The electric field in a region is radially outward with magnitude `E=Agamma_(0)`. The charge contained in a sphere of radius `gamma_(0)` centered at the origin is

Options

A. `1/(4piepsilon_(0)) Agamma_(0)^(3)`
B. `4 piepsilon_(0)Agamma_(0)^(3)`
C. `(4piepsilon_(0)A)/gamma_(0)`
D. `1/(4piepsilon_(0))A/(gamma_(0)^(3)`

Answer

To solve the problem, we need to determine the charge contained in a sphere of radius \( \gamma_0 \) centered at the origin, given that the electric field \( E \) in the region is radially outward with a magnitude of \( E = A \gamma_0 \).

### Step-by-step Solution:

1. **Understanding the Electric Field**:
The electric field \( E \) is given as \( E = A \gamma_0 \). This means that the electric field strength is proportional to the distance \( \gamma_0 \) from the origin, with \( A \) being a constant.

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

3. **Calculating the Electric Flux**:
The electric flux through a spherical surface of radius \( \gamma_0 \) is given by:
\[ \Phi_E = E \cdot A \]
where \( A \) is the surface area of the sphere. The surface area \( A \) of a sphere is given by \( 4\pi r^2 \). Therefore, for our sphere of radius \( \gamma_0 \):
\[ A = 4\pi (\gamma_0)^2 \]
Thus, the electric flux becomes:
\[ \Phi_E = E \cdot 4\pi (\gamma_0)^2 = (A \gamma_0) \cdot 4\pi (\gamma_0)^2 = 4\pi A \gamma_0^3 \]

4. **Relating Electric Flux to Charge**:
From Gauss's Law, we have:
\[ \Phi_E = \frac{Q}{\epsilon_0} \]
Setting the two expressions for electric flux equal gives:
\[ 4\pi A \gamma_0^3 = \frac{Q}{\epsilon_0} \]

5. **Solving for Charge \( Q \)**:
Rearranging the equation to solve for \( Q \):
\[ Q = 4\pi A \gamma_0^3 \epsilon_0 \]

6. **Identifying the Correct Option**:
Now we can compare this expression for \( Q \) with the options provided in the question. The expression can be rewritten as:
\[ Q = 4\pi \epsilon_0 A \gamma_0^3 \]
This matches with option 2, which states:
\[ Q = 4\pi \epsilon_0 A \gamma_0^3 \]

### Conclusion:
The charge contained in the sphere of radius \( \gamma_0 \) centered at the origin is:
\[ Q = 4\pi \epsilon_0 A \gamma_0^3 \]
Thus, the correct answer is **Option 2**.