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To solve the problem, we need to calculate the work required to move a charge \( q = -40 \, \mu\text{C} \) from infinity to point A in the given configuration. This involves using the concept of electric potential due to point charges.

The four charges and their positions create an electric potential at point A, which we need to calculate first.

1. **Charge positions and distances**:
    - \( q_1 = 30 \, \mu\text{C} \) at (0, 0)
    - \( q_2 = -30 \, \mu\text{C} \) at (0, 50 cm)
    - \( q_3 = 30 \, \mu\text{C} \) at (50 cm, 0)
    - \( q_4 = -30 \, \mu\text{C} \) at (50 cm, 50 cm)

2. **Distance from each charge to point A (50 cm, 50 cm)**:
    - \( r_1 \) from \( q_1 \) to A: \( \sqrt{(50^2 + 50^2)} = 50\sqrt{2} \) cm
    - \( r_2 \) from \( q_2 \) to A: 50 cm
    - \( r_3 \) from \( q_3 \) to A: 50 cm
    - \( r_4 \) from \( q_4 \) to A: \( \sqrt{(50^2 + 50^2)} = 50\sqrt{2} \) cm

3. **Electric potential at point A due to each charge**:
    The electric potential \( V \) due to a point charge is given by \( V = \frac{kq}{r} \), where \( k = 8.99 \times 10^9 \, \text{Nm}^2/\text{C}^2 \).

    - Potential due to \( q_1 \):
      \[
      V_1 = \frac{(8.99 \times 10^9) \times 30 \times 10^{-6}}{50\sqrt{2} \times 10^{-2}} = \frac{(8.99 \times 10^9) \times 30 \times 10^{-6}}{50 \times 1.414 \times 10^{-2}} 
      \]
      
    - Potential due to \( q_2 \):
      \[
      V_2 = \frac{(8.99 \times 10^9) \times (-30) \times 10^{-6}}{50 \times 10^{-2}} = \frac{(8.99 \times 10^9) \times (-30) \times 10^{-6}}{50 \times 10^{-2}}
      \]
      
    - Potential due to \( q_3 \):
      \[
      V_3 = \frac{(8.99 \times 10^9) \times 30 \times 10^{-6}}{50 \times 10^{-2}} = \frac{(8.99 \times 10^9) \times 30 \times 10^{-6}}{50 \times 10^{-2}}
      \]
      
    - Potential due to \( q_4 \):
      \[
      V_4 = \frac{(8.99 \times 10^9) \times (-30) \times 10^{-6}}{50\sqrt{2} \times 10^{-2}} = \frac{(8.99 \times 10^9) \times (-30) \times 10^{-6}}{50 \times 1.414 \times 10^{-2}} 
      \]

4. **Summing the potentials**:
    \[
    V_A = V_1 + V_2 + V_3 + V_4
    \]

    - Compute each term:
      \[
      V_1 = V_4 = \frac{(8.99 \times 10^9) \times 30 \times 10^{-6}}{70.71 \times 10^{-2}} \approx 3.81 \times 10^6 \, \text{V}
      \]
      \[
      V_2 = V_3 = \frac{(8.99 \times 10^9) \times (-30) \times 10^{-6}}{50 \times 10^{-2}} \approx -5.39 \times 10^6 \, \text{V}
      \]

    - Add these together:
      \[
      V_A = 3.81 \times 10^6 + (-5.39 \times 10^6) + 5.39 \times 10^6 + (-3.81 \times 10^6) = 0 \, \text{V}
      \]

5. **Calculate work done**:
    Work done \( W \) to move the charge from infinity to A:
    \[
    W = q \cdot V_A = (-40 \times 10^{-6}) \times 0 = 0 \, \text{J}
    \]

The work required to move the charge \( -40 \, \mu\text{C} \) from infinity to point A is \( 0 \, \text{J} \). 

Do you have any questions or need further details?

### Follow-Up Questions
1. How is electric potential due to a point charge calculated?
2. What is the significance of superposition in calculating electric potential?
3. How does the distance between charges affect the electric potential?
4. What is the principle of work done in moving a charge in an electric field?
5. How would the result change if one of the charges were removed?
6. What if the charge moved had a different magnitude?
7. How do different configurations of charges affect electric potential at a point?
8. Can you explain the concept of electric potential energy?

### Tip
Remember that the electric potential due to multiple charges is the algebraic sum of the potentials due to each individual charge, and distances play a crucial role in calculating these potentials.

Learn how to calculate electric potential and work done in moving charges due to point charges. Solve a specific physics problem involving electric potential and work calculation. Suitable for advanced high school and college students.

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