question_answer

    In Millikans oil drop experiment, an oil drop of mass\[16\times {{10}^{-6}}kg\]is balanced by an electric field of\[{{10}^{6}}V/m\]. The charge in coulomb on the drop is (assuming\[g=10m/{{s}^{2}}\])

A)  \[6.2\times {{10}^{-11}}\]           

B)  \[16\times {{10}^{-9}}\]

C)  \[16\times {{10}^{-11}}\]                            

D)  \[16\times {{10}^{-13}}\]

Correct Answer: C

Solution :

                Robert Millikan performed the experiment to determine the charge on an electron. When a drop is suspended, its weight mg is exactly equal to the electric force applied qE, where E is electric filed, q the charge, m the mass of drop and g the acceleration due to gravity. Hence, solving for q, we get                 \[q=\frac{mg}{E}\] Given,\[m=16\times {{10}^{-6}}kg,\text{ }g=10\text{ }m/{{s}^{2}},\] \[E={{10}^{6}}V/m\] Note: With the help of this experiment Millikan determined that there was a smallest unit charge or that charge is quantized. He received the noble prize for his work.