A sphere of charges of radius R carries a positive charge whose volume charge density depends only on the distance r from the ball's centre as `rho=rho_0(1-r/R)`, where` rho_0` is constant. Assume epsilon as theh permittivity of space.
The magnitude of the electric field as a functiion of the distance r outside the balll is given by

Options

A. `E=(rho_0R^3)/(8epsilonr^2)`
B. `E=(rho_0R^3)/(12epsilonr^2)`
C. `E=(rho_0R^2)/(8epsilonr^3)`
D. `E=(rho_0R^2)/(12epsilonr^3)`

Explanation

To find the magnitude of the electric field outside a uniformly charged sphere with a given volume charge density, we can follow these steps:

### Step 1: Understand the Charge Density
The charge density is given by:
\[ \rho(r) = \rho_0 \left(1 - \frac{r}{R}\right) \]
This means that the charge density varies with the distance \( r \) from the center of the sphere.

### Step 2: Determine the Total Charge Enclosed
To find the total charge \( Q \) enclosed within the sphere, we need to integrate the charge density over the volume of the sphere. The differential charge \( dq \) in a thin spherical shell of radius \( r \) and thickness \( dr \) is given by:
\[ dq = \rho(r) \cdot dV = \rho(r) \cdot 4\pi r^2 dr \]
Substituting the expression for \( \rho(r) \):
\[ dq = \rho_0 \left(1 - \frac{r}{R}\right) \cdot 4\pi r^2 dr \]

### Step 3: Integrate to Find Total Charge
Now we integrate \( dq \) from \( r = 0 \) to \( r = R \):
\[ Q = \int_0^R dq = \int_0^R \rho_0 \left(1 - \frac{r}{R}\right) \cdot 4\pi r^2 dr \]
This can be simplified as:
\[ Q = 4\pi \rho_0 \int_0^R \left( r^2 - \frac{r^3}{R} \right) dr \]
Calculating the integral:
\[ \int_0^R r^2 dr = \frac{R^3}{3}, \quad \int_0^R r^3 dr = \frac{R^4}{4} \]
Substituting these results back:
\[ Q = 4\pi \rho_0 \left( \frac{R^3}{3} - \frac{R^4}{4R} \right) = 4\pi \rho_0 \left( \frac{R^3}{3} - \frac{R^3}{4} \right) \]
Combining the fractions:
\[ Q = 4\pi \rho_0 R^3 \left( \frac{4}{12} - \frac{3}{12} \right) = 4\pi \rho_0 R^3 \cdot \frac{1}{12} = \frac{4\pi \rho_0 R^3}{12} \]

### Step 4: Calculate the Electric Field Outside the Sphere
For points outside the sphere (at a distance \( r > R \)), we can use Gauss's law:
\[ E \cdot 4\pi r^2 = \frac{Q}{\epsilon_0} \]
Substituting for \( Q \):
\[ E \cdot 4\pi r^2 = \frac{4\pi \rho_0 R^3}{12 \epsilon_0} \]
Solving for \( E \):
\[ E = \frac{\rho_0 R^3}{12 \epsilon_0 r^2} \]

### Final Answer
The magnitude of the electric field as a function of the distance \( r \) outside the sphere is:
\[ E(r) = \frac{\rho_0 R^3}{12 \epsilon_0 r^2} \]