Suppose the spheres A and B in Exercise - 12 have identical sizes. A third sphere of the same size but uncharged is brought in contact with the first, then brought in contact with second and finally removed from both. What is the new force of repulsion between A and B if the distance between them is 50 cm?

Explanation

Let us assume that both the spheres A and B carry charge Q and are kept at a separation r. The electric force between the two spheres can be written as follows:
`F_1=1/(4piepsilon_0)Q^2/r^2 =1.521xx10^(-2)` N (previous question )
We should be aware of the fact that when two identical metal objects are brought in contact with each other, they redistribute their total charge equally. Let us call the third uncharged sphere C. At first C is brought in contact with A. The new value of charges on both A and C become Q/2. When sphere C is brought in contact with B, the total charge 3Q/2 is divided equally between them. So the new value of charges on B and C becomes 3Q/4.
The new magnitude of the electrostatic force between the charges can be written as follows:
`F_2=1/(4piepsilon_0)((Q//2)(3Q//4))/r^2=3/8F_1`
`F_2=3/8xx1.521xx10^(-2)=5.7xx10^(-3)` N