In a region of space having a spherical symmetric distribution of charge , the electric flux enclosed by a concentric spherical Gaussian 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 relationship between electric flux and the radius of the spherical Gaussian surface in a region with a spherically symmetric charge distribution. The electric flux \(\Phi\) is given by:

\[ \Phi = \frac{\Phi_0 r^2}{R^3}, \quad \text{for } r \leq R \]

where \(\Phi_0\) and \(R\) are constants.

### Step 1: Understanding Electric Flux
Electric flux (\(\Phi\)) through a surface is defined as the integral of the electric field (\(E\)) over that surface. For a spherical surface, it can be expressed as:

\[ \Phi = E \cdot A \]

where \(A\) is the area of the spherical surface, given by \(A = 4\pi r^2\).

### Step 2: Relating Electric Field to Electric Flux
From Gauss's Law, we know that the electric flux through a closed surface is equal to the charge enclosed (\(Q_{\text{enc}}\)) divided by the permittivity of free space (\(\epsilon_0\)):

\[ \Phi = \frac{Q_{\text{enc}}}{\epsilon_0} \]

### Step 3: Finding Charge Enclosed
Given the expression for electric flux, we can set it equal to the charge enclosed divided by \(\epsilon_0\):

\[ \frac{\Phi_0 r^2}{R^3} = \frac{Q_{\text{enc}}}{\epsilon_0} \]

From this, we can express the charge enclosed as:

\[ Q_{\text{enc}} = \frac{\Phi_0 r^2 \epsilon_0}{R^3} \]

### Step 4: Analyzing the Charge Distribution
Since the charge distribution is spherically symmetric, the charge enclosed within the radius \(r\) can be related to the volume of the sphere of radius \(r\):

\[ Q_{\text{enc}} = \rho \cdot \frac{4}{3} \pi r^3 \]

where \(\rho\) is the charge density.

### Step 5: Equating the Two Expressions for Charge Enclosed
Now we can equate the two expressions for \(Q_{\text{enc}}\):

\[ \frac{\Phi_0 r^2 \epsilon_0}{R^3} = \rho \cdot \frac{4}{3} \pi r^3 \]

### Step 6: Solving for Charge Density
Rearranging the equation gives us:

\[ \rho = \frac{3 \Phi_0 \epsilon_0}{4 \pi R^3} \cdot \frac{1}{r} \]

This shows how the charge density varies with the radius \(r\).

### Final Expression
Thus, we have derived the relationship between the electric flux, the charge enclosed, and the charge density in a spherically symmetric charge distribution.

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