There are four charges, each with a magnitude of 4.25 C. Two are positive and two are negative. The charges are fixed to the corners of a 0.440-m square, one to a corner, in such a way that the net force on any charge is directed toward the center of the square. Find the magnitude of the net electrostatic force experienced by any charge.

In this exercise we ask for the net electric force. Let's start by writing the configuration of the charges, the charges of the same sign must be on the diagonal of the cube so that the net force is directed towards the interior of the cube, see in the attached numbering and sign of the charges

The net force is

F_ {net} = F₂₁ + F₂₃ + F₂₄

bold letters indicate vectors. The easiest method to solve this exercise is by using the components of each force.

let's use trigonometry

cos 45 = F₂₄ₓ / F₂₄

sin 45 = F_{24y) / F₂₄

F₂₄ₓ = F₂₄ cos 45

F_{24y} = F₂₄ sin 45

let's do the sum on each axis

X axis

Fₓ = -F₂₁ + F₂₄ₓ

Fₓ = -F₂₁₁ + F₂₄ cos 45

Y axis

F_y = - F₂₃ + F_{24y}

F_y = -F₂₃ + F₂₄ sin 45

They indicate that the magnitude of all charges is the same, therefore

F₂₁ = F₂₃

Let's use Coulomb's law

F₂₁ = k q₁ q₂ / r₁₂²

the distance between the two charges is

r = a

F₂₁ = k q² / a²

we calculate F₂₄

F₂₄ = k q₂ q₄ / r₂₄²

the distance is

r² = a² + a²

r² = 2 a²

we substitute

F₂₄ = k q² / 2 a²

we substitute in the components of the forces

Fx = [tex]- k \frac{q^2}{a^2} + k \frac{q^2}{2 a^2} \ cos 45[/tex]

Fx = [tex]k \frac{q^2}{a^2}[/tex] ( -1 + ½ cos 45)

F_y = k \frac{q^2}{a^2} ( -1 + ½ sin 45)

We calculate

F₀ = 9 10⁹ 4.25² / 0.440²

F₀ = 8.40 10¹¹ N

Fₓ = 8.40 10¹¹ (½ 0.707 - 1)

Fₓ = -5.43 10¹¹ N

remember cos 45 = sin 45

F_y = - 5.43 10¹¹ N

We can give the resultant force in two ways

a) F = Fₓ î + F_y ^j

F = -5.43 10¹¹ (i + j) N

b) In the form of module and angle.

For the module we use the Pythagorean theorem

F = [tex]\sqrt{F_x^2 + F_y^2}[/tex]

F = 5.43 10¹¹ √2

F = 7.68 10¹¹ N

in angle is

θ = 45º