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 find how the electric field varies for a uniformly charged spherical shell, we can apply Gauss's Law. Let's analyze the situation step by step.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We have a spherical shell of radius \( R \) with a uniformly distributed charge \( Q \). We need to determine how the electric field \( E \) varies inside and outside the shell.

2. **Case 1: Inside the Spherical Shell (\( r < R \))**:
- For any point inside the shell, the enclosed charge \( Q_{\text{enclosed}} \) is zero because all the charge is on the surface.
- According to Gauss's Law:
\[ \oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{\text{enclosed}}}{\epsilon_0} \]
- Since \( Q_{\text{enclosed}} = 0 \), we have:
\[ \oint \mathbf{E} \cdot d\mathbf{A} = 0 \]
- This implies that the electric field \( E \) inside the spherical shell is:
\[ E = 0 \quad \text{for } r < R \]

3. **Case 2: Outside the Spherical Shell (\( r > R \))**:
- For points outside the shell, we consider a Gaussian surface of radius \( r \) (where \( r > R \)).
- The total charge enclosed by this Gaussian surface is \( Q \).
- Applying Gauss's Law:
\[ \oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q}{\epsilon_0} \]
- The surface area of the Gaussian sphere is \( 4\pi r^2 \), and the electric field \( E \) is uniform over this surface:
\[ E \cdot 4\pi r^2 = \frac{Q}{\epsilon_0} \]
- Solving for \( E \):
\[ E = \frac{Q}{4\pi \epsilon_0 r^2} \quad \text{for } r > R \]

4. **Conclusion**:
- The electric field inside the spherical shell is zero:
\[ E = 0 \quad \text{for } r < R \]
- The electric field outside the spherical shell varies inversely with the square of the distance from the center:
\[ E = \frac{Q}{4\pi \epsilon_0 r^2} \quad \text{for } r > R \]

### Summary of Results:
- Inside the shell (\( r < R \)): \( E = 0 \)
- Outside the shell (\( r > R \)): \( E \propto \frac{1}{r^2} \)