Coulomb's Law, net electrostatic force

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SUMMARY

The discussion focuses on 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. Using Coulomb's Law, the forces exerted by each charge were calculated, yielding values of approximately 14.04 N and -983.01 N, respectively. The total force was incorrectly summed, leading to a magnitude of 997.05 N, which is not the correct approach for determining the net force on the 2 μC charge.

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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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