Question

A charged particle of mass 0.003 g 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 charge is

• A. $ 5\times 10^{ - 4} $ C
• B. $ 5\times 10^{ - 10} $ C
• C. $ 5\times 10^{ - 6} $ C
• D. $ 5\times 10^{ - 9} $ C

Answer 1

Answer: (B)

By using qE = mg
q = $ \frac{ mg }{ E } $
= $ \frac{ 3 \times 10^{ - 6} \times 10}{ 6 \times 10^4} = \frac{ 30 \times 10^{ 10}}{ 6 } $
= $ 5 \times 10^{ - 10} $ C