A spherical shell of radius R has a uniformly distributed charge ,then electric field varies as

Options

A. inside is zero and varies as `x^(-1)` outside it
B. inside is non zero constant and varies as `x^(2)` outside it
C. inside is zero and varies as `x^(-2)` outside it
D. inside is constant and varies as `x^(-1)` outside it

Explanation

To solve the problem regarding the electric field due to a uniformly charged spherical shell, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Configuration**:
We have a spherical shell of radius \( R \) with a uniformly distributed charge \( Q \). We need to analyze the electric field at different points relative to the shell.

2. **Electric Field Inside the Shell**:
According to Gauss's Law, the electric field inside a uniformly charged spherical shell is zero. This is because the symmetry of the charge distribution leads to cancellation of the electric field vectors at any point inside the shell.
\[ E_{\text{inside}} = 0 \quad \text{for} \quad r < R \]

3. **Electric Field on the Surface of the Shell**:
At the surface of the shell (where \( r = R \)), we can use Gauss's Law to find the electric field. The electric field \( E \) at the surface can be calculated as:
\[ E_{\text{surface}} = \frac{1}{4\pi \epsilon_0} \cdot \frac{Q}{R^2} \]

4. **Electric Field Outside the Shell**:
For points outside the shell (where \( r > R \)), the shell behaves like a point charge located at its center. Thus, the electric field at a distance \( r \) from the center is given by:
\[ E_{\text{outside}} = \frac{1}{4\pi \epsilon_0} \cdot \frac{Q}{r^2} \]

5. **Summary of Electric Field Behavior**:
- Inside the shell (\( r < R \)): \( E = 0 \)
- On the surface of the shell (\( r = R \)): \( E = \frac{1}{4\pi \epsilon_0} \cdot \frac{Q}{R^2} \)
- Outside the shell (\( r > R \)): \( E = \frac{1}{4\pi \epsilon_0} \cdot \frac{Q}{r^2} \)

6. **Conclusion**:
The electric field varies as follows:
- It is zero inside the shell.
- It behaves inversely with the square of the distance outside the shell.

### Final Answer:
The electric field varies as:
- \( E = 0 \) for \( r < R \)
- \( E \propto \frac{1}{r^2} \) for \( r > R \)