Charges of 4.0 µC and -6.0 µC are placed at two corners of an equilateral triangle with sides of 0.10 m. At the third corner, what is the electric field magnitude created by these two charges? (ke = 8.99 ´ 109 N·m2/C2)

To find the electric field magnitude created by the two charges, we need to use the principle of superposition. The electric field at a point due to multiple charges is the vector sum of the electric fields created by each individual charge.

Let's break down the problem into two parts:

1. Electric field due to the positive charge (4.0 µC):
- The magnitude of the electric field, E1, created by a point charge can be given by Coulomb's law: E1 = (k * |q1|) / r^2, where k is the electrostatic constant (8.99 × 10^9 N·m^2/C^2), |q1| is the magnitude of the charge (4.0 µC), and r is the distance from the charge to the point where we want to find the electric field.
- In this case, the distance is the length of one side of the equilateral triangle, which is 0.10 m.

2. Electric field due to the negative charge (-6.0 µC):
- Similarly, the magnitude of the electric field, E2, created by the negative charge can be calculated using Coulomb's law: E2 = (k * |q2|) / r^2, where |q2| is the magnitude of the charge (6.0 µC).

Once we have the magnitudes of the individual electric fields, we can find the resultant electric field at the third corner by adding them together vectorially.

Let's calculate the magnitudes:

E1 = (8.99 × 10^9 N·m^2/C^2 * 4.0 × 10^-6 C) / (0.10 m)^2
E1 ≈ 3.596 × 10^6 N/C

E2 = (8.99 × 10^9 N·m^2/C^2 * 6.0 × 10^-6 C) / (0.10 m)^2
E2 ≈ 5.394 × 10^6 N/C

Now, to find the resultant electric field E_total at the third corner, we need to add the two electric fields using vector addition. Since the triangle is equilateral, the direction of each field will be towards the third corner.

E_total = √(E1^2 + E2^2 + 2 * E1 * E2 * cos(60°))

Substituting the values:

E_total = √(3.596 × 10^6 N/C)^2 + (5.394 × 10^6 N/C)^2 + 2 * 3.596 × 10^6 N/C * 5.394 × 10^6 N/C * cos(60°)

E_total ≈ 9.79 × 10^6 N/C

So, the electric field magnitude created by the two charges at the third corner is approximately 9.79 × 10^6 N/C.