Coulomb's Law, net electrostatic force

AI Thread Summary
The discussion revolves around calculating the net electrostatic force on a third particle with a charge of 2 μC, positioned at -2 cm, due to two other charges: -9 μC at 8 cm and 5 μC at 6 cm. The user applied Coulomb's Law but received an incorrect total force. The calculations for the forces between the charges were performed, but the user misunderstood the direction of the forces and the resultant magnitude. The correct approach involves ensuring that the signs of the forces are accurately considered based on the nature of the charges involved. The final magnitude of the net electrostatic force should be verified for accuracy.
nn3568
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Homework Statement


A particle with charge −9 μC is located on the x-axis at the point 8 cm, and a second particle with charge 5 μC is placed on the x-axis at 6 cm. The Coulomb constant is 8.9875 × 109 N · m2/C2. What is the magnitude of the total electrostatic force on a third particle with charge 2 μC placed on the x-axis at −2 cm? Answer in units of N.


Homework Equations


Coulomb's Law
fe = (kq1q2)/(d2)


The Attempt at a Solution


(8.9875e9 * 2e-6 * 5e-6) / (0.08^2) = 14.04296875
(8.9875e9 * 5e-6 * -9e-6) / (0.02^2) = -983.0078125
14.04296875 + -983.0078125 = -997.0507813
magnitude 997.0507813

Why is this wrong? What can I do to make it right?
 
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nn3568 said:
(8.9875e9 * 5e-6 * -9e-6) / (0.02^2) = -983.0078125
You want the force on the 2μC charge.
 
Thank you so much!
 
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