Enter an expression for the horizontal component of the net force acting on the charge located ag the lower left corner of the rectangle in terms of the charges, the given distances, and the Coulomb constant. Use a coordinate system in which the positive direction to the right. The charges on the left side of the rectangle are positive while the charges on the right side of the rectangle are negative.

Four point charges of equal magnitude Q= 75nC

Corners of the rectangle: D1= 21cm , D2= 9cm

Fₓ = k q₁² [1 / D₁² + 0.919 / (D₁² + D₂²)]

Fₓ = 2.04 10⁻³ N

Explanation:

For this exercise we will use Coulomb's law and Newton's second law

               F = F₁₂ + F₁₃ + F₁₄

The values given are

              q₁ = q₄ = 75 10⁻⁹ C

              q₂ = q₃ = -75 10⁻⁹ C

              D₁ = 0.21 m      X axis

              D₂ = 0.09 m     Y axis

Let's look for each force

              F₁₂ = k q₁q₂ / r₁₂²

              F₁₂ = - k q₁² / D₁²

              F₁₃ = k q₁ q₃ / (D₁² + d₂²)

              F₁₃ = - k q₁² / (D₁² + D₂²)

As we are asked for the component on the x-axis of the force, let's decompose using trigonometry

              Tan θ = D₂ / D₁

              θ = tan⁻¹ D₂ / D₁

              θ = tan⁻¹ 0.09 / 0.21

              θ = 23.20º

            sin 23.20 = / F13

            cos 23.20 = F₁₃ₓ / F13

            F_{13y} = F13 sin23.20

            F₁₃ₓ = F13 cos 23.20

            F₁₃ₓ = F13 0.919

The force F₁₄ is in the direction y therefore does not contribute to the horizontal force

            Fₓ = F₁₂ + F₁₃ₓ

            Fₓ = k q₁² / D₁² + k q₁² / (D₁² + D₂²) cos 23.20º

            Fₓ = k q₁² [1 / D₁² + 0.919 / (D₁² + D₂²)]

Let's calculate

          Fₓ = 9 10⁹ (75 10⁻⁹)² [1 / 0.21² + 0.919 /(0.21² + 0.09²)]

          Fₓ = 5.0625 10⁻⁵ [22.676 + 17.605]

          Fₓ = 2.04 10⁻³ N

In the direction of the positive side of the x axis since the forces are attractive

the answer is a. at the equator

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