Sixty four conducting drops each of radius 0 . 02 m and each carrying a charge of 5 μ C are combined to form a bigger drop. The ratio of surface density of bigger drop to the smaller drop will be

Step-by-step solution

Total volume will be constant. Therefore, n 4 π 3 r 3 = 4 π 3 R 3 ⇒ 64 1 3 r = R ⇒ R = 4 r Final surface charge density σ ' = n σ 0 4 π r 2 4 π R 2 = 64 × σ 0 r 2 16 r 2 ⇒ σ ' σ 0 = 4 1