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} $
