If two charges of 1 coulomb are placed 1 km apart, the force between them will be:

9 \times 10^{3} N

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9 \times 10^{- 3}

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1.1 \times 10^{- 4} N

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10^{- 6} N

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Solution

The correct option is A $9 \times 10^{3} N$
We know that
$F = \frac{9 \times 10^{9} \times q \times q}{(r)^{2}}$
$F = \frac{9 \times 10^{9} \times 1 \times 1}{(1000)^{2}}$
$9 \times 10^{9 - 6} = 9 \times 10^{3} N$

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