The force between two small charged spheres having charges of $1 \times 10^{-7} \mathrm{C}$ and $2 \times…

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
  1. $4.5 \times 10^{-2} \mathrm{~N}$
  2. $4.5 \times 10^{-3} \mathrm{~N}$
  3. $5.4 \times 10^{-2} \mathrm{~N}$
  4. $5.4 \times 10^{-3} \mathrm{~N}$

Solution

Here, $q_{1}=1 \times 10^{-7} \mathrm{C}, q_{2}$ and $2 \times 10^{-1} \mathrm{C}$,
$r=20 \mathrm{~cm}=20 \times 10^{-2} \mathrm{~m}$
$F=\frac{q_{1} q_{2}}{4 \pi \varepsilon_{0} r^{2}}=\frac{9 \times 10^{9} \times 1 \times 10^{-7} \times 2 \times 10^{-7}}{\left(20 \times 10^{-2}\right)^{2}}$
$=4.5 \times 10^{-3} N$ ,

Asked in: JEE Mains - Electrostatics - Test 1

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