The problem states that four charges, 1mC, 1mC, -1mC, and 1mC, are placed at the vertices of a square with a side length of √2 cm. We need to determine the net force experienced by a 0.1mC charge placed at the center of the square. To solve this problem, we will first calculate the electric field due to each charge at the center and then add up the electric fields to find the net force.

Calculating Electric Field due to each charge:

The electric field at a point due to a single charge can be calculated using the formula:

E = (k * q) / r^2

Where:
E is the electric field,
k is the electrostatic constant (9 × 10^9 Nm^2/C^2),
q is the charge, and
r is the distance from the charge to the point.

Electric Field due to the 1mC charge:
The distance from the center of the square to any corner is √2/2 cm. Therefore, the electric field due to the 1mC charge at the center is:

E1 = (9 × 10^9 Nm^2/C^2 * 1mC) / (√2/2 cm)^2

Simplifying the equation:
E1 = (9 × 10^9 Nm^2/C^2 * 1 × 10^-3 C) / (2/√2 cm)^2
E1 = (9 × 10^6 Nm^2/C^2) / (2/√2 cm)^2
E1 = (9 × 10^6 Nm^2/C^2) / (2/2 cm)^2
E1 = (9 × 10^6 Nm^2/C^2) / 1 cm^2
E1 = 9 × 10^6 N/C

Electric Field due to the -1mC charge:
The distance from the center of the square to any side midpoint is √2/2 cm. Therefore, the electric field due to the -1mC charge at the center is:

E2 = (9 × 10^9 Nm^2/C^2 * (-1mC)) / (√2/2 cm)^2

Simplifying the equation:
E2 = (9 × 10^9 Nm^2/C^2 * -1 × 10^-3 C) / (2/√2 cm)^2
E2 = (-9 × 10^6 Nm^2/C^2) / (2/√2 cm)^2
E2 = (-9 × 10^6 Nm^2/C^2) / (2/2 cm)^2
E2 = (-9 × 10^6 Nm^2/C^2) / 1 cm^2
E2 = -9 × 10^6 N/C

Electric Field due to the 1mC charge:
The distance from the center of the square to any corner is √2/2 cm. Therefore, the electric field due to the 1mC charge at the center is:

E3 = (9 × 10

Zero.