JEE MAIN - Physics (2023 - 13th April Evening Shift - No. 19)
Three point charges $$\mathrm{q},-2 \mathrm{q}$$ and $$2 \mathrm{q}$$ are placed on $x$-axis at a distance $$x=0, x=\frac{3}{4} R$$ and $$x=R$$ respectively from origin as shown. If $$\mathrm{q}=2 \times 10^{-6} \mathrm{C}$$ and $$\mathrm{R}=2 \mathrm{~cm}$$, the magnitude of net force experienced by the charge $$-2 q$$ is ___________ N.
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Answer
5440
Explanation
$$
\begin{aligned}
& \mathrm{F}_{\mathrm{BA}}=\frac{\mathrm{Kq}(2 \mathrm{q})}{\left(\frac{3}{4} \mathrm{R}\right)^2}=\frac{32 \mathrm{Kq}^2}{9 \mathrm{R}^2} \\\\
& \mathrm{~F}_{\mathrm{BC}}=\frac{\mathrm{K}(2 \mathrm{q})(2 \mathrm{q})}{\left(\frac{\mathrm{R}}{4}\right)^2}=\frac{64 \mathrm{Kq}^2}{\mathrm{R}^2} \\\\
& \mathrm{~F}_{\mathrm{B}}=\mathrm{F}_{\mathrm{BC}}-\mathrm{F}_{\mathrm{BA}}=\frac{544 \mathrm{Kq}^2}{9 \mathrm{R}^2} \\\\
& =\frac{544 \times 9 \times 10^9 \times\left(2 \times 10^{-6}\right)^2}{9 \times\left(2 \times 10^{-2}\right)^2}=5440 \mathrm{~N}
\end{aligned}
$$
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