JEE MAIN - Physics (2024 - 1st February Evening Shift - No. 24)
Suppose a uniformly charged wall provides a uniform electric field of $2 \times 10^4 \mathrm{~N} / \mathrm{C}$ normally. A charged particle of mass $2 \mathrm{~g}$ being suspended through a silk thread of length $20 \mathrm{~cm}$ and remain stayed at a distance of $10 \mathrm{~cm}$ from the wall.
Then the charge on the particle will be $\frac{1}{\sqrt{x}} \mu \mathrm{C}$ where $x=$ ___________ . [use $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$ ]
Then the charge on the particle will be $\frac{1}{\sqrt{x}} \mu \mathrm{C}$ where $x=$ ___________ . [use $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$ ]
Answer
3
Explanation
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$\begin{aligned} & \sin \theta=\frac{10}{20}=\frac{1}{2} \\\\ & \theta=30^{\circ} \\\\ & \tan \theta=\frac{\mathrm{qE}}{\mathrm{mg}} \\\\ & \tan 30^{\circ}=\frac{\mathrm{q} \times 2 \times 10^4}{1 \times 10^{-3} \times 10}\end{aligned}$
$\begin{aligned} & \frac{1}{\sqrt{3}}=q \times 10^6 \\\\ & q=\frac{1}{\sqrt{3}} \times 10^{-6} C \\\\ & x=3\end{aligned}$
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