The electric field in a certain region is acting radially outwards and is given by `E=Ar. A` charge contained in a sphere of radius `'a'` centred at the origin of the field, will given by

Options

A. `Aepsilon_(0)a^(2)`
B. `4pi epsilon_(0)Aa^(2)`
C. `epsilon_(0)Aa^(3)`
D. `4pi epsilon_(0)Aa^(3)`

Explanation

To solve the problem, we need to find the charge contained within a sphere of radius 'a' centered at the origin, given the electric field \( E = Ar \), where \( A \) is a constant and \( r \) is the radial distance from the origin.

### Step-by-Step Solution:

1. **Understand the Electric Field**: The electric field is given as \( E = Ar \). This indicates that the electric field increases linearly with distance from the origin.

2. **Determine the Area of the Sphere**: The surface area \( A \) of a sphere with radius \( a \) is given by the formula:
\[ A = 4\pi a^2 \]

3. **Calculate the Electric Flux**: The electric flux \( \Phi_E \) through the surface of the sphere is given by:
\[ \Phi_E = E \cdot A \]
Substituting the values we have:
\[ \Phi_E = E \cdot (4\pi a^2) \]

4. **Substitute the Electric Field**: At the surface of the sphere where \( r = a \):
\[ E = Aa \]
Therefore, the electric flux becomes:
\[ \Phi_E = (Aa) \cdot (4\pi a^2) = 4\pi Aa^3 \]

5. **Use Gauss's Law**: According to Gauss's law, the electric flux through a closed surface is equal to the charge enclosed \( Q_{\text{inside}} \) divided by the permittivity of free space \( \epsilon_0 \):
\[ \Phi_E = \frac{Q_{\text{inside}}}{\epsilon_0} \]

6. **Set the Equations Equal**: From the previous steps, we have:
\[ 4\pi Aa^3 = \frac{Q_{\text{inside}}}{\epsilon_0} \]

7. **Solve for the Charge**: Rearranging the equation gives:
\[ Q_{\text{inside}} = 4\pi Aa^3 \epsilon_0 \]

### Final Answer:
The charge contained in the sphere of radius \( a \) is:
\[ Q_{\text{inside}} = 4\pi \epsilon_0 A a^3 \]