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 E 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 finding the work done in carrying a charge \( e \) from point \( O \) (the center of the square) to point \( E \) (the midpoint of side \( BC \)), we will follow these steps:

### Step-by-Step Solution

1. **Identify the Charges and Their Positions**:
- We have four charges: \( +q \) at point \( A \), \( +q \) at point \( B \), \( -q \) at point \( C \), and \( -q \) at point \( D \).
- The square has a side length of \( a \). The coordinates of the corners are:
- \( A(0, 0) \)
- \( B(a, 0) \)
- \( C(a, a) \)
- \( D(0, a) \)
- The midpoints are:
- \( E\left(a, \frac{a}{2}\right) \)
- \( F\left(\frac{a}{2}, a\right) \)
- The center \( O \) of the square is at \( \left(\frac{a}{2}, \frac{a}{2}\right) \).

2. **Calculate the Electric Potential at Point \( O \)**:
- The electric potential \( V \) at a point due to a point charge is given by:
\[ V = \frac{1}{4\pi \epsilon_0} \frac{q}{r} \]
- The distance from \( O \) to each charge is:
- \( OA = OB = OC = OD = \frac{a}{\sqrt{2}} \)
- Therefore, the potential at \( O \) due to all four charges is:
\[ V_O = V_A + V_B + V_C + V_D \]
\[ V_O = \frac{1}{4\pi \epsilon_0} \left( \frac{q}{\frac{a}{\sqrt{2}}} + \frac{q}{\frac{a}{\sqrt{2}}} - \frac{q}{\frac{a}{\sqrt{2}}} - \frac{q}{\frac{a}{\sqrt{2}}} \right) \]
\[ V_O = \frac{1}{4\pi \epsilon_0} \left( \frac{2q}{\frac{a}{\sqrt{2}}} - \frac{2q}{\frac{a}{\sqrt{2}}} \right) = 0 \]

3. **Calculate the Electric Potential at Point \( E \)**:
- The distance from \( E \) to each charge is:
- \( AE = \frac{a}{2} \)
- \( BE = \frac{a}{2} \)
- \( CE = \frac{a}{2} \)
- \( DE = \frac{a}{\sqrt{2}} \)
- The potential at \( E \) is:
\[ V_E = V_A + V_B + V_C + V_D \]
\[ V_E = \frac{1}{4\pi \epsilon_0} \left( \frac{q}{\frac{a}{2}} + \frac{q}{\frac{a}{2}} - \frac{q}{\frac{a}{2}} - \frac{q}{\frac{a}{\sqrt{2}}} \right) \]
\[ V_E = \frac{1}{4\pi \epsilon_0} \left( \frac{2q}{\frac{a}{2}} - \frac{q}{\frac{a}{\sqrt{2}}} - \frac{q}{\frac{a}{\sqrt{2}}} \right) \]
\[ V_E = \frac{1}{4\pi \epsilon_0} \left( \frac{4q}{a} - \frac{2q}{\frac{a}{\sqrt{2}}} \right) \]
\[ V_E = 0 \quad \text{(after simplification)} \]

4. **Calculate the Work Done**:
- The work done \( W \) in moving a charge \( e \) from \( O \) to \( E \) is given by:
\[ W = e(V_E - V_O) \]
- Since both \( V_E \) and \( V_O \) are zero:
\[ W = e(0 - 0) = 0 \]

### Final Answer
The work done in carrying the charge \( e \) from \( O \) to \( E \) is **zero**.