# Charges in a circle.

Tags:
1. Apr 27, 2017

### Buffu

1. The problem statement, all variables and given/known data

Two charges placed at circumference of a circle of radius $a$ at $\pi/2$ from each other. Find the relative magnitude of third charge kept on the circumference such that the system is at equilibrium.

2. Relevant equations

Coulombs law.

3. The attempt at a solution

Let $Q$ be the unknown charge and $x$ be the length of equal sides of the triangle . then I get,

By coulombs law,

$$F_{BA} = {-Qq \over x^2 } \left(\cos(135/2)\hat i + \hat j \sin(135/2) \right)$$

$$F_{CA} = {-q^2\over 4x^2(\cos (135/2))}(\hat i)$$

Now the force on $A$ should be zero for the system to be in equilibrium but clearly there is a net force in -y direction that is not balanced by anything, so how the system is in equilibrium for any value of $Q$ ?

Where did I go wrong ?

Last edited: Apr 27, 2017
2. Apr 27, 2017

### kuruman

What is the value of $x$?

3. Apr 27, 2017

### Buffu

Numerical value or what it represent in equations ?

4. Apr 27, 2017

### kuruman

It's OK, I figured it out from the way you used it. I think you misunderstood that the problem requires the charges to be constrained on the circle. Imagine three charged beads on a wire loop, for example. As you correctly discovered, the force on any one bead cannot be zero. However, if you constrained the charges on the (non-deformable) circle, equilibrium is reached if the net force on any one bead is directed radially out. Therefore, the task before you is to balance the tangential forces.

5. Apr 27, 2017

### kuruman

Upon doing the problem, I think you will benefit from expressing the two charge-to-charge distances in terms of the radius of the circle. To do that, draw the inner right triangle (with its apex at the center) and then use the law of sines.

6. Apr 27, 2017

### Buffu

Thank you, I will try and tell you the results.