JEE Main & Advanced
Physics
Electrostatics & Capacitance
Question Bank
Electric Field and Potential
question_answer
A charged particle of mass 0.003 gm is held stationary in space by placing it in a downward direction of electric field of \[6\times {{10}^{4}}N/C\]. Then the magnitude of the charge is [Orissa JEE 2002]
A) \[5\times {{10}^{-4}}C\]
B) \[5\times {{10}^{-10}}C\]
C) \[-18\times {{10}^{-6}}C\]
D) \[-5\times {{10}^{-9}}C\]
Correct Answer:
B
Solution :
By using Þ \[Q=\frac{mg}{E}=\frac{0.003\times {{10}^{-3}}\times 10}{6\times {{10}^{4}}}\]= \[5\times {{10}^{10}}C\]