Let there be a spherically symmetric charge distribution with charge density varying as $\rho(r)=\rho_0\left(\frac{5}{4}-\frac{r}{R}\right)$ upto $r=R$, and $\rho(r)=0$ for $rR$, where $r$ is the distance from the origin. The electric field at a distance $r(r

Question

Let there be a spherically symmetric charge distribution with charge density varying as $\rho(r)=\rho_0\left(\frac{5}{4}-\frac{r}{R}\right)$ upto $r=R$, and $\rho(r)=0$ for $r>R$, where $r$ is the distance from the origin. The electric field at a distance $r(r < R)$ from the origin is given by

Aicharya · Accepted Answer

Apply shell theorem the total charge upto distance $r$ can be calculated as followed $d q=4 \pi r^2 \cdot d r . ho$ $ \begin{aligned} & =4 \pi r^2 \cdot d r \cdot ho_0\left[\frac{5}{4}-\frac{r}{R}ight] \ & =4 \pi ho_0\left[\frac{5}{4} r^2 d r-\frac{r^3}{R} d right] \ & \int d q=q=4 \pi ho_0 \int_0^r\left(\frac{5}{4} r^2 d r-\frac{r^3}{R} d right) \ & =4 \pi ho_0\left[\frac{5}{4} \frac{r^3}{3}-\frac{1}{R} \frac{r^4}{4}ight] \ & E=\frac{k q}{r^2} \ & =\frac{1}{4 \pi \varepsilon_0} \frac{1}{r^2} \cdot 4 \pi ho_0\left[\frac{5}{4}\left(\frac{r^3}{3}ight)-\frac{r^4}{4 R}ight] \ & E=\frac{ho_0 r}{4 \varepsilon_0}\left[\frac{5}{3}-\frac{r}{R}ight] \ & \end{aligned} $

Let there be a spherically symmetric charge distribution with charge density varying as…

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