Four charges `+q, +q, -q, and -q` are placed, respectively, at the corners `A,B,C, and D` of a square of side `a`, arranged in the given order. If `O` is the center of square.
The electric field at `O` is.

Options

A. `q/(sqrt(2) pi epsilon_0 a^2)`
B. `q/(sqrt(3) pi epsilon_0 a^2)`
C. `(sqrt(3)q)/( pi epsilon_0 a^2)`
D. `(sqrt(2)q)/( pi epsilon_0 a^2)`

Explanation

To find the electric field at the center \( O \) of a square formed by four charges \( +q, +q, -q, -q \) at the corners \( A, B, C, \) and \( D \) respectively, we can follow these steps:

### Step 1: Identify the charges and their positions
- Place the charges at the corners of the square:
- Charge \( +q \) at corner \( A \) (top left)
- Charge \( +q \) at corner \( B \) (top right)
- Charge \( -q \) at corner \( C \) (bottom right)
- Charge \( -q \) at corner \( D \) (bottom left)

### Step 2: Determine the distance from the center \( O \) to each charge
- The distance from the center \( O \) to each corner of the square is given by:
\[ r = \frac{a}{\sqrt{2}} \]
where \( a \) is the side length of the square.

### Step 3: Calculate the electric field due to each charge at point \( O \)
- The electric field \( E \) due to a point charge \( q \) at a distance \( r \) is given by:
\[ E = \frac{k |q|}{r^2} \]
where \( k \) is Coulomb's constant \( \left( k = \frac{1}{4\pi \epsilon_0} \right) \).

- For each charge, the electric field at \( O \) can be calculated:
- For charges \( +q \) at \( A \) and \( B \):
\[ E_A = E_B = \frac{kq}{\left(\frac{a}{\sqrt{2}}\right)^2} = \frac{2kq}{a^2} \]
(directed away from the charges)

- For charges \( -q \) at \( C \) and \( D \):
\[ E_C = E_D = \frac{kq}{\left(\frac{a}{\sqrt{2}}\right)^2} = \frac{2kq}{a^2} \]
(directed towards the charges)

### Step 4: Determine the direction of the electric fields
- The electric field \( E_A \) and \( E_B \) are directed away from \( A \) and \( B \) respectively, while \( E_C \) and \( E_D \) are directed towards \( C \) and \( D \) respectively.

### Step 5: Resolve the electric fields into components
- The electric fields \( E_A \) and \( E_B \) will have components along the x-axis and y-axis:
- For \( E_A \) (upward and left):
\[ E_{Ax} = -E_A \cos(45^\circ) = -\frac{2kq}{a^2} \cdot \frac{1}{\sqrt{2}} = -\frac{2kq}{a^2\sqrt{2}} \]
\[ E_{Ay} = E_A \sin(45^\circ) = \frac{2kq}{a^2} \cdot \frac{1}{\sqrt{2}} = \frac{2kq}{a^2\sqrt{2}} \]

- For \( E_B \) (upward and right):
\[ E_{Bx} = E_B \cos(45^\circ) = \frac{2kq}{a^2} \cdot \frac{1}{\sqrt{2}} = \frac{2kq}{a^2\sqrt{2}} \]
\[ E_{By} = E_B \sin(45^\circ) = \frac{2kq}{a^2} \cdot \frac{1}{\sqrt{2}} = \frac{2kq}{a^2\sqrt{2}} \]

- For \( E_C \) and \( E_D \) (downward):
- The components will be:
\[ E_{Cx} = E_C \cos(45^\circ) = \frac{2kq}{a^2} \cdot \frac{1}{\sqrt{2}} = \frac{2kq}{a^2\sqrt{2}} \]
\[ E_{Cy} = -E_C \sin(45^\circ) = -\frac{2kq}{a^2} \cdot \frac{1}{\sqrt{2}} = -\frac{2kq}{a^2\sqrt{2}} \]

- For \( E_D \):
\[ E_{Dx} = -E_D \cos(45^\circ) = -\frac{2kq}{a^2} \cdot \frac{1}{\sqrt{2}} = -\frac{2kq}{a^2\sqrt{2}} \]
\[ E_{Dy} = -E_D \sin(45^\circ) = -\frac{2kq}{a^2} \cdot \frac{1}{\sqrt{2}} = -\frac{2kq}{a^2\sqrt{2}} \]

### Step 6: Sum the components of the electric fields
- The x-components:
\[ E_{net_x} = E_{Ax} + E_{Bx} + E_{Cx} + E_{Dx} = -\frac{2kq}{a^2\sqrt{2}} + \frac{2kq}{a^2\sqrt{2}} + \frac{2kq}{a^2\sqrt{2}} - \frac{2kq}{a^2\sqrt{2}} = 0 \]

- The y-components:
\[ E_{net_y} = E_{Ay} + E_{By} + E_{Cy} + E_{Dy} = \frac{2kq}{a^2\sqrt{2}} + \frac{2kq}{a^2\sqrt{2}} - \frac{2kq}{a^2\sqrt{2}} - \frac{2kq}{a^2\sqrt{2}} = 0 \]

### Final Result
- The net electric field at point \( O \) is:
\[ \mathbf{E_{net}} = E_{net_x} + E_{net_y} = 0 + 0 = 0 \]