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

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Explanation

To find the electric field \( \vec{E} \) due to a uniformly charged sphere of radius \( R \) as a function of the distance \( r \) from its center, we need to consider two cases: when the point of interest is inside the sphere (i.e., \( r < R \)) and when it is outside the sphere (i.e., \( r \geq R \)).

### Step 1: Electric Field Inside the Sphere (\( r < R \))

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

2. **Choose a Gaussian Surface**: For a point inside the sphere, we choose a Gaussian surface that is a sphere of radius \( r \) (where \( r < R \)).

3. **Charge Enclosed**: The charge enclosed by this Gaussian surface can be found using the volume charge density \( \rho \):
\[ Q_{\text{enc}} = \rho \cdot V = \rho \cdot \frac{4}{3} \pi r^3 \]
where \( \rho = \frac{Q}{\frac{4}{3} \pi R^3} \) (the total charge \( Q \) is uniformly distributed over the volume of the sphere).

4. **Substituting Charge into Gauss's Law**:
\[ \Phi_E = E \cdot 4\pi r^2 = \frac{\rho \cdot \frac{4}{3} \pi r^3}{\epsilon_0} \]
\[ E \cdot 4\pi r^2 = \frac{\left(\frac{Q}{\frac{4}{3} \pi R^3}\right) \cdot \frac{4}{3} \pi r^3}{\epsilon_0} \]

5. **Solving for \( E \)**:
\[ E = \frac{Q}{4\pi \epsilon_0 R^3} r \]
Thus, the electric field inside the uniformly charged sphere is:
\[ \vec{E} = \frac{Q}{4\pi \epsilon_0 R^3} r \hat{r} \]

### Step 2: Electric Field Outside the Sphere (\( r \geq R \))

1. **Gauss's Law**: Again, we apply Gauss's Law for a point outside the sphere.

2. **Choose a Gaussian Surface**: For \( r \geq R \), we choose a spherical Gaussian surface of radius \( r \).

3. **Charge Enclosed**: The total charge \( Q \) of the sphere is enclosed by this Gaussian surface:
\[ Q_{\text{enc}} = Q \]

4. **Substituting Charge into Gauss's Law**:
\[ \Phi_E = E \cdot 4\pi r^2 = \frac{Q}{\epsilon_0} \]

5. **Solving for \( E \)**:
\[ E = \frac{Q}{4\pi \epsilon_0 r^2} \]
Thus, the electric field outside the uniformly charged sphere is:
\[ \vec{E} = \frac{Q}{4\pi \epsilon_0 r^2} \hat{r} \]

### Summary of Results

- For \( r < R \):
\[ \vec{E} = \frac{Q}{4\pi \epsilon_0 R^3} r \hat{r} \]

- For \( r \geq R \):
\[ \vec{E} = \frac{Q}{4\pi \epsilon_0 r^2} \hat{r} \]