Five point charge (`+q` each) are placed at the five vertices of a regular hexagon of side `2a`.what is the magnitude of the net electric field at the centre of the hexazon?

Options

A. `1/(4piepsilon_0)q/a^2`
B. `q/(16 piepsilon_0a^2)`
C. `(sqrt2q)/(4piepsilon_0a^2)`
D. `(5q)/(16piepsilon_0a^2)`

Explanation

To find the magnitude of the net electric field at the center of a regular hexagon with five point charges of `+q` at its vertices, we can follow these steps:

### Step 1: Understand the Configuration
We have a regular hexagon with vertices labeled A, B, C, D, E, and F. Charges of `+q` are placed at vertices A, B, C, D, and E. Vertex F is empty. The side length of the hexagon is `2a`.

### Step 2: Determine the Distance from the Center to the Vertices
The distance from the center of the hexagon (point C) to any vertex (A, B, D, or E) can be calculated using the formula for the circumradius \( R \) of a regular hexagon:
\[ R = \frac{s}{\sqrt{3}} = \frac{2a}{\sqrt{3}} = \frac{2a\sqrt{3}}{3} \]

### Step 3: Calculate the Electric Field Due to Each Charge
The electric field \( E \) due to a single point charge \( +q \) at a distance \( r \) is given by:
\[ E = \frac{k \cdot q}{r^2} \]
where \( k = \frac{1}{4\pi\epsilon_0} \).

Substituting \( r = \frac{2a\sqrt{3}}{3} \):
\[ E = \frac{k \cdot q}{\left(\frac{2a\sqrt{3}}{3}\right)^2} = \frac{k \cdot q}{\frac{4a^2 \cdot 3}{9}} = \frac{9kq}{12a^2} = \frac{3kq}{4a^2} \]

### Step 4: Determine the Direction of Each Electric Field
The electric fields due to the charges at vertices A, B, C, D, and E will have specific directions:
- The electric fields from charges at A, B, D, and E will point away from the respective charges towards the center of the hexagon.
- The charge at vertex F is missing, which means the electric field due to the missing charge would have pointed towards it.

### Step 5: Calculate the Net Electric Field
The electric fields from the four charges (A, B, D, and E) will add vectorially. Due to symmetry, the horizontal components of the electric fields from charges at A and D will cancel out, as will the components from B and E.

The vertical components will add up. The resultant electric field \( E_{net} \) at the center due to the four charges can be calculated as:
\[ E_{net} = 4 \cdot E \cdot \sin(60^\circ) = 4 \cdot \frac{3kq}{4a^2} \cdot \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}kq}{a^2} \]

### Final Result
Thus, the magnitude of the net electric field at the center of the hexagon is:
\[ E_{net} = \frac{3\sqrt{3}kq}{a^2} \]