Question

two fixed point charges 4q. and 2q separated by a distance x where should the third point charge q is placed for it to be in equilibrium.

Answer 1

hope it will help you.....................

Answer 2

Answer:

The third charge should be placed at a distance  $\frac{x}{1+\sqrt{2} }$  for it to remain in equilibrium.

Explanation:

Given that there are 2 fixed charges of magnitude 2q and 4q which are fixed at a distance of x.

Since both charges have the same polarity so the third charge would be somewhere in between of both the charges where net electric field because of both the given charges is 0.

Let the net electric field be 0 at a point which is at a distance 'y' from 2q charge. So its distance from 4q charge is equal to (x-y)

Net electric field must be equal to 0 at that point,

$F_{net} =\frac{K2q}{y^{2} }- \frac{K4q}{(x-y)^{2} } =0$

On solving for y we get,

$y=\frac{x}{1+\sqrt{2} }$

Hence, the third charge should be placed at a distance  $\frac{x}{1+\sqrt{2} }$   for it to remain in equilibrium.