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A small block of mass $5 \mathrm{~g}$ and charge $5 \mu \mathrm{C}$ is placed on insulated, frictionless, inclined plane of angle $60^{\circ}$. An electric field is applied parallel to the inclined plane. If the block remains at rest then the magnitude of electric field is $\left[\right.$ Take $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$)
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Verified Answer
The correct answer is:
$\frac{\sqrt{3}}{2} \times 10^4 \mathrm{~N} / \mathrm{C}$
As block is at rest
So, $\mathrm{qE}=\mathrm{mg} \sin \theta$
$$
\begin{aligned}
& \Rightarrow \mathrm{E}=\frac{\mathrm{mg} \sin \theta}{\mathrm{q}} \\
& =\frac{5 \times 10^{-3} \times 10 \times \sqrt{3}}{2 \times 5 \times 10^{-6}} \\
& =\frac{5 \times 10^{-3} \times \sqrt{3}}{10^{-6}}=5 \sqrt{3} \times 10^3 \\
& =\frac{\sqrt{3}}{2} \times 10^4 \mathrm{~N} / \mathrm{C}
\end{aligned}
$$
So, $\mathrm{qE}=\mathrm{mg} \sin \theta$
$$
\begin{aligned}
& \Rightarrow \mathrm{E}=\frac{\mathrm{mg} \sin \theta}{\mathrm{q}} \\
& =\frac{5 \times 10^{-3} \times 10 \times \sqrt{3}}{2 \times 5 \times 10^{-6}} \\
& =\frac{5 \times 10^{-3} \times \sqrt{3}}{10^{-6}}=5 \sqrt{3} \times 10^3 \\
& =\frac{\sqrt{3}}{2} \times 10^4 \mathrm{~N} / \mathrm{C}
\end{aligned}
$$
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