Electrostatics Question 448
Question: 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]
Options:
A) $ 5\times {{10}^{-4}}C $
B) $ 5\times {{10}^{-10}}C $
C) $ -18\times {{10}^{-6}}C $
D) $ -5\times {{10}^{-9}}C $
Show Answer
Answer:
Correct Answer: B
Solution:
By using therefore $ Q=\frac{mg}{E}=\frac{0.003\times {{10}^{-3}}\times 10}{6\times 10^{4}} $ = $ 5\times 10^{-10}C $
BETA
