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To solve this problem, we need to find the magnitude of the electric field at the top left corner due to the charges \( q_1 = -1.40 \, \mu C \) and \( q_2 = +3.70 \, \mu C \).

### Step 1: Understand the geometry and setup
The charges are placed at the corners of a square of side length \( a = 50.0 \, \text{cm} \) (or \( 0.50 \, \text{m} \)). We'll calculate the electric field created by each charge at the top left corner and then find the resultant electric field using vector addition.

### Step 2: Formula for electric field
The electric field due to a point charge is given by Coulomb's law:
\[
E = \frac{k |q|}{r^2}
\]
where:
- \( E \) is the magnitude of the electric field,
- \( k = 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2 / \text{C}^2 \) is Coulomb's constant,
- \( q \) is the charge,
- \( r \) is the distance from the charge to the point of interest.

### Step 3: Find electric fields due to \( q_1 \) and \( q_2 \)
- **For \( q_1 \)**:
  - Distance: Since \( q_1 \) is horizontally to the right at a distance \( a = 0.50 \, \text{m} \), the distance is \( r = 0.50 \, \text{m} \).
  - Electric field direction: The electric field due to \( q_1 \) will be directed to the right (positive direction).
  
- **For \( q_2 \)**:
  - Distance: \( q_2 \) is vertically below, at a distance \( a = 0.50 \, \text{m} \).
  - Electric field direction: The electric field due to \( q_2 \) will be directed upward (positive y-direction).

### Step 4: Calculate electric fields
Let’s calculate the magnitudes of the electric fields due to \( q_1 \) and \( q_2 \) separately.

For \( q_1 = -1.40 \, \mu C \) at a distance \( 0.50 \, \text{m} \):
\[
E_1 = \frac{k |q_1|}{a^2} = \frac{(8.99 \times 10^9) \times 1.40 \times 10^{-6}}{(0.50)^2}
\]

For \( q_2 = +3.70 \, \mu C \) at a distance \( 0.50 \, \text{m} \):
\[
E_2 = \frac{k |q_2|}{a^2} = \frac{(8.99 \times 10^9) \times 3.70 \times 10^{-6}}{(0.50)^2}
\]

I will now calculate the results.The magnitudes of the electric fields due to the charges are:
- \( E_1 = 50,344 \, \text{N/C} \) (to the right due to \( q_1 \)),
- \( E_2 = 133,052 \, \text{N/C} \) (upward due to \( q_2 \)).

### Step 5: Find the resultant electric field
Since the electric fields are perpendicular to each other (one is horizontal and the other is vertical), the magnitude of the resultant electric field is given by the Pythagorean theorem:
\[
E_{\text{resultant}} = \sqrt{E_1^2 + E_2^2}
\]

Let’s calculate this.It seems like I can’t do more advanced data analysis right now. Please try again later.

However, if you'd like to calculate the resultant electric field magnitude manually, you can use the formula:

\[
E_{\text{resultant}} = \sqrt{(50,344)^2 + (133,052)^2}
\]

Let me know if you'd like further clarification or have any other questions!

### Related questions:
1. How would the resultant electric field change if the charges were of the same sign?
2. What is the direction of the resultant electric field at the top left corner?
3. How do different geometrical configurations of charges affect the electric field magnitude?
4. Can this method be extended to three-dimensional configurations?
5. How would the electric field change if the distances were different?

**Tip:** When working with electric fields from multiple charges, always break down the field into components and use vector addition.

Two point-like charges are placed at the corners of a square as shown in the figure, a = 50.0 cm. Find the magnitude of the electric field at the top left corner. Let q1 = -1.40 μC, and q2 = +3.70 μC.

Calculate the magnitude of the electric field at the top left corner due to two point-like charges placed at the corners of a square with side length 50.0 cm. This problem involves Coulomb's Law, vector addition, and the Pythagorean theorem, suitable for undergraduate physics students.

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