The electric field is a region (relative permittivity = 1) varies with distance "r" from origin according to relation `E = Ar^2hatr`, where A is a positive constant. The volume charge density in the region is proportional to

Options

A. r
B. `r^2`
C. `1//r`
D. `1//r^2`

Answer

To find the volume charge density in the region where the electric field varies according to the relation \( E = Ar^2 \hat{r} \), we can use Gauss's law. Here’s the step-by-step solution:

### Step 1: Understanding the Electric Field
The electric field is given by:
\[ E = Ar^2 \hat{r} \]
where \( A \) is a positive constant and \( r \) is the distance from the origin.

### Step 2: Applying Gauss's Law
According to Gauss's law, the electric flux through a closed surface is equal to the charge enclosed divided by the permittivity of free space (\( \epsilon_0 \)):
\[ \Phi_E = \frac{Q_{\text{enc}}}{\epsilon_0} \]
The electric flux \( \Phi_E \) through a spherical surface of radius \( r \) is given by:
\[ \Phi_E = E \cdot A = E \cdot (4\pi r^2) \]
Substituting the expression for \( E \):
\[ \Phi_E = (Ar^2) \cdot (4\pi r^2) = 4\pi A r^4 \]

### Step 3: Charge Enclosed
The charge enclosed \( Q_{\text{enc}} \) can be expressed in terms of the volume charge density \( \rho \):
\[ Q_{\text{enc}} = \int \rho \, dV \]
For a spherical shell of thickness \( dr \), the volume element \( dV \) is:
\[ dV = 4\pi r^2 \, dr \]
Thus, the total charge enclosed is:
\[ Q_{\text{enc}} = \int_0^r \rho \cdot 4\pi r'^2 \, dr' \]

### Step 4: Equating the Two Expressions
From Gauss's law, we have:
\[ 4\pi A r^4 = \frac{Q_{\text{enc}}}{\epsilon_0} \]
Substituting the expression for \( Q_{\text{enc}} \):
\[ 4\pi A r^4 = \frac{1}{\epsilon_0} \int_0^r \rho \cdot 4\pi r'^2 \, dr' \]

### Step 5: Simplifying the Equation
We can simplify this to:
\[ A r^4 = \frac{1}{\epsilon_0} \int_0^r \rho \cdot r'^2 \, dr' \]

### Step 6: Differentiating Both Sides
Differentiating both sides with respect to \( r \):
\[ \frac{d}{dr}(A r^4) = \frac{d}{dr}\left(\frac{1}{\epsilon_0} \int_0^r \rho \cdot r'^2 \, dr'\right) \]
This gives:
\[ 4A r^3 = \frac{1}{\epsilon_0} \rho \cdot r^2 \]

### Step 7: Solving for Volume Charge Density
Rearranging the equation, we find:
\[ \rho = 4\epsilon_0 A r \]
This shows that the volume charge density \( \rho \) is proportional to \( r \):
\[ \rho \propto r \]

### Final Result
Thus, the volume charge density in the region is proportional to \( r \).

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