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Q.

A charged water drop whose radius is $0 .1 μm$ is in equilibrium in an electric field. If charge on it is equal to charge of an electron, then intensity of electric field will be $(g = 10 {ms}^{- 2})$

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a

262 N/C

b

1.61 N/C

c

26.2 N/C

d

1610 N/C

answer is C.

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Detailed Solution

In balance condition $QE = mg = (\frac{4}{3} {πr}^{3} ρ) g$

$\Rightarrow E = \frac{4 \times (3 .14) {(0 .1 \times 10^{- 6})}^{3} \times 10^{3} \times 10}{3 \times 1 .6 \times 10^{- 19}} = 262 N / C$

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