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

  • question_answer
    A charged water drop whose radius is \[0.1\,\mu m\] is in equilibrium in an electric field. If charge on it is equal to charge of an electron, then intensity of electric field will be \[(g=10\,m{{s}^{-1}})\]                                          [RPET 1997]

    A)                    \[1.61\,N/C\]               

    B)            \[26.2\,N/C\]

    C)                    \[262\,N/C\]

    D)                                      \[1610\,N/C\]

    Correct Answer: C

    Solution :

                         In balance condition \[QE=mg=\left( \frac{4}{3}\pi {{r}^{3}}\rho  \right)\,g\] Þ \[E=\frac{4\times (3.14)\,{{(0.1\times {{10}^{-6}})}^{3}}\times {{10}^{3}}\times 10}{3\times 1.6\times {{10}^{-19}}}\]\[=262\,N/C\]


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