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.

Answer:

 F = 7.68 10¹¹ N,  θ = 45º

Explanation:

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

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