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12th Class Physics Electric Charges and Fields Question Bank Case Based (MCQs) - Electric Charges and Fields

  • question_answer
    A negatively charged oil drop is prevented from falling under gravity by applying a vertical electric field \[V\,{{m}^{-1}}\]. If the mass of the drop is \[1\centerdot 6\times {{10}^{-3}}g\], the number of electrons carried by the drop is \[\left( g=10\text{ }m{{s}^{-2}} \right)\]

    A) \[{{10}^{18}}\]

    B) \[{{10}^{15}}\]

    C) \[{{10}^{12}}\]

    D) \[{{10}^{9}}\]

    Correct Answer: C

    Solution :

    For the drop to be stationary, Force on the drop due to elecrtic field = Weight of the drop \[qE\text{ }=\text{ }mg\] \[q=\frac{mg}{E}=\frac{1\,.\,6\times {{10}^{-6}}\times 10}{100}\] \[=\,1\,.\,6\times {{10}^{-7}}C\] Number of electrons carried by the drop is \[n=\frac{q}{e}=\frac{1.6\times {{10}^{-7}}C}{1\,.\,6\times {{10}^{-19}}C}={{10}^{12}}\]


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