A solid ball of radius R has a charge density $$\rho$$ given by $$\rho = \rho_0\left(1 - \frac{r}{R}\right)$$ for $$0 \leq r \leq R$$. The electric field outside the ball is:

We are given a solid sphere of radius $$R$$ whose volume-charge density varies with the distance $$r$$ from the centre as

$$\rho(r)=\rho_0\!\left(1-\frac{r}{R}\right),\qquad 0\le r\le R.$$

To find the electric field at an external point $$P$$ situated at a distance $$r$$ (with $$r>R$$) from the centre, we first need the total charge $$Q$$ contained in the sphere, because outside the sphere the entire charge can be treated as if it were concentrated at the centre (this is a consequence of Gauss’s law combined with spherical symmetry).

The infinitesimal charge of a thin spherical shell of radius $$r$$ and thickness $$dr$$ is

$$dq=\rho(r)\,dV=\rho(r)\,4\pi r^{2}dr.$$

Substituting $$\rho(r)=\rho_0\left(1-\dfrac{r}{R}\right)$$, we get

$$dq=4\pi r^{2}\rho_0\left(1-\frac{r}{R}\right)dr.$$

The total charge is obtained by integrating $$dq$$ from the centre $$r=0$$ to the surface $$r=R$$:

$$\begin{aligned} Q &=\int_{0}^{R}dq \\ &=4\pi\rho_0\int_{0}^{R}r^{2}\left(1-\frac{r}{R}\right)dr. \end{aligned}$$

We separate the integrand:

$$\begin{aligned} \int_{0}^{R}r^{2}\left(1-\frac{r}{R}\right)dr &=\int_{0}^{R}\left(r^{2}-\frac{r^{3}}{R}\right)dr\\ &=\int_{0}^{R}r^{2}dr-\frac{1}{R}\int_{0}^{R}r^{3}dr. \end{aligned}$$

Using the standard power-integral formulas $$\int r^{n}dr=\frac{r^{n+1}}{n+1},$$ we evaluate the two parts:

$$\int_{0}^{R}r^{2}dr=\left[\frac{r^{3}}{3}\right]_{0}^{R}=\frac{R^{3}}{3},$$

$$\int_{0}^{R}r^{3}dr=\left[\frac{r^{4}}{4}\right]_{0}^{R}=\frac{R^{4}}{4}.$$

Hence,

$$\begin{aligned} \int_{0}^{R}r^{2}\left(1-\frac{r}{R}\right)dr &=\frac{R^{3}}{3}-\frac{1}{R}\cdot\frac{R^{4}}{4}\\ &=\frac{R^{3}}{3}-\frac{R^{3}}{4}\\ &=R^{3}\!\left(\frac{1}{3}-\frac{1}{4}\right)\\ &=R^{3}\!\left(\frac{4-3}{12}\right)\\ &=\frac{R^{3}}{12}. \end{aligned}$$

Substituting this result into the expression for $$Q$$, we have

$$\begin{aligned} Q &=4\pi\rho_0\left(\frac{R^{3}}{12}\right)\\ &=\frac{4\pi\rho_0 R^{3}}{12}\\ &=\frac{\pi\rho_0 R^{3}}{3}. \end{aligned}$$

Now we turn to the electric field at the external point. From Gauss’s law, for a spherical Gaussian surface of radius $$r$$ (with $$r>R$$) concentric with the sphere, the outward flux is

$$\Phi=E\,(4\pi r^{2}),$$

and Gauss’s law states

$$\Phi=\frac{Q_{\text{enc}}}{\varepsilon_0},$$

where $$Q_{\text{enc}}=Q$$, the total charge calculated above. Therefore,

$$E(4\pi r^{2})=\frac{Q}{\varepsilon_0}.$$

Solving for $$E$$ gives

$$\begin{aligned} E &=\frac{Q}{4\pi\varepsilon_0\,r^{2}}\\ &=\frac{1}{4\pi\varepsilon_0\,r^{2}}\left(\frac{\pi\rho_0 R^{3}}{3}\right)\\ &=\frac{\pi\rho_0 R^{3}}{12\pi\varepsilon_0\,r^{2}}\\ &=\frac{\rho_0 R^{3}}{12\varepsilon_0 r^{2}}. \end{aligned}$$

The electric field outside the charged ball thus decreases as $$1/r^{2}$$ and has the magnitude

$$E(r)=\frac{\rho_0 R^{3}}{12\varepsilon_0 r^{2}}.$$

Comparing with the given alternatives, we see that this matches Option D.

Hence, the correct answer is Option D.