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JEE Main & Advanced Physics Electrostatics & Capacitance Question Bank Electric Field and Potential

  • question_answer
    In Millikan's oil drop experiment an oil drop carrying a charge Q is held stationary by a potential difference \[2400\,V\] between the plates. To keep a drop of half the radius stationary the potential difference had to be made \[600\,V\]. What is the charge on the second drop            [MP PET 1997]

    A)                    \[\frac{Q}{4}\]

    B)                                      \[\frac{Q}{2}\]

    C)                    \[Q\]

    D)                                      \[\frac{3Q}{2}\]

    Correct Answer: B

    Solution :

                         In balance condition Þ \[QE=mg\] Þ \[Q\frac{V}{d}=\left( \frac{4}{3}\pi {{r}^{3}}\rho  \right)\,g\] Þ \[Q\propto \frac{{{r}^{3}}}{V}\] Þ \[\frac{{{Q}_{1}}}{{{Q}_{2}}}={{\left( \frac{{{r}_{1}}}{{{r}_{2}}} \right)}^{3}}\times \frac{{{V}_{2}}}{{{V}_{1}}}\] Þ \[\frac{Q}{{{Q}_{2}}}={{\left( \frac{r}{r/2} \right)}^{3}}\times \frac{600}{2400}=2\] Þ Q2 = Q / 2


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