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

Options

A. `(4q)/(3piepsi_(0)x)`
B. `(4q)/(sqrt3pi epsi_(0)x)`
C. `(3q)/(4 pi epsi_(0)x)`
D. `(2q)/(sqrt3piepsi_(0)x)`

Explanation

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

### Step 1: Understand the Geometry of the Cube
A cube has 8 vertices, and we need to find the potential at the center of the cube due to the charges at these vertices.

### Step 2: Calculate the Distance from the Center to a Vertex
The distance from the center of the cube to any vertex can be calculated using the formula for the diagonal of the cube. The length of the diagonal \( d \) of a cube with side length \( x \) is given by:
\[ d = \sqrt{x^2 + x^2 + x^2} = \sqrt{3x^2} = x\sqrt{3} \]
Since we want the distance from the center to a vertex, we take half of this diagonal:
\[ r = \frac{d}{2} = \frac{x\sqrt{3}}{2} \]

### Step 3: Formula for Electric Potential
The electric potential \( V \) due to a point charge \( q \) at a distance \( r \) is given by:
\[ V = k \frac{q}{r} \]
where \( k = \frac{1}{4\pi\epsilon_0} \).

### Step 4: Calculate the Total Potential at the Center
Since there are 8 charges at the vertices, the total potential \( V_{\text{total}} \) at the center of the cube is the sum of the potentials due to each charge:
\[ V_{\text{total}} = 8 \times V = 8 \times k \frac{q}{r} \]
Substituting for \( r \):
\[ V_{\text{total}} = 8 \times k \frac{q}{\frac{x\sqrt{3}}{2}} = 8 \times k \frac{2q}{x\sqrt{3}} = \frac{16kq}{x\sqrt{3}} \]

### Step 5: Substitute the Value of \( k \)
Now substituting \( k = \frac{1}{4\pi\epsilon_0} \):
\[ V_{\text{total}} = \frac{16 \cdot \frac{1}{4\pi\epsilon_0} \cdot q}{x\sqrt{3}} = \frac{4q}{\pi\epsilon_0 x\sqrt{3}} \]

### Final Answer
Thus, the potential at the center of the cube due to the charges at its vertices is:
\[ V = \frac{4q}{\pi\epsilon_0 x\sqrt{3}} \]