Question

Point charges + 4q, -q and +4q are kept on the X-axis at points x = 0, x = a and x = 2a respectively,Then What will happen?

1. Only - q is in stable equilibrium
2. None of the charges is in equilibrium
3. All the charges are in unstable equilibrium
4. All the charges are in stable equilibrium.

Answer 1

`3.`

Explanation:

Let's see what is the force acting on charge `-q`, it is `(-k 4q^2)/a^2` by both the `4q` charges but on opposite direction,so `-q` is at equilibrium.

Again, on each of `=4q` charge,force acting is `+k (16q^2)/(2a)^2 - k((4q^2)/a^2)=k(4q^2)/a^2 - k(4q^2)/a^2=0`

So,they are at equlibrium as well,

Now,if we move charge `-q` a bit to either side,due to decreased distance,the side on to which we pushed,the force acting on that direction will increase (as `F prop 1/r^2`),so the charge `-q` will start moving towards that charge as on moving further towards it,the force will increase,so `-q` is in unstable equilibrium.

Now,doing the same thing, for `4q`,we can clearly call it is as well in unstable equilibrium,as the magnitude of the two other charges are different,it was made stable by balancing the charge inequality with similar inequality of distance between them.