In a region of space having a spherical symmetic distribution of charge , the electric flux enclosed by a concentric spherical Gaussion surface, varies with radius
`r" as "phi={(phi_(0)r^(2))/R^(3), r le R
phi_(0)`, `ltR` where R and `phi_(0)` are the constants.

Options

A. `E=(phi_(0))/(4pir^(2))`,for `rleR`
B. `E=(phi_(0))/(4pir^(2))`,for `rgtR`
C. `E=(phi_(0)r)/(4pir^(2))`,for ` rleR`
D. `E=(phi_(0)r)/(4pir^(2))`,for `rgtR`

Answer

To solve the problem, we need to analyze the electric flux enclosed by a concentric spherical Gaussian surface in a region with a spherical symmetric distribution of charge. The electric flux is given by the equations for two cases based on the radius \( r \) relative to \( R \).

### Step-by-Step Solution:

1. **Understanding Electric Flux**:
The electric flux \( \Phi \) through a closed surface is given by the equation:
\[ \Phi = E \cdot A \]
where \( E \) is the electric field and \( A \) is the area of the surface. For a sphere of radius \( r \), the area \( A \) is:
\[ A = 4\pi r^2 \]

2. **Case 1: When \( r \leq R \)**:
The electric flux is given by:
\[ \Phi = \frac{\Phi_0 r^2}{R^3} \]
Setting this equal to the expression for flux:
\[ \Phi = E \cdot 4\pi r^2 \]
We can equate the two expressions:
\[ E \cdot 4\pi r^2 = \frac{\Phi_0 r^2}{R^3} \]
Dividing both sides by \( 4\pi r^2 \) (assuming \( r \neq 0 \)):
\[ E = \frac{\Phi_0}{4\pi R^3} \]

3. **Case 2: When \( r > R \)**:
The electric flux is constant and equal to \( \Phi_0 \):
\[ \Phi = \Phi_0 \]
Again, using the flux formula:
\[ \Phi = E \cdot 4\pi r^2 \]
Setting these equal gives:
\[ E \cdot 4\pi r^2 = \Phi_0 \]
Dividing both sides by \( 4\pi r^2 \):
\[ E = \frac{\Phi_0}{4\pi r^2} \]

4. **Summary of Results**:
- For \( r \leq R \):
\[ E = \frac{\Phi_0}{4\pi R^3} \]
- For \( r > R \):
\[ E = \frac{\Phi_0}{4\pi r^2} \]

### Final Answer:
The electric field \( E \) varies with radius \( r \) as follows:
- For \( r \leq R \): \( E = \frac{\Phi_0}{4\pi R^3} \)
- For \( r > R \): \( E = \frac{\Phi_0}{4\pi r^2} \)