For spherically symmetrical charge distribution, electric field at a distance r from the centre of sphere is `vec(E)=kr^(7) vec(r)`, where k is a constant. What will be the volume charge density at a distance r from the centre of sphere?

Options

A. `rho=9k epsilon_(0)r^(6)`
B. `rho=5k epsilon_(0)r^(3)`
C. `rho=3k epsilon_(0)r^(6)`
D. `rho=9k epsilon_(0)r^(0)`

Explanation

To find the volume charge density at a distance \( r \) from the center of a sphere with a given electric field \( \vec{E} = k r^7 \vec{r} \), we can follow these steps:

### Step 1: Understand the Electric Field and Gauss's Law
The electric field is given as \( \vec{E} = k r^7 \vec{r} \). According to Gauss's law, the electric flux \( \Phi \) through a closed surface is equal to the charge \( Q_{\text{enc}} \) enclosed by that surface divided by the permittivity of free space \( \epsilon_0 \):
\[ \Phi = \frac{Q_{\text{enc}}}{\epsilon_0} \]

### Step 2: Calculate the Electric Flux
The electric flux \( \Phi \) through a spherical Gaussian surface of radius \( r \) is given by:
\[ \Phi = E \cdot A \]
where \( A \) is the surface area of the sphere, \( A = 4\pi r^2 \). Thus,
\[ \Phi = E \cdot 4\pi r^2 = (k r^7) \cdot (4\pi r^2) = 4\pi k r^9 \]

### Step 3: Relate Electric Flux to Charge Enclosed
From Gauss's law, we have:
\[ 4\pi k r^9 = \frac{Q_{\text{enc}}}{\epsilon_0} \]
The charge enclosed \( Q_{\text{enc}} \) can also be expressed in terms of the volume charge density \( \rho \):
\[ Q_{\text{enc}} = \rho \cdot V = \rho \cdot \frac{4}{3}\pi r^3 \]

### Step 4: Set the Equations Equal
Equating the two expressions for \( Q_{\text{enc}} \):
\[ 4\pi k r^9 = \frac{\rho \cdot 4}{3}\pi r^3 \]

### Step 5: Simplify the Equation
Cancel \( 4\pi \) from both sides:
\[ k r^9 = \frac{\rho}{3} r^3 \]
Now, multiply both sides by 3:
\[ 3k r^9 = \rho r^3 \]

### Step 6: Solve for Volume Charge Density \( \rho \)
Now, divide both sides by \( r^3 \):
\[ \rho = 3k r^6 \]

### Final Answer
Thus, the volume charge density at a distance \( r \) from the center of the sphere is:
\[ \rho = 3k \epsilon_0 r^6 \]