A sphere of charges of radius R carries a positive charge whose volume chasrge 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 electric field as a function of the distance r inside the sphere is given by

Options

A. `E=rho_0/epsilon[r/3-r^2/(4R)]`
B. `E=rho_0/epsilon[r/4-r^3/(3R)]`
C. `E=rho_0/epsilon[r/3+r^2(4R)]`
D. `E=rho_0/epsilon[r/4+r^2/(3R)]`

Explanation

To find the electric field inside a sphere with a varying charge density, we can follow these steps:

### Step 1: Understand the charge density
The volume charge density is given by:
\[ \rho(r) = \rho_0 \left(1 - \frac{r}{R}\right) \]
where \( \rho_0 \) is a constant, \( r \) is the distance from the center of the sphere, and \( R \) is the radius of the sphere.

### Step 2: Use Gauss's Law
To find the electric field \( E \) at a distance \( r \) from the center, we apply Gauss's Law:
\[ \Phi_E = \frac{Q_{\text{enc}}}{\epsilon_0} \]
where \( \Phi_E \) is the electric flux through a Gaussian surface of radius \( r \) and \( Q_{\text{enc}} \) is the charge enclosed within that surface.

### Step 3: Calculate the electric flux
The electric flux through the Gaussian surface is given by:
\[ \Phi_E = E \cdot 4\pi r^2 \]
Thus, we have:
\[ E \cdot 4\pi r^2 = \frac{Q_{\text{enc}}}{\epsilon_0} \]

### Step 4: Determine the enclosed charge \( Q_{\text{enc}} \)
To find \( Q_{\text{enc}} \), we need to integrate the charge density over the volume of the sphere up to radius \( r \):
\[ Q_{\text{enc}} = \int_0^r \rho(r') \, dV \]
The differential volume element \( dV \) in spherical coordinates is:
\[ dV = 4\pi r'^2 dr' \]
Substituting the charge density:
\[ Q_{\text{enc}} = \int_0^r \rho_0 \left(1 - \frac{r'}{R}\right) 4\pi r'^2 dr' \]

### Step 5: Perform the integration
Now, we can expand and integrate:
\[ Q_{\text{enc}} = 4\pi \rho_0 \int_0^r \left(r'^2 - \frac{r'^3}{R}\right) dr' \]
Calculating the integrals separately:
\[ \int_0^r r'^2 dr' = \frac{r^3}{3}, \quad \text{and} \quad \int_0^r r'^3 dr' = \frac{r^4}{4} \]
So,
\[ Q_{\text{enc}} = 4\pi \rho_0 \left( \frac{r^3}{3} - \frac{r^4}{4R} \right) \]

### Step 6: Substitute \( Q_{\text{enc}} \) back into Gauss's Law
Now substitute \( Q_{\text{enc}} \) into Gauss's Law:
\[ E \cdot 4\pi r^2 = \frac{4\pi \rho_0 \left( \frac{r^3}{3} - \frac{r^4}{4R} \right)}{\epsilon_0} \]
Cancelling \( 4\pi \) from both sides:
\[ E r^2 = \frac{\rho_0}{\epsilon_0} \left( \frac{r^3}{3} - \frac{r^4}{4R} \right) \]

### Step 7: Solve for \( E \)
Finally, we can solve for \( E \):
\[ E = \frac{\rho_0}{\epsilon_0} \left( \frac{r}{3} - \frac{r^2}{4R} \right) \]

### Final Result
The magnitude of the electric field as a function of the distance \( r \) inside the sphere is:
\[ E(r) = \frac{\rho_0}{\epsilon_0} \left( \frac{r}{3} - \frac{r^2}{4R} \right) \]