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}} $



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