A hexagon of side 8 cm has a charge `4 muC` at each of its vertices. The potential at the centre of the hexagon is

Options

A. `2.7xx10^(6)V`
B. `7.2xx10^(11)V`
C. `2.5xx10^(12)V`
D. `3.4xx10^(4)V`

Explanation

To find the electric potential at the center of a hexagon with a charge of \(4 \, \mu C\) at each vertex, we can follow these steps:

### Step 1: Understand the Geometry of the Hexagon
A regular hexagon has all sides equal, and the distance from the center to any vertex is equal to the length of a side. In this case, the side length \(a\) is given as \(8 \, cm\).

### Step 2: Convert Units
Convert the side length from centimeters to meters for consistency in SI units:
\[ a = 8 \, cm = 0.08 \, m \]

### 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:
\[ V = \frac{kQ}{r} \]
where \(k\) is Coulomb's constant, approximately \(9 \times 10^9 \, N \cdot m^2/C^2\).

### Step 4: Calculate the Potential Due to One Charge
For one charge \(Q = 4 \, \mu C = 4 \times 10^{-6} \, C\) located at a distance \(r = a = 0.08 \, m\):
\[ V_{one} = \frac{(9 \times 10^9) \times (4 \times 10^{-6})}{0.08} \]

### Step 5: Calculate the Total Potential at the Center
Since there are 6 charges at the vertices of the hexagon, the total potential \(V_{total}\) at the center is:
\[ V_{total} = 6 \times V_{one} \]
Substituting the expression for \(V_{one}\):
\[ V_{total} = 6 \times \frac{(9 \times 10^9) \times (4 \times 10^{-6})}{0.08} \]

### Step 6: Simplify the Expression
Calculating \(V_{one}\):
\[ V_{one} = \frac{(9 \times 10^9) \times (4 \times 10^{-6})}{0.08} = \frac{36 \times 10^3}{0.08} = 450000 \, V = 4.5 \times 10^5 \, V \]
Now substituting back to find \(V_{total}\):
\[ V_{total} = 6 \times 4.5 \times 10^5 = 27 \times 10^5 \, V = 2.7 \times 10^6 \, V \]

### Final Answer
The potential at the center of the hexagon is:
\[ V_{total} = 2.7 \times 10^6 \, V \]