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A charge distribution with spherical symmetry has volume charge density given by $\rho(r)=\rho_0\left(1-\frac{r}{R}\right)$ for $r<R$, and $\rho(r)=0$ for $r\ge R$. Here $r$ denotes the distance from the centre and $\rho_0$ is a constant. The electric field at a point inside the distribution $(r<R)$ is

1. $\dfrac{\rho_0}{4\varepsilon_0}\left(r-\dfrac{r^2}{4R}\right)$
2. $\dfrac{\rho_0}{\varepsilon_0}\left(r-\dfrac{r^2}{4R}\right)$
3. $\dfrac{\rho_0}{3\varepsilon_0}\left(r-\dfrac{r^2}{4R}\right)$
4. $\dfrac{\rho_0}{12\varepsilon_0}\left(r-\dfrac{r^2}{4R}\right)$

Correct answer: $\dfrac{\rho_0}{3\varepsilon_0}\left(r-\dfrac{r^2}{4R}\right)$

Solution

By spherical symmetry, $E$ is radial and constant on a Gaussian sphere of radius $r$. Integrating the charge density up to $r$ and applying Gauss’s law gives the stated expression.

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