A cube of side b has a charge q at each of its vertices. Determine the electric field due to this charge array at the center of the cube.

Options

A. `q//b^(2)`
B. `q//2b^(2)`
C. `32q//b^(2)`
D. zero

Answer

To determine the electric field at the center of a cube with a charge \( q \) at each of its vertices, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Geometry of the Cube**:
- A cube has 8 vertices, and we have a charge \( q \) located at each vertex.
- The side length of the cube is given as \( b \).

2. **Identify the Center of the Cube**:
- The center of the cube can be found at the coordinates \( \left( \frac{b}{2}, \frac{b}{2}, \frac{b}{2} \right) \).

3. **Calculate the Distance from the Center to a Vertex**:
- The distance \( r \) from the center of the cube to any vertex can be calculated using the Pythagorean theorem:
\[ r = \sqrt{\left(\frac{b}{2}\right)^2 + \left(\frac{b}{2}\right)^2 + \left(\frac{b}{2}\right)^2} = \sqrt{\frac{3b^2}{4}} = \frac{b\sqrt{3}}{2} \]

4. **Determine the Electric Field due to a Single Charge**:
- The electric field \( E \) due to a single charge \( q \) at a distance \( r \) is given by:
\[ E = \frac{k \cdot q}{r^2} \]
- Substituting for \( r \):
\[ E = \frac{k \cdot q}{\left(\frac{b\sqrt{3}}{2}\right)^2} = \frac{k \cdot q}{\frac{3b^2}{4}} = \frac{4kq}{3b^2} \]

5. **Direction of the Electric Field**:
- The electric field due to each charge points away from the charge if it is positive.
- For each pair of opposite charges, the electric fields will have components that cancel each other out due to symmetry.

6. **Analyze the Contribution of All Charges**:
- Since there are 8 charges symmetrically placed at the vertices, the contributions to the electric field from opposite pairs will cancel out.
- For every charge at a vertex, there is an equal and opposite charge at the diagonally opposite vertex, leading to cancellation of the electric fields.

7. **Conclusion**:
- The net electric field at the center of the cube due to all the charges at the vertices is:
\[ E_{\text{net}} = 0 \]

### Final Answer:
The electric field at the center of the cube is zero.