Four charges `+q, -q, +q` and `-q` are placed in order on the four consecutive corners of a square of side a. The work done in interchanging the positions of any two neighboring charges of the opposite sign is

Options

A. `q^2/(4piepsilon_0a)(-4+sqrt2)`
B. `q^2/(4piepsilon_0a)(4+2sqrt2)`
C. `q^2/(4piepsilon_0a)(4-2sqrt2)`
D. `q^2/(4piepsilon_0a)(4+sqrt2)`

Explanation

To find the work done in interchanging the positions of any two neighboring charges of opposite signs in the given configuration, we will calculate the initial and final potential energy of the system and then determine the work done based on the change in potential energy.

### Step-by-Step Solution:

1. **Identify the Initial Configuration:**
The charges are arranged at the corners of a square:
- Corner 1: +q
- Corner 2: -q
- Corner 3: +q
- Corner 4: -q

2. **Calculate the Initial Potential Energy (U_initial):**
- The potential energy between two charges \( q_1 \) and \( q_2 \) separated by a distance \( r \) is given by:
\[ U = \frac{k q_1 q_2}{r} \]
- For neighboring charges (e.g., +q and -q), the distance is \( a \).
- For diagonal charges (e.g., +q and +q), the distance is \( a\sqrt{2} \).
- The contributions to the initial potential energy are:
- For the pairs of neighboring charges:
\[ U_{\text{neighboring}} = -\frac{kq^2}{a} - \frac{kq^2}{a} - \frac{kq^2}{a} - \frac{kq^2}{a} = -4\frac{kq^2}{a} \]
- For the pairs of diagonal charges:
\[ U_{\text{diagonal}} = \frac{kq^2}{a\sqrt{2}} + \frac{kq^2}{a\sqrt{2}} = 2\frac{kq^2}{a\sqrt{2}} \]
- Therefore, the total initial potential energy is:
\[ U_{\text{initial}} = -4\frac{kq^2}{a} + 2\frac{kq^2}{a\sqrt{2}} \]

3. **Identify the Final Configuration After Interchanging Charges:**
After interchanging the positions of two neighboring charges of opposite signs, the new arrangement will be:
- Corner 1: +q
- Corner 2: +q
- Corner 3: -q
- Corner 4: -q

4. **Calculate the Final Potential Energy (U_final):**
- The contributions to the final potential energy are:
- For the pairs of neighboring charges (now both +q):
\[ U_{\text{neighboring}} = 0 \quad (\text{since they are like charges}) \]
- For the pairs of neighboring charges (now both -q):
\[ U_{\text{neighboring}} = 0 \quad (\text{since they are like charges}) \]
- For the pairs of diagonal charges (one +q and one -q):
\[ U_{\text{diagonal}} = -\frac{kq^2}{a\sqrt{2}} - \frac{kq^2}{a\sqrt{2}} = -2\frac{kq^2}{a\sqrt{2}} \]
- Therefore, the total final potential energy is:
\[ U_{\text{final}} = 0 + 0 - 2\frac{kq^2}{a\sqrt{2}} = -2\frac{kq^2}{a\sqrt{2}} \]

5. **Calculate the Work Done (W):**
The work done by the external agent is equal to the change in potential energy:
\[ W = U_{\text{final}} - U_{\text{initial}} \]
Substituting the values:
\[ W = \left(-2\frac{kq^2}{a\sqrt{2}}\right) - \left(-4\frac{kq^2}{a} + 2\frac{kq^2}{a\sqrt{2}}\right) \]
Simplifying this expression:
\[ W = -2\frac{kq^2}{a\sqrt{2}} + 4\frac{kq^2}{a} - 2\frac{kq^2}{a\sqrt{2}} = 4\frac{kq^2}{a} - 4\frac{kq^2}{a\sqrt{2}} \]
Factoring out common terms:
\[ W = \frac{kq^2}{a}(4 - 2\sqrt{2}) \]

### Final Answer:
The work done in interchanging the positions of any two neighboring charges of opposite sign is:
\[ W = \frac{kq^2}{a}(4 - 2\sqrt{2}) \]