Four particles each having a charge q are placed on the four vertices of a regular pentagon. The distance of each corner from the centre is 'a'. Find the electric field at the centre of pentagon.

Explanation

To find the electric field at the center of a regular pentagon with four charges \( q \) placed at its vertices, we can follow these steps:

### Step 1: Understand the Configuration
We have a regular pentagon with four charges \( q \) located at four of its vertices. The distance from the center of the pentagon to each vertex is \( a \).

### Step 2: Calculate the Electric Field Due to One Charge
The electric field \( E \) due to a point charge \( q \) at a distance \( r \) is given by the formula:
\[ E = \frac{k \cdot q}{r^2} \]
where \( k \) is Coulomb's constant. Since the distance from the center to each vertex is \( a \), the electric field due to one charge at the center will be:
\[ E = \frac{k \cdot q}{a^2} \]

### Step 3: Determine the Direction of the Electric Fields
The electric field vectors due to each charge will point away from the charge if the charge is positive. Since all four charges are equal and positive, the electric fields will point outward from each charge toward the center.

### Step 4: Analyze the Symmetry
Due to the symmetry of the pentagon, the electric fields produced by the charges at the vertices will have components that cancel each other out. Specifically, for every charge, there is another charge opposite to it that will produce an electric field in the opposite direction.

### Step 5: Calculate the Net Electric Field
Since the charges are symmetrically placed, the horizontal components of the electric fields will cancel out, and the vertical components will also cancel out. Therefore, the net electric field at the center of the pentagon due to the four charges will be:
\[ E_{\text{net}} = 0 \]

### Conclusion
The electric field at the center of the pentagon, due to the four charges placed at its vertices, is zero.