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The force of attraction between two charges $8 \mu \mathrm{C}$ and $-4 \mu \mathrm{C}$ is 0.2 N . Find the distance of separation
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The correct answer is:
$1.2 m$
Given that, $F=0.2 \mathrm{~N}, \mathrm{Q}_1=8 \mu \mathrm{C}, \mathrm{Q}_2=4 \mu \mathrm{C}$
$\because$ Force between two charges, $\mathrm{F}=\mathrm{k} \frac{\mathrm{Q}_1 \mathrm{Q}_2}{\mathrm{r}^2}$
or $\quad 0.2=9 \times 10^9 \times \frac{8 \times 10^{-6} \times 4 \times 10^{-6}}{\mathrm{r}^2}$
$\therefore \quad r^2=\frac{9 \times 10^9 \times 8 \times 10^{-6} \times 4 \times 10^{-6}}{0.2}$
or $\quad \mathrm{r}=1.2 \mathrm{~m}$
$\because$ Force between two charges, $\mathrm{F}=\mathrm{k} \frac{\mathrm{Q}_1 \mathrm{Q}_2}{\mathrm{r}^2}$
or $\quad 0.2=9 \times 10^9 \times \frac{8 \times 10^{-6} \times 4 \times 10^{-6}}{\mathrm{r}^2}$
$\therefore \quad r^2=\frac{9 \times 10^9 \times 8 \times 10^{-6} \times 4 \times 10^{-6}}{0.2}$
or $\quad \mathrm{r}=1.2 \mathrm{~m}$
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