Three point charges are located at the corners of an equilateral triangle as in the figure below. Find the magnitude and direction of the net electric force on the 1.65 μC charge. (Let A = 1.65 μC, B = 6.60 μC, and

Question

C = −4.20 μC.)

Answer 1

To find the net electric force on charge A, we will first calculate the electric force between charges A and B, and the electric force between charges A and C, separately. The net electric force on charge A will be the vector sum of these two forces.

We can use the formula for electric force between two point charges: F = k|q₁q₂|/r², where k is the electrostatic constant, q₁ and q₂ are the two charges, and r is the distance between them.

Given, k = 8.99 × 10^9 N·m²/C².

First, let's calculate the electric force between charges A and B.

Charge A (q₁): 1.65 μC = 1.65 × 10^(-6) C
Charge B (q₂): 6.60 μC = 6.60 × 10^(-6) C
Distance (r): Given that they are located at the vertices of an equilateral triangle, let's assume the side length of the triangle to be s.

F_AB = k|q₁q₂|/r²
F_AB = 8.99 × 10^9 × |(1.65 × 10^(-6))(6.60 × 10^(-6))|/(s²)

Let's keep the value of F_AB in terms of 's' for now, and move on to calculate the electric force between charges A and C.

Charge C (q₂): -4.20 μC = -4.20 × 10^(-6) C

F_AC = k|q₁q₂|/r²
F_AC = 8.99 × 10^9 × |(1.65 × 10^(-6))(-4.20 × 10^(-6))|/(s²)

Now we have the magnitudes of the forces F_AB and F_AC acting on charge A. To find the net force, we need to find the sum of these two forces as vectors. Since charges A and B as well as A and C are located at the vertices of an equilateral triangle, we know that the angle between the forces F_AB and F_AC is 120°.

Let F_AN be the net force:

F_AN^2 = F_AB^2 + F_AC^2 + 2F_AB × F_AC × cos(120°)

Plugging in our values for F_AB and F_AC, we get:

F_AN^2 = (8.99 × 10^9 × |(1.65 × 10^(-6))(6.60 × 10^(-6))|/(s²))^2 + (8.99 × 10^9 × |(1.65 × 10^(-6))(-4.20 × 10^(-6))|/(s²))^2 - (8.99 × 10^9 × |(1.65 × 10^(-6))(6.60 × 10^(-6))|/(s²)) × (8.99 × 10^9 × |(1.65 × 10^(-6))(-4.20 × 10^(-6))|/(s²))

Solving for F_AN^2, we get:

F_AN = √(s² × (1.089 × 10^(-3))^2 + s² × (1.226 × 10^(-3))^2 - 2 × s² × 1.089 × 10^(-3) × 1.226 × 10^(-3))
F_AN = √(5.970 × 10^(-5) + 8.074 × 10^(-5) - 4.799 × 10^(-5))
F_AN = √(9.245 × 10^(-5) s²)

Now, we cannot solve for F_AN without the value of 's'. However, if we are given the side length of the triangle or the distance between the charges, we can find the magnitude of the net electric force.

As for the direction, we can find the angle of F_AN with respect to F_AB since we know the magnitudes and angle between F_AB and F_AC.

Let θ be the angle between F_AB and F_AN.

Using the law of cosines,

cos(θ) = [(1.089 × 10^(-3))^2 + (9.245 × 10^(-5))^2 - (1.226 × 10^(-3))^2]/(2 × 1.089 × 10^(-3) × 9.245 × 10^(-5))
cos(θ) ≈ 0.455

θ ≈ arccos(0.455) ≈ 63°

So, the direction of the net electric force F_AN is approximately 63° with respect to the direction of F_AB.