Electrostatics Question 42
Question: $ N $ identical spherical drops charged to the same potential $ V $ are combined to form a big drop. The potential of the new drop will be [MP PMT 1990, 2001; KCET 2000; Kerala PET 2002]
Options:
A) $ V $
B) $ V/N $
C) $ V\times N $
D) $ V\times {{N}^{2/3}} $
Show Answer
Answer:
Correct Answer: D
Solution:
If the drops are conducting, then $ \frac{4}{3}\pi R^{3}=N( \frac{4}{3}\pi r^{3} ) $
therefore $ R={{N}^{1/3}}r $ .
Final charge Q = Nq So final potential $ V=\frac{Q}{R} $
$ =\frac{Nq}{{{N}^{1/3}}r}=V\times {{N}^{2/3}} $
BETA
