Three Charges in a Triangle problem

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The problem involves three identical charged balls forming an equilateral triangle, each experiencing electrostatic repulsion and gravitational forces. The forces acting on each ball must be analyzed, considering both the horizontal and vertical components. The symmetry of the setup allows for simplification, focusing on one charge and its interactions with the others. The correct approach involves using the formula F=kq^2/r^2 for the electrostatic force while also accounting for gravitational force and tension in the strings. A comprehensive understanding of all forces and their directions is essential for solving the problem accurately.
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Homework Statement


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Three 3.0 g balls are tied to 80-cm-long threads and hung from a single fixed point. Each of the balls is given the same charge q. At equilibrium, the three balls form an equilateral triangle in a horizontal plane with 20 cm sides.

Homework Equations



F=kq1q2/r^2

F (of 1 on 2) = F (of 2 on 1)

The Attempt at a Solution



Since they all have the same charge, I figured that the forces would be equal at each point. All points are in the same plane, so I assumed they only have x and y components. I tried using F=kq1q2/r^2=kq2q3/r^2=kq1q3/r^2 but it ended up a huge mess with all the charges canceling. But they all have the same charge? So should I use F=k2q/r^2 and do something with that?
 
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The forces are equal in magnitude, but not in direction.
Due to the symmetry, it is sufficient to consider the position of a single charge. What are all forces acting on it, including their directions?

What is "k2q"?
 
Well on each charge, since they are positioned in an equilateral triangle, they all have 2 forces acting on them, both at 30 degrees (I think?) and one would be at +30 degrees and one would be at -30 degrees.

It was supposed to be kq^2, from the F=(k*q1*q2)/r^2 equation, but since q1=q2 we could just use q^2.
 
Panda_Doll said:
and one would be at +30 degrees and one would be at -30 degrees.
Okay.
Panda_Doll said:
It was supposed to be kq^2, from the F=(k*q1*q2)/r^2 equation, but since q1=q2 we could just use q^2.
Right, but that is just the magnitude of the charge.
 
Sure, so I have the magnitude. And each charge is sitting in a triangle with sides 20cm. So, if we put the first point at say, what I'll make my origin, then I have another point 20 cm to the right, so now its coordinates are (20,0) and the third charge would be at about... (10, 17.3)? Okay. So how do their relative positions help?
 
mfb said:
What are all forces acting on it, including their directions?
You listed the two from the electrostatic repulsion, but there are two more. Those objects are not floating in free space.
 
OH! Gravity and the tension in the strings!
 
Right. And if you add all together, ...
 
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