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A charge of $1 \mu \mathrm{C}$ is divided into two parts such that their charges are in the ratio of $2: 3$. These two charges are kept at a distance $1 \mathrm{~m}$ apart in vacuum. Then, the electric force between them $($ in $N)$ is
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Verified Answer
The correct answer is:
$0.00216$
Ratio of charges $=2: 3$
$$
\therefore q_1=\frac{2}{5} \times 1 \mu \mathrm{C} \text { and } q_2=\frac{3}{5} \times 1 \mu \mathrm{C}
$$
Electrostatic force between the two charges
$$
\begin{aligned}
F & =\frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r^2} \\
& =\frac{9 \times 10^9 \times 2 \times 10^{-6} \times 3 \times 10^{-6}}{5 \times 5 \times(1)^2} \\
& =2.16 \times 10^{-3} \mathrm{~N}
\end{aligned}
$$
$$
\therefore q_1=\frac{2}{5} \times 1 \mu \mathrm{C} \text { and } q_2=\frac{3}{5} \times 1 \mu \mathrm{C}
$$
Electrostatic force between the two charges
$$
\begin{aligned}
F & =\frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r^2} \\
& =\frac{9 \times 10^9 \times 2 \times 10^{-6} \times 3 \times 10^{-6}}{5 \times 5 \times(1)^2} \\
& =2.16 \times 10^{-3} \mathrm{~N}
\end{aligned}
$$
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