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Q.

The force between two electrons when placed in air is equal to 0.5 times the weight of an electron. Find the distance between two electrons (mass of electron= 9.1×10^–31kg)

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a

720 m

b

72 cm

c

7.2 m

d

72 m

answer is A.

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Detailed Solution

$\begin{array}{l} F = 0.5 mg = 0.5 \times 9.1 \times 10^{- 31} \times 9.8 N \\ 4.9 \times 9.1 \times 10^{- 31} = 9 \times 10^{9} \times \frac{1.6 \times 10^{- 19} \times 1.6 \times 10^{- 19}}{r^{2}} \\ \Rightarrow r^{2} = \frac{1.6 \times 10^{- 19} \times 1.6 \times 10^{- 19}}{4.9 \times 10^{- 31}} \times 10^{9} \\ \Rightarrow r = 0.72 \times 10^{- 19 + 20} = 7.2 m \end{array}$

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