Question

Suppose the spheres A and B in Exercise 1.12 have identical sizes.A third sphere of the same size but uncharged is brought in contactwith the first, then brought in contact with the second, and finallyremoved from both. What is the new force of repulsion between Aand B?

Answer 1

Given: The distance between spheres is 50 cm, initial charges on each sphere A and B is 6.5× 10 −7  C.

When sphere C is bought in contact with A then the charge qwill be equally distributed in A and C, thus q 2 amount of charge will be transferred from A to C. So charges on sphere Aand Cwill be q 2 .

When sphere C is bought in contact with sphere B, total charge ( q+ q 2 ) will be divided into two equal halves. So the charge on sphere B and C will be 3q 4 .

The force of repulsion between sphere A and B is given as,

F= 1 4π ε 0 × q A q B r 2 = 1 4π ε 0 ⋅ q 2 ⋅ 3q 4 r 2 = 1 4π ε 0 ⋅ 3 q 2 8 r 2

Where, q A is the charge on sphere A, q B is the charge on the sphere B and r is the distance between the two spheres.

By substituting the values in the above equation, we get

F=9× 10 9 × 3 ( 6.5× 10 −7 ) 2 8 ( 0.5 ) 2 =5.7× 10 −3  N

Thus, the force of repulsion between two spheres will be 5.7× 10 −3  N.