Coulomb's Law, net electrostatic force

  • Thread starter nn3568
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  • #1
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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?
 

Answers and Replies

  • #2
Doc Al
Mentor
45,245
1,591
(8.9875e9 * 5e-6 * -9e-6) / (0.02^2) = -983.0078125
You want the force on the 2μC charge.
 
  • #3
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Thank you so much!
 

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