A charged particle of mass 0.003g is held stationary in a space by placing it in a downward direction of electric field of \(6×10^4NC^{-1}\). Then, the magnitude of charge is:

(A) \(5×10^4 C\)

(B) \(5×10^{-10}C\)

(C) \(5×10^{-6} C\)

(D) \(5×10^{-9} C\)

The correct option of this question is (B)

Solution :

We know that,

\(qE=mg\)

\(So, q = \frac{mg}{E}\)

\(\\=\frac{3×10^-6×10}{6×10^4}\)

\(=\frac{30×10^{-10}}{6}\)

\(=5×10^{-10}C\)

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