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JEE Main & Advanced Physics Electrostatics & Capacitance Question Bank Capacitance

  • question_answer
    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]

    A)            1 : 64                                        

    B)            64 : 1

    C)            4 : 1                                          

    D)            1 : 4

    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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