Four equal charges `Q` are placed at the four corners of a square of each side is 'a'. Work done in removing a charge `-Q` from its centre to infinity is

Options

A. zero
B. `(sqrt2q^(2))/(pi epsi_(0)a)`
C. `(sqrt2q)/(pi epsi_(0)a)`
D. `(q^(2))/(pi epsi_(0)a)`

Explanation

To solve the problem of calculating the work done in removing a charge `-Q` from the center of a square with four equal charges `Q` at its corners, we can follow these steps:

### Step 1: Understand the Configuration
We have four equal charges `Q` placed at the corners of a square with side length `a`. We need to find the work done in moving a charge `-Q` from the center of the square to infinity.

### Step 2: Determine the Initial Potential Energy
The potential energy \( U \) at a point due to a charge is given by the formula:
\[ U = k \frac{Q_1 Q_2}{r} \]
where \( k \) is Coulomb's constant, \( Q_1 \) and \( Q_2 \) are the magnitudes of the charges, and \( r \) is the distance between them.

### Step 3: Calculate the Distance from the Center to a Corner
The distance from the center of the square to any corner can be calculated using the Pythagorean theorem. The distance \( r \) from the center to a corner is:
\[ r = \frac{a}{\sqrt{2}} \]

### Step 4: Calculate the Total Potential at the Center
Since there are four charges, the total potential \( V \) at the center due to these four charges is:
\[ V = k \left( \frac{Q}{r} + \frac{Q}{r} + \frac{Q}{r} + \frac{Q}{r} \right) = 4k \frac{Q}{r} \]
Substituting \( r = \frac{a}{\sqrt{2}} \):
\[ V = 4k \frac{Q}{\frac{a}{\sqrt{2}}} = \frac{4\sqrt{2}kQ}{a} \]

### Step 5: Calculate the Initial Potential Energy
The initial potential energy \( U_i \) of the charge `-Q` at the center is given by:
\[ U_i = V \cdot (-Q) = -Q \cdot \frac{4\sqrt{2}kQ}{a} = -\frac{4\sqrt{2}kQ^2}{a} \]

### Step 6: Determine the Work Done
The work done \( W \) in moving the charge `-Q` from the center to infinity is equal to the negative of the initial potential energy (since potential energy at infinity is zero):
\[ W = -U_i = \frac{4\sqrt{2}kQ^2}{a} \]

### Step 7: Substitute the Value of \( k \)
Using \( k = \frac{1}{4\pi\epsilon_0} \):
\[ W = \frac{4\sqrt{2} \cdot \frac{1}{4\pi\epsilon_0} Q^2}{a} = \frac{\sqrt{2}Q^2}{\pi\epsilon_0 a} \]

### Final Answer
Thus, the work done in removing the charge `-Q` from the center of the square to infinity is:
\[ W = \frac{\sqrt{2}Q^2}{\pi\epsilon_0 a} \]