Two spheres of electric charges + 2 n C and - 8 n C are placed at a distance d apart. If they are allowed to touch each other, what is the new distance between them to get a repulsive force of same magnitude as before?

Step-by-step solution

In the first condition, Given, q 1 = + 2 n C = 2 × 10 - 9 q 2 = - 8 C = - 8 × 10 - 9 C F = k q 1 q 2 r 2 F = k 2 × 10 - 9 × 8 × 10 - 9 d 2 F = k 16 × 10 - 18 d 2 .....(i) In the second condition, F 1 = k q 1 q 2 d ′ 2 After touching of sphere to each other the total charge = 2 - 8 = - 6 n C charge on each sphere = 3 n C = 3 × 10 - 9 C F 1 = k 3 × 10 - 9 3 × 10 - 9 d ′ 2 F 1 = k 9 × 10 - 18 d ′ 2 ...(ii) From Equations. (i) and (ii), we get k 16 × 10 - 18 d 2 = k 9 × 10 - 18 d ′ 2 d 2 d ′ 2 = 16 × 10 - 18 9 × 10 - 1