Electrostatic force problem in Newtons

AI Thread Summary
The discussion focuses on calculating the net electrostatic force on a charge located at x = +3 m in a system with three fixed charges. The user initially calculated the forces between the charge at x and the origin, as well as between the charge at x and the charge on the y-axis, using Coulomb's law. They reported a net force of -0.3287 N and speculated that the direction might be 45 degrees. However, other participants emphasized the importance of breaking the forces into x and y components to accurately determine the total net force. The conversation highlights the need for careful vector analysis in electrostatic force problems.
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Three charges are fixed to an xy coordinate system. A charge of +14 uC is on the y-axis at y= +3 m. A charge of -14 uC is at the origin. Lastly a charge of +47 is on the x-axis at x= +3 m. Determine the magnitude and direction of the net electrostatic force on the charge at x= +3 m. Specify the direction relative to the -x axis.


Magnitude is in Newtons; direction is in degrees.

State whether the direction is above or below the -x axis.
 
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Welcome to the forums. There's this policy that you have to show your own work when you ask for help (https://www.physicsforums.com/showthread.php?t=94379): So, what have you tried?
 
If tried to get the force between points x any y, and also between x and the origin, by using the formula F= Kq1q1)/r^2. The radius between x and the origin I assumed is 3, and by using the pythagoreon theorem, I used the square root of 18 for the radius between x and y. The force between x and y came out to be .3286 N. The force between x and the origin came out to be -.65727 N. I added them together to get the net electrostatic force, which came out to be -.3287 N. I'm not sure if I did this problem right. The only thing I could think to come up with for the direction is 45 degrees.
 
Force is a vector, and in this case the vectors are not pointing in the same direction. Break the vectors into x and y components.
 
I did cos45= (x/.3286) to get the magnitude of the force between x and y going in the same direction as the force between x and the origin. Does addin them together then give me the correct net force?
 
Supposing I understood what you were about to do: yes, you get the net force in x direction. To get the total net force though, you have to take the y component into account as well.
 
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