A charge Q is placed at each corner of a cube of side a. The potential at the centre of the cube is

Options

A. `(8Q)/(piin_(a)a)`
B. `(4Q)/(4piin_(0)a)`
C. `(4Q)/(sqrt(3)piin_(0)a)`
D. `(4Q)/(piin_(a)a)`

Answer

To find the potential at the center of a cube with charge \( Q \) placed at each corner, we can follow these steps:

### Step 1: Understand the Geometry of the Cube
The cube has a side length \( a \). Each corner of the cube has a charge \( Q \). The center of the cube is equidistant from all corners.

### Step 2: Calculate the Distance from a Corner to the Center
The distance \( R \) from a corner of the cube to the center can be found using the body diagonal of the cube. The length of the body diagonal \( d \) of a cube with side \( a \) is given by:
\[ d = \sqrt{a^2 + a^2 + a^2} = \sqrt{3a^2} = a\sqrt{3} \]
Since the center is at half the length of the body diagonal, the distance \( R \) is:
\[ R = \frac{d}{2} = \frac{a\sqrt{3}}{2} \]

### Step 3: Use the 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 corners of the cube, the total potential \( V_{\text{net}} \) at the center is the sum of the potentials due to each charge:
\[ V_{\text{net}} = 8 \times k \frac{Q}{R} \]

### Step 5: Substitute the Expression for R
Substituting \( R = \frac{a\sqrt{3}}{2} \) into the equation gives:
\[ V_{\text{net}} = 8 \times k \frac{Q}{\frac{a\sqrt{3}}{2}} = 8 \times k \frac{2Q}{a\sqrt{3}} = \frac{16kQ}{a\sqrt{3}} \]

### Step 6: Substitute the Value of k
Now substituting \( k = \frac{1}{4\pi \epsilon_0} \):
\[ V_{\text{net}} = \frac{16 \cdot \frac{1}{4\pi \epsilon_0} Q}{a\sqrt{3}} = \frac{4Q}{\pi \epsilon_0 a\sqrt{3}} \]

### Final Answer
Thus, the potential at the center of the cube is:
\[ V_{\text{net}} = \frac{4Q}{\pi \epsilon_0 a\sqrt{3}} \]