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The force between two small charged spheres having charges of $1 \times 10^{-7} \mathrm{C}$ and $2 \times 10^{-7} \mathrm{C}$ placed $20 \mathrm{~cm}$ apart in air is
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The correct answer is:
$4.5 \times 10^{-3} \mathrm{~N}$
Here, $q_1=1 \times 10^{-7} \mathrm{C}, q_2$ and $2 \times 10^{-7} \mathrm{C}$, $r=20 \mathrm{~cm}=20 \times 10^{-2} \mathrm{~m}$
$F=\frac{q_1 q_2}{4 \pi \varepsilon_0 r^2}=\frac{9 \times 10^9 \times 1 \times 10^{-7} \times 2 \times 10^{-7}}{\left(20 \times 10^{-2}\right)^2}$
$=4.5 \times 10^{-3} N$
$F=\frac{q_1 q_2}{4 \pi \varepsilon_0 r^2}=\frac{9 \times 10^9 \times 1 \times 10^{-7} \times 2 \times 10^{-7}}{\left(20 \times 10^{-2}\right)^2}$
$=4.5 \times 10^{-3} N$
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