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Question

Physics Question on Electric Charge

A charged particle of mass 0.003 g is held stationary in space by placing it in a downward direction of electric field of 6×104 6 \times 10^4 N/C. Then the magnitude of charge is

A

5×10−45\times 10^{ - 4} C

B

5×10−105\times 10^{ - 10} C

C

5×10−65\times 10^{ - 6} C

D

5×10−95\times 10^{ - 9} C

Answer

5×10−105\times 10^{ - 10} C

Explanation

Solution

By using qE = mg q = mgE\frac{ mg }{ E } = 3×10−6×106×104=30×10106\frac{ 3 \times 10^{ - 6} \times 10}{ 6 \times 10^4} = \frac{ 30 \times 10^{ 10}}{ 6 } = 5×10−105 \times 10^{ - 10} C