There is a solid sphere of radius R having uniformly distributed charge throughout it. What is the relation between electric field E and distance r from the centre (r is less than R) ?

Options

A. `Epropr^(-2)`
B. `Epropr^(-1)`
C. `Epropr`
D. `Epropr^(2)`

Answer

To find the relation between the electric field \( E \) and the distance \( r \) from the center of a uniformly charged solid sphere (where \( r < R \)), we can follow these steps:

### Step-by-Step Solution

1. **Understand the Setup**:
We have a solid sphere of radius \( R \) with a uniform charge distribution. We want to find the electric field \( E \) at a distance \( r \) from the center of the sphere, where \( r < R \).

2. **Use Gauss's Law**:
According to Gauss's Law, the electric flux through a closed surface is equal to the charge enclosed by that surface divided by the permittivity of free space \( \epsilon_0 \):
\[ \Phi_E = \frac{Q_{\text{enc}}}{\epsilon_0} \]
where \( \Phi_E = \oint \vec{E} \cdot d\vec{A} \).

3. **Choose a Gaussian Surface**:
We choose a Gaussian surface that is a sphere of radius \( r \) (where \( r < R \)). The electric field \( E \) is uniform over this surface and directed radially outward.

4. **Calculate the Electric Flux**:
The electric flux through the Gaussian surface is:
\[ \Phi_E = E \cdot 4\pi r^2 \]
where \( 4\pi r^2 \) is the surface area of the sphere.

5. **Determine the Charge Enclosed**:
The charge enclosed \( Q_{\text{enc}} \) within the Gaussian surface can be calculated using the charge density \( \rho \):
\[ Q_{\text{enc}} = \rho \cdot V_{\text{enc}} = \rho \cdot \frac{4}{3}\pi r^3 \]
where \( V_{\text{enc}} \) is the volume of the sphere of radius \( r \).

6. **Relate Charge Density to Total Charge**:
The charge density \( \rho \) can be expressed in terms of the total charge \( Q \) of the sphere:
\[ \rho = \frac{Q}{\frac{4}{3}\pi R^3} \]

7. **Substitute Charge Enclosed into Gauss's Law**:
Substitute \( Q_{\text{enc}} \) into Gauss's Law:
\[ E \cdot 4\pi r^2 = \frac{\rho \cdot \frac{4}{3}\pi r^3}{\epsilon_0} \]

8. **Simplify the Equation**:
Substitute \( \rho \) from step 6:
\[ E \cdot 4\pi r^2 = \frac{\left(\frac{Q}{\frac{4}{3}\pi R^3}\right) \cdot \frac{4}{3}\pi r^3}{\epsilon_0} \]
This simplifies to:
\[ E \cdot 4\pi r^2 = \frac{Q \cdot r^3}{3\epsilon_0 R^3} \]

9. **Solve for Electric Field \( E \)**:
Rearranging gives:
\[ E = \frac{Q \cdot r}{4\pi \epsilon_0 R^3} \]

10. **Conclusion**:
The electric field \( E \) inside a uniformly charged solid sphere (for \( r < R \)) is directly proportional to the distance \( r \) from the center:
\[ E \propto r \]