μC = ‒2×10‒6 C.

You have written them as 9‒6 C, ‒4‒6 C, and ‒2‒6 C .

(Maybe you are simply using a non-standard notation.)

Yeah, I just write it like that as a kind of short hand.

Anyways, I figured out what I was doing wrong. The equation should only be used to calculate the magnitude of the force, so I shouldn't put the negative in the equation for the -4 micro Coulomb charge. That takes care of it. Thanks for the reply though.
 
  • #5


Your logic is correct. The net force on the charge would be zero when it is located at a distance of 6 units from the -1 micro Coulomb charge, making its position on the x-axis 7 units. However, when solving for the distance using the equation F = k*(q1q2)/d^2, you should not be taking the square root of a negative number. The equation should be set up as follows:

F = k*(q1q2)/d^2
0 = k*(-1 micro C)(-2 micro C)/(d-7)^2
0 = (9x10^9)(2x10^-12)/(d-7)^2
0 = (1.8x10^-2)/(d-7)^2

Solving for d, we get d = 7, which is the correct answer. There is no need to take the square root of a negative number in this case. If you are getting a negative number when solving for d, it is likely that you made a calculation error. Double check your calculations to ensure that you are getting the correct answer.
 

1. What is the definition of "net force" in terms of charges on the x-axis?

Net force is the overall force acting on an object due to the combination of all individual forces acting on that object. In the context of charges on the x-axis, it refers to the resultant force experienced by a third charge due to the presence of two other charges.

2. How is the direction of net force determined on the x-axis?

The direction of net force on the x-axis is determined by the relative positions and charges of the two other charges. If the two charges have the same sign, the net force will be in the same direction as the individual forces. If the two charges have opposite signs, the net force will be in the opposite direction.

3. What factors affect the magnitude of net force on the x-axis?

The magnitude of net force on the x-axis is affected by the distance between the two other charges and the charges themselves. As the distance between the two charges increases, the magnitude of net force on the x-axis decreases. Similarly, increasing the magnitude of one or both charges will also increase the net force.

4. How can the location of a third charge be determined to have no net force on the x-axis?

The location of a third charge where there is no net force on the x-axis can be determined by using the principle of superposition. This means that the net force on the third charge will be zero when the individual forces from the two other charges cancel each other out. This occurs when the third charge is located at the midpoint between the two other charges.

5. Can a third charge ever have no net force on the x-axis?

Yes, it is possible for a third charge to have no net force on the x-axis. This occurs when the third charge is located at the midpoint between the two other charges, as the individual forces from the two charges will cancel each other out. This is known as a point of equilibrium and it is a stable position for the third charge.

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