Magnitude of Force on Q1 with Three Point Charges

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The discussion revolves around calculating the net force on a point charge Q1 due to two other charges, q1 and q2, using Coulomb's law. The user initially calculated the forces from both charges but struggled to obtain the correct net force when three charges are involved. They expressed confusion over the signs of the forces and how to determine the direction of the resultant force. The user noted that attractive forces from opposite charges lead to a positive direction, while repulsive forces from like charges result in a negative direction. Overall, the thread highlights the complexities of force calculations in multi-charge systems and the importance of understanding vector directions.
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Homework Statement


A point charge q1 = -3.00 µC is situated at the origin. Another point charge q2 = 5.00 µC is located at the location x = 1.00 m. If another charge Q1 = 10.0 µC is placed at x = -0.100 m, what is the magnitude of the force on it?


Homework Equations


F=[k(q1)(q2)]/r^2
E=F/Q (I'm not sure if this one is relevant, considering how I have no idea how to do this).



The Attempt at a Solution



For two particles, I did the same thing and it worked. However, when I tried with 3 particles, obviously it didn't give me the right answer or else I wouldn't be here?

F1= [(9x10^9)(10x10^-6)(-3x10^-6)]/(.1^2) = -27 i-hat
F2= [(9x10^9)(10x10^-6)(5x10^-6)]/(1.1^2) = -.37 i-hat
Fnet = -27 i-hat + -.37 i-hat = -27.37 i-hat.

Meaning the magnitude would be 27.37 N, and it would be going in the negative i-hat direction. Of course, that was not the right answer, and I was really hoping for some help because I have no clue what to do.

Thank you!
 
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Welcome to PF.

Just curious how both forces ended up with the same sign?
 
That's pretty interesting actually. I think what I did was that the F1 ended up being negative because of the -3x10^-6 force and because the i-hat is positive for that, so that one stayed the same.

As for F2, I wasn't exactly sure about this one because I have a problem with figuring out when to use - or + but. I figured the i-hat was negative because of a method our teacher taught us. But I'm not really sure if the i-hats even change based on direction because of a previous example he showed us. Having three charges just really throws me off. I feel it shouldn't be much different than two...
 
If you have a + & - sign then the charges are attractive and that means the force is toward the charge that you are figuring the force for.

In this case both q1 and q2 charges affecting the force are located in the + X direction, so a + Force toward those charges (if -) on the Q1 will be positive X.

The opposite is true for the same sign charges. Repulsive force at Q1 will be - X.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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