Question

Consider 4 charges placed at the corners of a square with side 1.25m as shown. What are the magnitude and direction of the electrostatic force on Q1 resulring from other three charges?
(Note: Please put your final answer in 4 decimal places.). pls answer:)​

Answer 1

Answer:

 F = 2,8289 i ^ + 1,0909 j ^) 10⁻² N

F = 3.0226 10⁻² N ,  θ  = 21.16º

Explanation:

For this exercise we use Coulomb's law

         F = k q₁q₂ / r₁₂²

We also use that the force is a vector magnitude, so we must calculate each component of the force , see the adjoint for the direction of the vectors

X axis

          Fₓ = -F₁₄ + F₁₃ₓ

         

Y axis

        $F_{y}$ = F₁₂ -F_{13y}

let's look for the expression for each force

where the side of the square is a = 1.25 m

  F₁₂ = k Q₁Q₂ / a²

  F₁₄ = k Q₁Q₄ / a²

the distance between 1 and 3 is

         d = √(a² + a²) = a √2

   F₁₃ = k Q₁Q₃ / d²

let's use trigonometry to find the components

              cos 45 = F₁₃ₓ / F₁₃

              F₁₃ₓ = F₁₃ cos 45

              F₁₃ₓ = k Q₁Q₃ / 2a²

              sin 45 = F_{13y} / F₁₃

              F_{13y} = F₁₃ sin 45

              F_{13y} = k Q₁Q₃ / 2a²  sin 45

 

Taking all terms, we substitute in the force for each axis

X axis

          Fₓ = - k Q₁Q₄ / a² + k Q₁Q₃ / 2a₂ cos 45

          Fₓ = k Q₁ / a² ( -Q₄ + Q₃ /2   cos 45)

          Fₓ = 9 10⁹ 1.5 10⁻⁶ / 1.25²   (- 4.5 10⁻⁶ + 3.5/2  cos 45  10⁻⁶)

          Fₓ = 8.64 10³ (3.2626 10⁻⁶)

          Fₓ = 2.8189 10⁻² N

Y axis

          F_{y} = k Q₁Q₂ / a² - k Q₁Q₃ /2a²   sin 45

          F_{y} = k Q₁ / a² (Q₂ - Q₃ /2 sin45)      

          F_{y} = 9 10⁹ 1.5 10⁻⁶/ 1.25²    (2.5 10⁻⁶ - 3.5/2   sin 45  10⁻⁶)

          F_{y} = 8.64 10³ (1.26256 10⁻⁶)

          F_{y} = 1.0909 10⁻² N

The result can be given in two ways

1) F = Fₓ i ^ + F_{y} j ^

     F = 2,8289 i ^ + 1,0909 j ^) 10⁻² N

2) in the form of a module and an angle, for which we use the Pythagorean theorem and trigonometry

       F = √ (Fₓ² + F_{y}²)

       F = 10⁻² √ (2,8189² + 1,0909²)

       F = 3.0226 10⁻² N

   

       tan θ = F_{y} / Fx

       θ = tan⁻¹ (F_{y} / Fₓ)

       θ = tan⁻¹ (1.0909 / 2.8189)

        θ  = 21.16º