Help with Electric charges using coulomb's law

AI Thread Summary
The discussion focuses on calculating the net force exerted on a third charge (q3=+5.0nC) by two point charges (q1=-2.5nC and q2=+3.5nC) using Coulomb's law. The participants emphasize the importance of treating forces as vectors and removing absolute values to account for direction. The distances from q3 to q1 and q2 are noted as critical for the calculations. The formula for force is provided, with the constant k specified as 9x10^9 N*M^2/C^2. The conversation seeks clarification on applying these principles correctly to find the net force on q3.
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Homework Statement



Two point charges are located on the y-axis as follows: charge q1=-2.5nC at y=-0.6m, and charge q2=+3.5nC at the origin. What is the net force (x, y components) exerted by these two charges on a third charge q3=+5.0nC located at y=-0.4m?

Homework Equations


F=k (abs(q1*q2)/r^2)
k is constant k=9x10^9 N*M^2/C^2

The Attempt at a Solution


Like Charges repel, opposite charges attract.
q2 is at the origin, q1 is -0.6m far from q2. q3 -0.4m far from q2.
Not sure if i quite got it right...
if any1 can help me out...thx a lot for your time.
 
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im pretty sure you would do Fnet = F1 + F2

where F1 is the distance from the third charge to one charge

F2 is the distance from the third charge to the second point charge

oh and the equation you gave is the magnitude
remove the absolute value and treat each force as a vector
 
SSJ2 said:
im pretty sure you would do Fnet = F1 + F2

where F1 is the distance from the third charge to one charge

F2 is the distance from the third charge to the second point charge

oh and the equation you gave is the magnitude
remove the absolute value and treat each force as a vector

so what is gne be net force using the coulombs law...
Fq2 to 3 =(9*10^9)(3.5nC)(5.0nC)/-0.4^2
Fq1 to 3 = (9*10^9)( -2.5nC)(5.0nC)/-0.6^2
 
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