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 work done in carrying a charge `e` from `O` to `F` is.

Options

A. `(sqrt(2) q e)/(pi epsilon_0 a)`
B. `(q e)/(pi epsilon_0 a)[(1)/(sqrt(5)) - 1]`
C. `(q e)/(pi epsilon_0 a)[(1)/(sqrt(5)) + 1]`
D. zero

Explanation

To solve the problem of calculating the work done in carrying a charge \( e \) from point \( O \) (the center of the square) to point \( F \) (the midpoint of side \( CD \)), we will follow these steps:

### Step 1: Understand the Configuration
We have four charges placed at the corners of a square:
- Charge \( +q \) at point \( A \)
- Charge \( +q \) at point \( B \)
- Charge \( -q \) at point \( C \)
- Charge \( -q \) at point \( D \)

The square has a side length of \( a \). Points \( E \) and \( F \) are the midpoints of sides \( BC \) and \( CD \), respectively, and point \( O \) is the center of the square.

### Step 2: Calculate the Potential at Point \( F \)
The potential \( V_F \) at point \( F \) is the sum of the potentials due to all four charges:
\[ V_F = V_A + V_B + V_C + V_D \]
Where:
- \( V_A = \frac{1}{4\pi \epsilon_0} \frac{q}{r_{AF}} \)
- \( V_B = \frac{1}{4\pi \epsilon_0} \frac{q}{r_{BF}} \)
- \( V_C = \frac{1}{4\pi \epsilon_0} \frac{-q}{r_{CF}} \)
- \( V_D = \frac{1}{4\pi \epsilon_0} \frac{-q}{r_{DF}} \)

### Step 3: Calculate Distances
To find the distances \( r_{AF}, r_{BF}, r_{CF}, \) and \( r_{DF} \):
- The distance from \( A \) to \( F \) and from \( B \) to \( F \) can be calculated using the Pythagorean theorem. Since \( F \) is the midpoint of \( CD \), the coordinates of \( F \) are \( \left(\frac{a}{2}, a\right) \) and those of \( A \) and \( B \) are \( (0, a) \) and \( (a, a) \) respectively.
- Thus, \( r_{AF} = r_{BF} = \sqrt{\left(\frac{a}{2}\right)^2 + \left(0\right)^2} = \frac{a}{\sqrt{2}} \).
- The distances \( r_{CF} \) and \( r_{DF} \) are \( \frac{a}{2} \).

### Step 4: Substitute Distances into Potential Formula
Now substituting the distances into the potential equations:
\[ V_F = \frac{1}{4\pi \epsilon_0} \left( \frac{q}{\frac{a}{\sqrt{2}}} + \frac{q}{\frac{a}{\sqrt{2}}} - \frac{q}{\frac{a}{2}} - \frac{q}{\frac{a}{2}} \right) \]
This simplifies to:
\[ V_F = \frac{1}{4\pi \epsilon_0} \left( \frac{2q\sqrt{2}}{a} - \frac{2q}{a} \right) = \frac{q}{4\pi \epsilon_0 a} \left( 2\sqrt{2} - 2 \right) \]
\[ V_F = \frac{q( \sqrt{2} - 1)}{2\pi \epsilon_0 a} \]

### Step 5: Calculate the Potential at Point \( O \)
The potential at the center \( O \) of the square is zero since the contributions from the positive and negative charges cancel each other out:
\[ V_O = 0 \]

### Step 6: Calculate Work Done
The work done \( W \) in moving the charge \( e \) from \( O \) to \( F \) is given by:
\[ W = e(V_F - V_O) = eV_F \]
Substituting \( V_F \):
\[ W = e \cdot \frac{q( \sqrt{2} - 1)}{2\pi \epsilon_0 a} \]

### Final Answer
Thus, the work done in carrying the charge \( e \) from \( O \) to \( F \) is:
\[ W = \frac{eq( \sqrt{2} - 1)}{2\pi \epsilon_0 a} \]