Find out electric field intensity due to uniformly charged solid non-conducting sphere of volume charge density `rho` and radius R at following points :
(i) At a distance r from surface of sphere (inside)
(ii) At a distance r from the surface of sphere (outside)

Answer

To find the electric field intensity due to a uniformly charged solid non-conducting sphere of volume charge density \( \rho \) and radius \( R \) at two different points, we can use Gauss's law.

### Step-by-Step Solution:

**(i) At a distance \( r \) from the surface of the sphere (inside)**

1. **Identify the distance from the center:**
Since we are considering a point inside the sphere, the distance from the center of the sphere will be \( x = R - r \).

2. **Calculate the charge enclosed:**
The volume of the sphere of radius \( x \) is given by:
\[ V = \frac{4}{3} \pi x^3 = \frac{4}{3} \pi (R - r)^3 \]
The charge enclosed within this volume is:
\[ Q_{\text{enc}} = \rho \cdot V = \rho \cdot \frac{4}{3} \pi (R - r)^3 \]

3. **Apply Gauss's Law:**
According to Gauss's law:
\[ \oint E \cdot dS = \frac{Q_{\text{enc}}}{\epsilon_0} \]
For a spherical Gaussian surface of radius \( x \), the left side becomes:
\[ E \cdot 4 \pi x^2 \]
Thus, we have:
\[ E \cdot 4 \pi (R - r)^2 = \frac{\rho \cdot \frac{4}{3} \pi (R - r)^3}{\epsilon_0} \]

4. **Solve for \( E \):**
Cancel \( 4 \pi \) from both sides:
\[ E (R - r)^2 = \frac{\rho (R - r)^3}{3 \epsilon_0} \]
Therefore,
\[ E = \frac{\rho (R - r)}{3 \epsilon_0} \]

**(ii) At a distance \( r \) from the surface of the sphere (outside)**

1. **Identify the distance from the center:**
For a point outside the sphere, the distance from the center is \( x = R + r \).

2. **Calculate the charge enclosed:**
The total charge \( Q \) of the sphere is:
\[ Q = \rho \cdot \frac{4}{3} \pi R^3 \]

3. **Apply Gauss's Law:**
For a spherical Gaussian surface of radius \( x \):
\[ \oint E \cdot dS = \frac{Q}{\epsilon_0} \]
Thus, we have:
\[ E \cdot 4 \pi (R + r)^2 = \frac{\rho \cdot \frac{4}{3} \pi R^3}{\epsilon_0} \]

4. **Solve for \( E \):**
Cancel \( 4 \pi \) from both sides:
\[ E (R + r)^2 = \frac{\rho R^3}{3 \epsilon_0} \]
Therefore,
\[ E = \frac{\rho R^3}{3 \epsilon_0 (R + r)^2} \]

### Final Results:
- **Electric field intensity inside the sphere (at distance \( r \) from the surface):**
\[ E = \frac{\rho (R - r)}{3 \epsilon_0} \]

- **Electric field intensity outside the sphere (at distance \( r \) from the surface):**
\[ E = \frac{\rho R^3}{3 \epsilon_0 (R + r)^2} \]