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

Answer

To solve the problem, we need to determine the charge contained within a sphere of radius 'a' centered at the origin, given the electric field \( E = Ar \) which acts radially outwards.

### Step-by-Step Solution:

1. **Understanding the Electric Field**:
The electric field is given by \( E = Ar \), where \( A \) is a constant and \( r \) is the distance from the origin. This indicates that the electric field increases linearly with distance from the origin.

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

3. **Calculating the Electric Flux**:
To find the electric flux through a spherical surface of radius \( a \), we first calculate the area of the sphere:
\[ \text{Area} = 4\pi a^2 \]
The electric field at the surface of the sphere (where \( r = a \)) is:
\[ E = Aa \]
Therefore, the electric flux through the surface is:
\[ \Phi_E = E \times \text{Area} = (Aa)(4\pi a^2) = 4\pi Aa^3 \]

4. **Relating Flux to Charge**:
According to Gauss's Law:
\[ 4\pi Aa^3 = \frac{Q_{\text{inside}}}{\epsilon_0} \]
Rearranging gives:
\[ Q_{\text{inside}} = 4\pi Aa^3 \epsilon_0 \]

5. **Final Expression for Charge**:
Thus, the charge contained within the sphere of radius \( a \) is:
\[ Q_{\text{inside}} = 4\pi \epsilon_0 A a^3 \]

### Conclusion:
The charge contained in a sphere of radius \( a \) centered at the origin is given by:
\[ Q_{\text{inside}} = 4\pi \epsilon_0 A a^3 \]