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 )

 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

        = 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

d is the answer i came up with

electromagnetic shielding is the practice of reducing the electromagnetic field in a space by blocking the field with barriers made of conductive or magnetic materials. shielding is typically applied to enclosures to isolate electrical devices from their surroundings, and to cables to isolate wires from the environment through which the cable runs. electromagnetic shielding that blocks radio frequency electromagnetic radiation is also known as rf shielding.

the shielding can reduce the coupling of radio waves, electromagnetic fields and electrostatic fields. a conductive enclosure used to block electrostatic fields is also known as a faraday cage. the amount of reduction depends very much upon the material used, its thickness, the size of the shielded volume and the frequency of the fields of interest and the size, shape and orientation of apertures in a shield to an incident electromagnetic

from wikipedia