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Question: 64 small drops of mercury, each of radius \(r\) and charge \(q\) coalesce to form a big drop. The ra...

64 small drops of mercury, each of radius rr and charge qq coalesce to form a big drop. The ratio of the surface density of charge of each small drop with that of the big drop is

A

1 : 64

B

64 : 1

C

4 : 1

D

1 : 4

Answer

1 : 4

Explanation

Solution

σSmallσBig=q/4πr2Q/4πR2=(qQ)(Rr)2;\frac{\sigma_{Small}}{\sigma_{Big}} = \frac{q/4\pi r^{2}}{Q/4\pi R^{2}} = \left( \frac{q}{Q} \right)\left( \frac{R}{r} \right)^{2}; since R = n1/3r and Q = nq

So σSmallσBig=1n1/3\frac{\sigma_{Small}}{\sigma_{Big}} = \frac{1}{n^{1/3}} ⇒ σSmallσBig=14\frac{\sigma_{Small}}{\sigma_{Big}} = \frac{1}{4}