A regular pentagon has four charges each +q at four of its vertices. At the centre of the pentagon, a charge +q is kept. If the distance of a vertex from the center is a the magnitude of the net force acting on the charge at the center is

Options

A. `(q^(2))/(4 p e_(0)a^(2))`
B. zero
C. `(4q^(2))/( 4 p e_(0)a^(2))`
D. `(2q^(2))/(4 p e_(0)a^(2))`

Explanation

To solve the problem, we need to find the net force acting on a charge +q placed at the center of a regular pentagon with four other charges +q located at four of its vertices. The distance from the center to a vertex is given as 'a'.

### Step-by-Step Solution:

1. **Understanding the Configuration**:
- We have a regular pentagon with vertices labeled A, B, C, D, and E.
- Charges +q are placed at vertices A, B, C, and D.
- A charge +q is placed at the center O of the pentagon.
- The distance from the center O to any vertex (A, B, C, or D) is 'a'.

2. **Electric Field Due to One Charge**:
- The electric field \( E \) due to a point charge \( +q \) at a distance \( a \) is given by:
\[ E = \frac{k \cdot q}{a^2} \]
where \( k \) is Coulomb's constant.

3. **Direction of Electric Fields**:
- The electric field at point O due to each of the charges at the vertices A, B, C, and D will point away from the charges since they are all positive.
- Due to the symmetry of the pentagon, the horizontal components of the electric fields from charges at A, B, C, and D will cancel out, while the vertical components will add up.

4. **Calculating the Net Electric Field**:
- The vertical components of the electric fields from the four charges can be calculated. Each charge contributes an electric field \( E \) at an angle of \( 72^\circ \) (the angle between adjacent vertices of a pentagon).
- The vertical component of the electric field from one charge is:
\[ E_{y} = E \cdot \sin(72^\circ) = \frac{k \cdot q}{a^2} \cdot \sin(72^\circ) \]
- Since there are four charges, the total vertical component of the electric field at the center is:
\[ E_{net} = 4 \cdot E_{y} = 4 \cdot \frac{k \cdot q}{a^2} \cdot \sin(72^\circ) \]

5. **Calculating the Net Force**:
- The net force \( F \) acting on the charge at the center due to the electric field is given by:
\[ F = Q \cdot E_{net} = q \cdot \left( 4 \cdot \frac{k \cdot q}{a^2} \cdot \sin(72^\circ) \right) \]
- Simplifying this, we get:
\[ F = \frac{4kq^2 \sin(72^\circ)}{a^2} \]

6. **Final Result**:
- The magnitude of the net force acting on the charge at the center is:
\[ F = \frac{4kq^2 \sin(72^\circ)}{a^2} \]