To solve the problem, we need to find the charge enclosed inside the cube placed in an electric field defined by \(\vec{E} = 150 y^2 \, \text{N/C}\). We will use Gauss's Law, which states that the electric flux through a closed surface is equal to the charge enclosed divided by the permittivity of free space (\(\epsilon_0\)).
### Step-by-Step Solution:
1. **Understand the Geometry of the Cube**:
The cube has a side length of \(0.5 \, \text{m}\) and is positioned such that one corner is at the origin (0, 0, 0) and extends along the positive y-axis. Therefore, the coordinates of the cube's corners will be:
- (0, 0, 0)
- (0.5, 0, 0)
- (0, 0.5, 0)
- (0, 0, 0.5)
- (0.5, 0.5, 0)
- (0.5, 0, 0.5)
- (0, 0.5, 0.5)
- (0.5, 0.5, 0.5)
2. **Calculate the Electric Field at the Cube**:
The electric field is given by \(\vec{E} = 150 y^2 \, \text{N/C}\). Since the cube extends from \(y = 0\) to \(y = 0.5\), we need to evaluate the electric field at the average position of the cube along the y-axis. The average \(y\) value is:
\[
y_{\text{avg}} = \frac{0 + 0.5}{2} = 0.25 \, \text{m}
\]
Now, substituting \(y_{\text{avg}}\) into the electric field equation:
\[
E = 150 (0.25)^2 = 150 \times 0.0625 = 9.375 \, \text{N/C}
\]
3. **Calculate the Area of the Cube Face**:
The area \(A\) of one face of the cube is given by:
\[
A = \text{side}^2 = (0.5)^2 = 0.25 \, \text{m}^2
\]
4. **Calculate the Electric Flux**:
The electric flux \(\Phi_E\) through the face of the cube in the direction of the electric field is given by:
\[
\Phi_E = E \cdot A = 9.375 \, \text{N/C} \times 0.25 \, \text{m}^2 = 2.34375 \, \text{N m}^2/\text{C}
\]
5. **Use Gauss's Law to Find Charge Enclosed**:
According to Gauss's Law:
\[
\Phi_E = \frac{Q_{\text{enc}}}{\epsilon_0}
\]
Rearranging gives:
\[
Q_{\text{enc}} = \Phi_E \cdot \epsilon_0
\]
Where \(\epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2\). Thus:
\[
Q_{\text{enc}} = 2.34375 \, \text{N m}^2/\text{C} \times 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2
\]
\[
Q_{\text{enc}} \approx 2.08 \times 10^{-11} \, \text{C}
\]
### Final Answer:
The charge enclosed inside the cube is approximately \(2.08 \times 10^{-11} \, \text{C}\).
