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Coulomb's Law, net electrostatic force

by nn3568
Tags: coulomb, electrostatic, force
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nn3568
#1
Mar28-09, 07:10 PM
P: 14
1. The problem statement, all variables and given/known data
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.


2. Relevant equations
Coulomb's Law
fe = (kq1q2)/(d2)


3. 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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Doc Al
#2
Mar28-09, 07:23 PM
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Quote Quote by nn3568 View Post
(8.9875e9 * 5e-6 * -9e-6) / (0.02^2) = -983.0078125
You want the force on the 2μC charge.
nn3568
#3
Mar28-09, 08:55 PM
P: 14
Thank you so much!


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