2. Questions Bank
  3. JEE Main & Advanced
  4. Physics
  5. Electrostatics & Capacitance

JEE Main & Advanced Physics Electrostatics & Capacitance Question Bank Mock Test - Electrostatics

  • question_answer
    A charged oil drop is suspended in a uniform field of \[3\times {{10}^{4}}V/m\] so that it neither falls nor rises. The charge on the drop will be (take the mass of the charge as \[9.9\times {{10}^{-15}}kg\,and\,g\,as\,10m/{{s}^{2}}\]

    A) \[3.3\times {{10}^{-18}}C\]     

    B) \[3.2\times {{10}^{-18}}C\]

    C) \[1.6\times {{10}^{-18}}C\]     

    D) \[4.8\times {{10}^{-18}}C\]

    Correct Answer: A

    Solution :

    [a] In equilibrium, \[qE=mg\] \[\Rightarrow q=3.3\times {{10}^{-18}}C\]


You need to login to perform this action.
You will be redirected in 3 sec spinner