A hallow metal sphere of radius R is uniformly charged. The electric field due to the sphere at a distance r from the centre:

Options

A. decreases as r increases for `rltR` and `rgtR`
B. increases as r increases for `rltR` and `rgtR`
C. zero as r increases for `rltR`, decreases as r increases for `rgtR`
D. zero as r increases for `rltR`, increases for `rgtR`

Explanation

To determine the electric field due to a hollow metal sphere of radius \( R \) at a distance \( r \) from its center, we will analyze two cases: when \( r < R \) (inside the sphere) and when \( r > R \) (outside the sphere).

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a hollow metal sphere with radius \( R \) that is uniformly charged with total charge \( Q \).
- We need to find the electric field \( E \) at a distance \( r \) from the center of the sphere.

2. **Case 1: Inside the Sphere (\( r < R \))**:
- According to Gauss's law, the electric field inside a conductor in electrostatic equilibrium is zero.
- Since there is no charge enclosed within a Gaussian surface that lies inside the hollow region of the sphere, we can write:
\[ E \cdot A = \frac{Q_{\text{enclosed}}}{\epsilon_0} \]
- Here, \( A \) is the area of the Gaussian surface and \( Q_{\text{enclosed}} = 0 \).
- Therefore, we have:
\[ E \cdot 0 = 0 \implies E = 0 \]
- Thus, for \( r < R \), the electric field \( E = 0 \).

3. **Case 2: Outside the Sphere (\( r > R \))**:
- For a point outside the hollow sphere, we can consider the entire charge \( Q \) to be concentrated at the center of the sphere.
- Applying Gauss's law again, we take a Gaussian surface of radius \( r \) (where \( r > R \)):
\[ E \cdot A = \frac{Q_{\text{enclosed}}}{\epsilon_0} \]
- The area \( A \) of the Gaussian surface is \( 4\pi r^2 \) and \( Q_{\text{enclosed}} = Q \).
- Thus, we have:
\[ E \cdot 4\pi r^2 = \frac{Q}{\epsilon_0} \]
- Solving for \( E \):
\[ E = \frac{Q}{4\pi \epsilon_0 r^2} \]
- This shows that the electric field \( E \) outside the sphere decreases with the square of the distance \( r \) from the center.

4. **Conclusion**:
- For \( r < R \), \( E = 0 \).
- For \( r > R \), \( E \) decreases as \( r \) increases.
- Therefore, the correct option is: "0 as \( r \) increases for \( r < R \) and decreases as \( r \) increases for \( r > R \)."