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

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Homework Help Overview

The original poster is investigating the position along the x-axis where a third charge, specifically a (-2) micro Coulomb charge, experiences no net force due to two other charges, one being 9 μC and the other -4 μC. The problem involves understanding the forces acting on the charge and their magnitudes.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to solve for the position by applying the formula for electric force but encounters a mathematical issue when trying to take the square root of a negative number. They question what they might be missing in their reasoning.
  • Some participants clarify the notation used for the charges and suggest ensuring that the signs are correctly accounted for in the force equations.
  • One participant emphasizes the need for the forces to balance, indicating that the forces from the two charges must be equal in magnitude but opposite in direction.

Discussion Status

The discussion is active with participants providing clarifications on notation and force calculations. There is an acknowledgment of the need to correctly interpret the signs of the charges in the equations. The original poster seems to have identified an error in their approach, but the conversation is ongoing without a definitive resolution.

Contextual Notes

There is a mention of a picture attached to the original post, which may provide additional context for the charge positions. The original poster's notation has been noted as potentially non-standard, which could affect clarity in communication.

bbraves7
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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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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.)
 
You need F1 + F2 = 0.

So that F1 = ‒ F2

That should take care of your signs.
 
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.
 

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