Electrostatics Question 123

Question: 64 small drops of mercury, each of radius r and charge q 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 [KCET 2002]

Options:

A) 1 : 64

B) 64 : 1

C) 4 : 1

D) 1 : 4

Show Answer

Answer:

Correct Answer: D

Solution:

$ \frac{{\sigma _{small}}}{{\sigma _{Big}}}=\frac{q}{Q}\times \frac{R^{2}}{r^{2}}=\frac{q}{(nq)}\times \frac{{{({{n}^{1/3}}r)}^{2}}}{r^{2}} $

$ ={{n}^{-1/3}}={{(64)}^{-1/3}}=\frac{1}{4} $



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