First slide

Electric field

Question

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

Easy

A

3.3 x 10^-18 C
B

3.2 x10 ^-18 C
C

1.6 x 10^-18C
D

4.8 x10^-18C

Solution

In equilibrium position m g = g 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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