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A charged oil drop is suspended in a uniform field of $3 \times 10^4 V/m$ so that it neither falls nor rises. The charge on the drop will be (take the mass of the charge = $9.9 \times 10^{-15} kg$ and $g = 10\;m/s^2$)


( A ) $1.6 \times 10^{-18} C$
( B ) $4.8 \times 10^{-18} C$
( C ) $3.3 \times 10^{-18} C$
( D ) $3.2 \times 10^{-18} C$

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