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A sphere of radius $R$ has a volume density of charge $\rho=k r,$ where $r$ is the distance from the centre of the sphere and $k$ is constant. The magnitude of the electric field which exists at the surface of the sphere is given by $\left(\varepsilon_{0}=\right.$ permittivity of free space)
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The correct answer is:
$\frac{k R^{2}}{4 \varepsilon_{0}}$
By Gauss's theorem
\(\begin{aligned} \mathrm{E}\left(4 \pi \mathrm{r}^{2}\right) &=\frac{\int \mathrm{p} \times 4 \pi \mathrm{r}^{2} \mathrm{dr}}{\varepsilon_{0}} \\ &=\frac{\int \mathrm{kr} \times 4 \pi \mathrm{r}^{2} \mathrm{dr}}{\varepsilon_{0}} \\ &=\mathrm{E}=\frac{\mathrm{Kr}^{2}}{4 \varepsilon_{0}} \end{aligned}\)
\(\begin{aligned} \mathrm{E}\left(4 \pi \mathrm{r}^{2}\right) &=\frac{\int \mathrm{p} \times 4 \pi \mathrm{r}^{2} \mathrm{dr}}{\varepsilon_{0}} \\ &=\frac{\int \mathrm{kr} \times 4 \pi \mathrm{r}^{2} \mathrm{dr}}{\varepsilon_{0}} \\ &=\mathrm{E}=\frac{\mathrm{Kr}^{2}}{4 \varepsilon_{0}} \end{aligned}\)
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