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 $g = 10\,m/{s^2}$)
$1.6 \times {10^{ - 18}}\,C$
$3.2 \times {10^{ - 18}}\,C$
$3.3 \times {10^{ - 18}}\,C$
$4.8 \times {10^{ - 18}}\,C$
Solution
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