Let there be a spherical symmetric charge density varying as `p(r )=p_(0)(r )/(R )` upto r = R and `rho(r )=0` for `r gt R`, where r is the distance from the origin. The electric field at on a distance `r(r lt R)` from the origin is given by -

Options

A. `(rho_(0)r^(2))/(4epsilon_(0)R)`
B. `(rho_(0)r)/(4epsilon_(0)R)`
C. `(rho_(0)r^(4))/(epsilon_(0)R)`
D. `(rho_(0)r^(2))/(epsilon_(0)R)`

Answer

To find the electric field at a distance \( r \) from the origin (where \( r < R \)) for a spherical symmetric charge density given by \( \rho(r) = \frac{\rho_0 r}{R} \) for \( r \leq R \) and \( \rho(r) = 0 \) for \( r > R \), we can use Gauss's law.

### Step-by-Step Solution:

1. **Identify the Charge Density**:
The charge density is given as:
\[ \rho(r) = \frac{\rho_0 r}{R} \quad \text{for } r < R \]
and
\[ \rho(r) = 0 \quad \text{for } r > R. \]

2. **Use Gauss's Law**:
Gauss's law states that:
\[ \Phi_E = \frac{Q_{\text{enc}}}{\epsilon_0} \]
where \( \Phi_E \) is the electric flux through a closed surface and \( Q_{\text{enc}} \) is the charge enclosed by that surface.

3. **Choose a Gaussian Surface**:
We choose a spherical Gaussian surface of radius \( r \) (where \( r < R \)). The electric field \( E \) is uniform over this surface due to symmetry.

4. **Calculate the Area of the Gaussian Surface**:
The area \( A \) of the spherical surface is:
\[ A = 4\pi r^2. \]

5. **Calculate the Enclosed Charge \( Q_{\text{enc}} \)**:
To find \( Q_{\text{enc}} \), we integrate the charge density over the volume of the sphere of radius \( r \):
\[ Q_{\text{enc}} = \int_0^r \rho(r') dV, \]
where \( dV = 4\pi r'^2 dr' \). Thus,
\[ Q_{\text{enc}} = \int_0^r \left( \frac{\rho_0 r'}{R} \right) (4\pi r'^2) dr' = \frac{4\pi \rho_0}{R} \int_0^r r'^3 dr'. \]

6. **Evaluate the Integral**:
The integral \( \int_0^r r'^3 dr' \) evaluates to:
\[ \int_0^r r'^3 dr' = \frac{r^4}{4}. \]
Therefore,
\[ Q_{\text{enc}} = \frac{4\pi \rho_0}{R} \cdot \frac{r^4}{4} = \frac{\pi \rho_0 r^4}{R}. \]

7. **Substitute into Gauss's Law**:
Now, substituting \( Q_{\text{enc}} \) into Gauss's law:
\[ E \cdot 4\pi r^2 = \frac{\frac{\pi \rho_0 r^4}{R}}{\epsilon_0}. \]

8. **Solve for the Electric Field \( E \)**:
Rearranging gives:
\[ E = \frac{\frac{\pi \rho_0 r^4}{R}}{4\pi \epsilon_0 r^2} = \frac{\rho_0 r^2}{4 \epsilon_0 R}. \]

### Final Result:
Thus, the electric field at a distance \( r \) from the origin (for \( r < R \)) is:
\[ E = \frac{\rho_0 r^2}{4 \epsilon_0 R}. \]