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A charge $17.7 \times 10^{-4} \mathrm{C}$ is distributed uniformly over a large sheet of area $200 \mathrm{~m}^2$. The electric field intensity at a distance $20 \mathrm{~cm}$ from it in air will be $\left[\varepsilon_0=8.85 \times 10^{-12} \mathrm{C}^2 / \mathrm{Nm}^2\right]$
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Verified Answer
The correct answer is:
$5 \times 10^5 \mathrm{~N} / \mathrm{C}$
The surface charge density is given by, $\sigma=\frac{\mathrm{q}}{\mathrm{A}}=\frac{17.7 \times 10^{-4}}{200}=8.85 \times 10^{-6} \mathrm{C}_{\mathrm{m}} \mathrm{m}^2$ The electric field intensity at a distance of $20 \mathrm{~cm}$ in air is,
$\mathrm{E}=\frac{\sigma}{2 \varepsilon_0}=\frac{8.85 \times 10^{-6}}{2 \times 8.85 \times 10^{-12}}=5 \times 10^5 \mathrm{~N} / \mathrm{C}$
Hence, option (A).
$\mathrm{E}=\frac{\sigma}{2 \varepsilon_0}=\frac{8.85 \times 10^{-6}}{2 \times 8.85 \times 10^{-12}}=5 \times 10^5 \mathrm{~N} / \mathrm{C}$
Hence, option (A).
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