A square of side a charge Q at its centre and charge q at one of the corners. The work required to be done in moving the charge q from the corner to the diagonally opposite corner is

Options

A. zero
B. `(Qq)/(4 pi epsilo_(0)a)`
C. `(Qqsqrt(2))/(4piepsilon_(0)a)`
D. `(Qq)/(2piepsilon_(0)a)`

Explanation

To solve the problem of calculating the work required to move a charge \( q \) from one corner of a square to the diagonally opposite corner, with a charge \( Q \) placed at the center of the square, we can follow these steps:

### Step 1: Understand the Configuration
We have a square with side length \( a \). The charge \( Q \) is located at the center of the square, and we are moving charge \( q \) from one corner (let's say corner A) to the diagonally opposite corner (corner C).

### Step 2: Calculate the Distance from Charge \( Q \) to Each 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 given by:
\[ R = \frac{a}{\sqrt{2}} \]

### Step 3: Calculate the Electric Potential at Each Corner
The electric potential \( V \) at a point due to a point charge is given by:
\[ V = \frac{kQ}{r} \]
where \( k \) is Coulomb's constant and \( r \) is the distance from the charge to the point where we are calculating the potential.

- **Potential at Corner A (where charge \( q \) starts):**
\[ V_A = \frac{kQ}{\frac{a}{\sqrt{2}}} = \frac{kQ \sqrt{2}}{a} \]

- **Potential at Corner C (where charge \( q \) moves to):**
Since the distance from the center to corner C is the same as to corner A, we have:
\[ V_C = \frac{kQ}{\frac{a}{\sqrt{2}}} = \frac{kQ \sqrt{2}}{a} \]

### Step 4: Calculate the Change in Electric Potential
The change in electric potential \( \Delta V \) when moving from corner A to corner C is:
\[ \Delta V = V_C - V_A = 0 \]

### Step 5: Calculate the Work Done
The work done \( W \) in moving a charge in an electric field is given by:
\[ W = q \Delta V \]
Substituting the value of \( \Delta V \):
\[ W = q \cdot 0 = 0 \]

### Conclusion
The work required to move the charge \( q \) from one corner of the square to the diagonally opposite corner is:
\[ \boxed{0} \]