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A spherically symmetric charge distribution is characterised by a charge density having the following variations:
$\rho(r)=\rho_o\left(1-\frac{r}{R}\right)$ for $r < R$
$\rho(\mathrm{r})=0$ for $r \geq \mathrm{R}$
Where $r$ is the distance from the centre of the charge distribution $\rho_{\mathrm{o}}$ is a constant. The electric field at an internal point $(r < R)$ is:
Options:
$\rho(r)=\rho_o\left(1-\frac{r}{R}\right)$ for $r < R$
$\rho(\mathrm{r})=0$ for $r \geq \mathrm{R}$
Where $r$ is the distance from the centre of the charge distribution $\rho_{\mathrm{o}}$ is a constant. The electric field at an internal point $(r < R)$ is:
Solution:
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Verified Answer
The correct answer is:
$\frac{\rho_0}{\varepsilon_0}\left(\frac{r}{3}-\frac{r^2}{4 R}\right)$
$\frac{\rho_0}{\varepsilon_0}\left(\frac{r}{3}-\frac{r^2}{4 R}\right)$
Let us consider a spherical shell of radius $x$ and thickness $\mathrm{dx}$.
Charge on this shell
$$
d q=\rho .4 \pi x^2 d x=\rho_0\left(1-\frac{x}{R}\right) \cdot 4 \pi x^2 d x
$$
$\therefore$ Total charge in the spherical region from centre to $r(r < R)$ is
$$
\begin{aligned}
&q=\int d q=4 \pi \rho_0 \int_0^r\left(1-\frac{x}{R}\right) x^2 d x \\
&=4 \pi \rho_0\left[\frac{x^3}{3}-\frac{x^4}{4 R}\right]_0^r \\
&=4 \pi \rho_0\left[\frac{r^3}{3}-\frac{r^4}{4 R}\right] \\
&=4 \pi \rho_0 r^3\left[\frac{1}{3}-\frac{r}{4 R}\right]
\end{aligned}
$$
$\therefore$ Electric field at $\mathrm{r}, \mathrm{E}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\mathrm{q}}{\mathrm{r}^2}$
$$
\begin{aligned}
&=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{4 \pi \rho_0 r^3}{r^2}\left[\frac{1}{3}-\frac{r}{4 R}\right] \\
&=\frac{\rho_0}{\varepsilon_0}\left[\frac{r}{3}-\frac{r^2}{4 R}\right]
\end{aligned}
$$
Charge on this shell
$$
d q=\rho .4 \pi x^2 d x=\rho_0\left(1-\frac{x}{R}\right) \cdot 4 \pi x^2 d x
$$
$\therefore$ Total charge in the spherical region from centre to $r(r < R)$ is
$$
\begin{aligned}
&q=\int d q=4 \pi \rho_0 \int_0^r\left(1-\frac{x}{R}\right) x^2 d x \\
&=4 \pi \rho_0\left[\frac{x^3}{3}-\frac{x^4}{4 R}\right]_0^r \\
&=4 \pi \rho_0\left[\frac{r^3}{3}-\frac{r^4}{4 R}\right] \\
&=4 \pi \rho_0 r^3\left[\frac{1}{3}-\frac{r}{4 R}\right]
\end{aligned}
$$
$\therefore$ Electric field at $\mathrm{r}, \mathrm{E}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\mathrm{q}}{\mathrm{r}^2}$
$$
\begin{aligned}
&=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{4 \pi \rho_0 r^3}{r^2}\left[\frac{1}{3}-\frac{r}{4 R}\right] \\
&=\frac{\rho_0}{\varepsilon_0}\left[\frac{r}{3}-\frac{r^2}{4 R}\right]
\end{aligned}
$$
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