A spherically symmetric charge distribution is characterised by the following charge density variations: $$\rho(r) = \rho_o\left(1 - \frac{r}{R}\right)$$ for $$r < R$$, $$\rho(r) = 0$$ for $$r \geq R$$. Where r is the distance from the centre of the charge distribution and $$\rho_o$$ is a constant. The electric field at an internal point ($$r < R$$) is:

To find the electric field at an internal point where $$ r < R $$, we use Gauss's law due to the spherical symmetry. Gauss's law states that the flux through a closed surface is equal to the charge enclosed divided by $$ \varepsilon_0 $$. We choose a spherical Gaussian surface of radius $$ r $$ centered at the same point as the charge distribution.

The electric field is radial and constant in magnitude on this surface, so the flux is $$ E \times 4\pi r^2 $$. By Gauss's law:

$$ E \times 4\pi r^2 = \frac{Q_{\text{enc}}}{\varepsilon_0} $$

Solving for $$ E $$:

$$ E = \frac{Q_{\text{enc}}}{4\pi \varepsilon_0 r^2} $$

Now, we need to find $$ Q_{\text{enc}} $$, the charge enclosed within the sphere of radius $$ r $$. The charge density is given by $$ \rho(r') = \rho_o \left(1 - \frac{r'}{R}\right) $$ for $$ r' < R $$. Since $$ r < R $$, we integrate the charge density over the volume inside radius $$ r $$. Using spherical coordinates, the volume element is $$ dV' = 4\pi r'^2 dr' $$, so:

$$ Q_{\text{enc}} = \int_0^r \rho(r') dV' = \int_0^r \rho_o \left(1 - \frac{r'}{R}\right) \times 4\pi r'^2 dr' $$

Factor out the constants:

$$ Q_{\text{enc}} = 4\pi \rho_o \int_0^r \left(1 - \frac{r'}{R}\right) r'^2 dr' $$

Expand the integrand:

$$ Q_{\text{enc}} = 4\pi \rho_o \int_0^r \left( r'^2 - \frac{r'^3}{R} \right) dr' $$

Integrate term by term:

$$ \int_0^r r'^2 dr' = \left[ \frac{r'^3}{3} \right]_0^r = \frac{r^3}{3} $$

$$ \int_0^r \frac{r'^3}{R} dr' = \frac{1}{R} \left[ \frac{r'^4}{4} \right]_0^r = \frac{r^4}{4R} $$

So the integral becomes:

$$ \int_0^r \left( r'^2 - \frac{r'^3}{R} \right) dr' = \frac{r^3}{3} - \frac{r^4}{4R} $$

Therefore:

$$ Q_{\text{enc}} = 4\pi \rho_o \left( \frac{r^3}{3} - \frac{r^4}{4R} \right) $$

Substitute $$ Q_{\text{enc}} $$ back into the expression for $$ E $$:

$$ E = \frac{4\pi \rho_o \left( \frac{r^3}{3} - \frac{r^4}{4R} \right)}{4\pi \varepsilon_0 r^2} = \frac{\rho_o \left( \frac{r^3}{3} - \frac{r^4}{4R} \right)}{\varepsilon_0 r^2} $$

Simplify by dividing each term by $$ r^2 $$:

$$ E = \frac{\rho_o}{\varepsilon_0} \left( \frac{r^3}{3r^2} - \frac{r^4}{4R r^2} \right) = \frac{\rho_o}{\varepsilon_0} \left( \frac{r}{3} - \frac{r^2}{4R} \right) $$

Comparing with the options, this matches option B:

$$ \frac{\rho_o}{\varepsilon_o}\left(\frac{r}{3} - \frac{r^2}{4R}\right) $$

Hence, the correct answer is Option B.