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 (ρ) at a distance r from the center of a sphere with a given electric field \( \vec{E} = k r^7 \hat{r} \), we will use Gauss's law.

### Step-by-Step Solution:

1. **Understand the Electric Field**:
The electric field is given as \( \vec{E} = k r^7 \hat{r} \), where \( k \) is a constant, \( r \) is the distance from the center of the sphere, and \( \hat{r} \) is the unit vector in the radial direction.

2. **Apply Gauss's Law**:
Gauss's law states that the electric flux through a closed surface is equal to the charge enclosed divided by the permittivity of free space (\( \epsilon_0 \)):
\[ \Phi_E = \frac{Q_{\text{in}}}{\epsilon_0} \]
The electric flux \( \Phi_E \) through a spherical surface of radius \( r \) is given by:
\[ \Phi_E = E \cdot A = E \cdot 4\pi r^2 \]
where \( A = 4\pi r^2 \) is the surface area of the sphere.

3. **Substitute the Electric Field**:
Substitute the expression for \( E \):
\[ \Phi_E = (k r^7) \cdot (4\pi r^2) = 4\pi k r^9 \]

4. **Set Up the Charge Enclosed**:
The charge enclosed \( Q_{\text{in}} \) can be expressed in terms of the volume charge density \( \rho \):
\[ Q_{\text{in}} = \int_0^r \rho \, dV = \int_0^r \rho \cdot 4\pi r'^2 \, dr' \]
where \( r' \) is a variable of integration from 0 to \( r \).

5. **Express the Volume Element**:
The volume element \( dV \) for a spherical shell at radius \( r' \) is \( 4\pi r'^2 \, dr' \).

6. **Combine the Equations**:
From Gauss's law, we equate the two expressions:
\[ 4\pi k r^9 = \frac{Q_{\text{in}}}{\epsilon_0} \]
Therefore,
\[ Q_{\text{in}} = 4\pi k r^9 \epsilon_0 \]

7. **Integrate to Find Charge Density**:
Now we substitute \( Q_{\text{in}} \):
\[ 4\pi k r^9 \epsilon_0 = \int_0^r \rho \cdot 4\pi r'^2 \, dr' \]
This simplifies to:
\[ 4\pi k r^9 \epsilon_0 = 4\pi \int_0^r \rho r'^2 \, dr' \]
Dividing both sides by \( 4\pi \):
\[ k r^9 \epsilon_0 = \int_0^r \rho r'^2 \, dr' \]

8. **Differentiate with Respect to r**:
To find \( \rho \), differentiate both sides with respect to \( r \):
\[ \frac{d}{dr}(k r^9 \epsilon_0) = \rho r^2 \]
The left-hand side becomes:
\[ 9k r^8 \epsilon_0 \]
Thus, we have:
\[ \rho r^2 = 9k r^8 \epsilon_0 \]
Therefore, solving for \( \rho \):
\[ \rho = \frac{9k \epsilon_0}{r^2} r^2 = 9k \epsilon_0 r^6 \]

### Final Result:
The volume charge density at a distance \( r \) from the center of the sphere is:
\[ \rho = 9k \epsilon_0 r^6 \]