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Q.

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)

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a

10 V/m, downward

b

0.1 V /m, upward

c

0.1 V/m, downward

d

10 V/m, upward

answer is A.

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Detailed 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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