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A conducting sphere of radius $0.1 \mathrm{~m}$ has uniform charge dénsity $1.8 \mu \mathrm{C} / \mathrm{m}^2$ on its surface. The electric field in free space at radial distance $0.2 \mathrm{~m}$ from $\alpha$ point on the surface is ( $\varepsilon_0=$ permittivity of free space)
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$\frac{2 \times 10^{-7}}{\varepsilon_0} \mathrm{Vm}^{-1}$
$\begin{aligned} \mathrm{q} & =\sigma \mathrm{A}=\sigma \pi \mathrm{r}^2 \\ \mathrm{E} & =\frac{1}{4 \pi \varepsilon_0} \times \frac{\mathrm{q}}{\mathrm{d}^2}=\frac{1}{4 \pi \varepsilon_0} \times \frac{\sigma \pi \mathrm{r}^2}{\mathrm{~d}^2} \\ & =\frac{1}{4 \pi \varepsilon_0} \times \frac{1.8 \times 10^{-6} \times \pi \times 10^{-2}}{4 \times 10^{-2}} \\ \therefore \quad \mathrm{E} & =\frac{1.125 \times 10^{-7}}{\varepsilon_0} \approx \frac{2 \times 10^{-7}}{\varepsilon_0} \mathrm{Vm}^{-1}\end{aligned}$
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