A charged oil drop is suspended in a uniform filed 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 $= 9.9 \times 10^{- 15} k g$ and $g = 10 m / s^{2}$ )

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$3.3 \times 10^{- 18} C$

$1.6 \times 10^{- 18} C$

$4.8 \times 10^{- 18} C$

$3.2 \times 10^{- 18} C$

answer is C.

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Detailed Solution

Given, Electric field, $E = 3 \times 10^{4}$
Mass of the drop, $m = 9.9 \times 10^{- 15} k g$
At equilibrium coulomb force on drop balances weight of drop. $q E = m g$

$\Rightarrow q = \frac{m g}{E} \Rightarrow q = \frac{9.9 \times 10^{- 15} \times 10}{3 \times 10^{4}} = 3.3 \times 10^{- 18} C$

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