Question

An electric charge q exerts a force F on a similar electric charge q separated by a distance r. A third charge q/4 is placed mid way between the two charges . Now, the force F will

Answer 1

Answer:

The force will remain the same.

Explanation:

The force between the two charges is given by Coulomb's Law

Thus

$F=k\frac{q\times q}{r^2}$

$\implies F=k\frac{q^2}{r^2}$

Since charges are similar, this force will be that of repulsion and will act away from the charges

A third similar charge q/4 is placed at distance 1/2 between them

Now due to this charge, if force on either of the charges q is F' then

$F'=k\frac{q\times q/4}{(r/2)^2}$

$\implies F'=k\frac{q^2/4}{(r^2/4}$

$\implies F'=k\frac{q^2}{(r^2}$

$\implies F'=F$

Thus total force on the charge will be $F+F=2F$

Therefore, the net force F wil be doubled.

However if we are only interested in the force between the two charges q then it will remain the same because the charges will still exert the force F between them.

Hope this answer is useful.