Four point charges are respectively placed at the corners A, B, C and D of a square of side a. If F is the mid-point of side CD, the work done in carrying an electron of charge from O [the centre of the square] to F fill be:

Options

A. `(qe)/(pi epsi_(0)a)(1-(1)/sqrt(5))`
B. `(qe)/(pi epsi_(0) a) (1+(1)/(sqrt(5)))`
C. `(qe)/(pi epsi_(0)a)`
D. Zero

Explanation

To solve the problem of calculating the work done in carrying an electron from the center of a square to the midpoint of one of its sides, we will follow these steps:

### Step 1: Understand the Configuration
We have four point charges placed at the corners of a square ABCD with side length \( a \). Let's denote the charges as follows:
- Charge at A: \( -Q \)
- Charge at B: \( -Q \)
- Charge at C: \( +Q \)
- Charge at D: \( +Q \)

The center of the square is denoted as \( O \), and the midpoint of side \( CD \) is denoted as \( F \).

### Step 2: Calculate the Initial Potential at Point O
The potential \( V \) at a point due to a point charge is given by the formula:
\[ V = \frac{kQ}{r} \]
where \( k \) is Coulomb's constant, \( Q \) is the charge, and \( r \) is the distance from the charge to the point.

At point \( O \) (the center of the square), the distances from \( O \) to each of the charges (A, B, C, D) are equal and can be calculated using the Pythagorean theorem:
\[ r = \frac{a}{\sqrt{2}} \]

The potential at \( O \) due to each charge is:
- Due to \( -Q \) at A: \( V_A = \frac{-kQ}{\frac{a}{\sqrt{2}}} = \frac{-\sqrt{2}kQ}{a} \)
- Due to \( -Q \) at B: \( V_B = \frac{-kQ}{\frac{a}{\sqrt{2}}} = \frac{-\sqrt{2}kQ}{a} \)
- Due to \( +Q \) at C: \( V_C = \frac{kQ}{\frac{a}{\sqrt{2}}} = \frac{\sqrt{2}kQ}{a} \)
- Due to \( +Q \) at D: \( V_D = \frac{kQ}{\frac{a}{\sqrt{2}}} = \frac{\sqrt{2}kQ}{a} \)

Adding these potentials together:
\[ V_O = V_A + V_B + V_C + V_D = \frac{-\sqrt{2}kQ}{a} + \frac{-\sqrt{2}kQ}{a} + \frac{\sqrt{2}kQ}{a} + \frac{\sqrt{2}kQ}{a} = 0 \]

### Step 3: Calculate the Potential at Point F
Point \( F \) is the midpoint of side \( CD \). The distances from \( F \) to the charges at C and D are:
- Distance from F to C: \( \frac{a}{2} \)
- Distance from F to D: \( \frac{a}{2} \)
- Distance from F to A: \( \frac{a}{\sqrt{2}} \)
- Distance from F to B: \( \frac{a}{\sqrt{2}} \)

The potential at \( F \) is:
\[ V_F = V_C + V_D + V_A + V_B \]
Calculating each potential:
- \( V_C = \frac{kQ}{\frac{a}{2}} = \frac{2kQ}{a} \)
- \( V_D = \frac{kQ}{\frac{a}{2}} = \frac{2kQ}{a} \)
- \( V_A = \frac{-kQ}{\frac{a}{\sqrt{2}}} = \frac{-\sqrt{2}kQ}{a} \)
- \( V_B = \frac{-kQ}{\frac{a}{\sqrt{2}}} = \frac{-\sqrt{2}kQ}{a} \)

Adding these together:
\[ V_F = \frac{2kQ}{a} + \frac{2kQ}{a} - \frac{\sqrt{2}kQ}{a} - \frac{\sqrt{2}kQ}{a} = \frac{(4 - 2\sqrt{2})kQ}{a} \]

### Step 4: Calculate the Work Done
The work done \( W \) in moving a charge \( q \) from point \( O \) to point \( F \) is given by:
\[ W = q(V_F - V_O) \]
Since \( V_O = 0 \):
\[ W = qV_F = q \cdot \frac{(4 - 2\sqrt{2})kQ}{a} \]
For an electron, \( q = -e \):
\[ W = -e \cdot \frac{(4 - 2\sqrt{2})kQ}{a} \]

### Final Answer
Thus, the work done in carrying an electron of charge from \( O \) to \( F \) is:
\[ W = -\frac{e(4 - 2\sqrt{2})kQ}{a} \]