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Q.

A charged oil drop is suspended in a uniform field of 3x10 ⁴ v/m so that it neither falls nor rises .The charge on the drop will be (take mass of the charge =9.9 x10-¹⁵kg and g = 10m /s²

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a

4.8 x10^-18C

b

3.2 x10 ^-18 C

c

1.6 x 10^-18C

d

3.3 x 10^-18 C

answer is A.

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Detailed Solution

In equilibrium position m g = q E
or

$q = \frac{mg}{E} = \frac{(9 \cdot 9 \times 10^{- 15}) \times 10}{3 \times 10^{4}} = 3 \cdot 3 \times 10^{- 18} C$

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