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

  • question_answer
    An oil drop having charge \[2e\] is kept stationary between two parallel horizontal plates 2.0 cm apart when a potential difference of 12000 volts is applied between them. If the density of oil is\[900kg/{{m}^{3}}\], the radius of the drop will be [AMU 1999]

    A)            \[2.0\times {{10}^{-6}}m\]

    B)                                      \[1.7\times {{10}^{-6}}m\]

    C)            \[1.4\times {{10}^{-6}}m\]

    D)                                      \[1.1\times {{10}^{-6}}m\]

    Correct Answer: B

    Solution :

                         In equilibrium QE = mg  Þ \[Q.\frac{V}{d}=mg=\left( \frac{4}{3}\pi {{r}^{3}}\rho  \right)\,g\] Þ \[2\times 1.6\times {{10}^{-19}}\times \frac{12000}{2\times {{10}^{-2}}}=\frac{4}{3}\pi {{r}^{3}}\times 900\times 10\] Þ\[~r=\text{ }1.7\times {{10}^{6}}m\]


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