The force between two small charged spheres having charges of $1 \times 10^{- 7} C$ and $2 \times 10^{- 7} C$ placed $20 c m$ apart in air is

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$4.5 x 10^{- 2} N$

$4.5 x 10^{- 3} N$

$5.4 x 10^{- 2} N$

$5.4 x 10^{- 3} N$

answer is B.

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Detailed Solution

Here, $q_{1} = 1 \times 10^{- 7} C, q_{2} = 2 \times 10^{- 7} C, r = 20 c m = 20 \times 10^{- 2} m$

As $F = \frac{q_{1} q_{2}}{4 π ε_{0} r^{2}} = \frac{9 \times 10^{9} \times 1 \times 10^{- 7} \times 2 \times 10^{- 7}}{{(20 \times 10^{- 2})}^{2}} = 4.5 \times 10^{- 3} N$

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