`vec (E)` due to uniformly charged sphere of radius R as a function of distance from its centre is represented by:

Options

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D.

Explanation

To determine the electric field \(\vec{E}\) due to a uniformly charged sphere of radius \(R\) as a function of distance from its center, we can analyze the situation in two distinct regions: inside the sphere (for \(r < R\)) and outside the sphere (for \(r \geq R\)).

### Step-by-Step Solution:

1. **Identify the Charge Distribution**:
- Consider a uniformly charged sphere with total charge \(Q\) and radius \(R\).

2. **Electric Field Inside the Sphere (\(r < R\))**:
- According to Gauss's Law, the electric field inside a uniformly charged sphere is zero. This is because any Gaussian surface inside the sphere encloses no charge.
- Therefore, for \(r < R\):
\[ E = 0 \]

3. **Electric Field Outside the Sphere (\(r \geq R\))**:
- For points outside the sphere, we can treat the entire charge \(Q\) as if it were concentrated at the center of the sphere.
- Using Gauss's Law, the electric field at a distance \(r\) from the center (where \(r \geq R\)) is given by:
\[ E = \frac{kQ}{r^2} \]
- Here, \(k\) is Coulomb's constant.

4. **Summary of the Electric Field**:
- The electric field \(\vec{E}\) as a function of distance \(r\) from the center of the uniformly charged sphere can be summarized as:
\[ E(r) = \begin{cases} 0 & \text{if } r < R \\ \frac{kQ}{r^2} & \text{if } r \geq R \end{cases} \]

### Final Answer:
The electric field \(\vec{E}\) due to a uniformly charged sphere of radius \(R\) is:
\[ E(r) = \begin{cases} 0 & \text{if } r < R \\ \frac{kQ}{r^2} & \text{if } r \geq R \end{cases} \]