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. `E and F` are the midpoints of sides `BC and CD`, respectively, `O` is the center of square.
The electric potential at `O` is.

Options

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

Explanation

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

### Step 1: Understand the Configuration
We have a square with side length \( a \). The charges are placed at the corners:
- Charge at \( A \) (top left) = \( +q \)
- Charge at \( B \) (top right) = \( +q \)
- Charge at \( C \) (bottom right) = \( -q \)
- Charge at \( D \) (bottom left) = \( -q \)

The center \( O \) of the square is equidistant from all four corners.

### Step 2: Calculate the Distance from Center to Corners
The distance from the center \( O \) to any corner (say \( A \)) can be calculated using the Pythagorean theorem:
\[ OA = \sqrt{\left(\frac{a}{2}\right)^2 + \left(\frac{a}{2}\right)^2} = \sqrt{\frac{a^2}{4} + \frac{a^2}{4}} = \sqrt{\frac{a^2}{2}} = \frac{a}{\sqrt{2}} \]
Thus, the distance \( OA = OB = OC = OD = \frac{a}{\sqrt{2}} \).

### 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:
\[ V = \frac{1}{4 \pi \epsilon_0} \frac{Q}{R} \]

### Step 4: Calculate the Electric Potential at Point \( O \)
The total electric potential at point \( O \) due to all four charges is the sum of the potentials due to each charge:
\[ V_O = V_A + V_B + V_C + V_D \]

Calculating each term:
- \( V_A = \frac{1}{4 \pi \epsilon_0} \frac{+q}{\frac{a}{\sqrt{2}}} = \frac{q \sqrt{2}}{4 \pi \epsilon_0 a} \)
- \( V_B = \frac{1}{4 \pi \epsilon_0} \frac{+q}{\frac{a}{\sqrt{2}}} = \frac{q \sqrt{2}}{4 \pi \epsilon_0 a} \)
- \( V_C = \frac{1}{4 \pi \epsilon_0} \frac{-q}{\frac{a}{\sqrt{2}}} = -\frac{q \sqrt{2}}{4 \pi \epsilon_0 a} \)
- \( V_D = \frac{1}{4 \pi \epsilon_0} \frac{-q}{\frac{a}{\sqrt{2}}} = -\frac{q \sqrt{2}}{4 \pi \epsilon_0 a} \)

### Step 5: Combine the Potentials
Now, we can combine these potentials:
\[ V_O = \left( \frac{q \sqrt{2}}{4 \pi \epsilon_0 a} + \frac{q \sqrt{2}}{4 \pi \epsilon_0 a} - \frac{q \sqrt{2}}{4 \pi \epsilon_0 a} - \frac{q \sqrt{2}}{4 \pi \epsilon_0 a} \right) \]
This simplifies to:
\[ V_O = 0 \]

### Conclusion
The electric potential at the center \( O \) of the square is \( 0 \).

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