A drop of 10-6 kg water carries 10-6 C charge. What electric field should be applied to balance its weight? (assume, g = 10 ms-2)

A drop of 10^-6 kg water carries 10^-6 C charge. What electric field should be applied to balance its weight? (assume, g = 10 ms^-2)

A
10 V/m, upward
B
10 V/m, downward
C
0.1 V/m, downward
D
0.1 V /m, upward

Solution:

By using, QE = mg

$\Rightarrow E = \frac{m g}{Q} = \frac{10^{- 6} \times 10}{10^{- 6}} = 10 V / m, upward because$ charge is positive.

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