Question

Two identical metal balls with charges +2q and -q are separated by some distance and exert a force F on each other. They are joined by a conducting wire which is then removed. What will be the force between them?
Please solve with proper details

Answer 1

After joining and removing them... the charge on each metallic ball is q/2.
Now, 
F=(q/2*q/2)/4π∈r^2
  =(q^2)/16∈r^2

Answer 2

Answer:

The new force between them will be $F'=\dfrac{F}{-8}$

Explanation:

Given that,

First charge$q_{1} = +2 q$

Second charge$q_{2} = -q$

Let the distance between the charges is r.

The force between the charges is

$F= \dfrac{k(-q)\times(2q)}{r^{2}}$

They are joined by a conducting wire which is then removed.

When the charges on the metal ball will be same after contact.

The charge will be

$Q= \dfrac{q_{1}+q_{2}}{2}$

$Q=\dfrac{2q-q}{2}$

$Q=\dfrac{q}{2}$

The new force between them

$F'= \dfrac{k(Q)^2}{r^2}$

$F'=\dfrac{k(\dfrac{q}{2})^2}{r^2}$

The ratio of the F and F'

$\dfrac{F}{F'}=\dfrac{\dfrac{k(-q)\times(2q)}{r^{2}}}{\dfrac{k(\dfrac{q}{2})^2}{r^2}}$

$\dfrac{F}{F'}=\dfrac{-8}{1}$

$F'=\dfrac{F}{-8}$

Negative sign indicates the attraction between them.

Hence, The new force between them will be $F'=\dfrac{F}{-8}$