At each corner of a square of side l there are point chargesof
magnitude q, 2q, 3q, and 4q. determine the force on thecharge
2q?

 F = k Q² / L² (4.82 i ^ - 8.82 j ^ N  ,  F = k Q² / L² 10.05 N ,   θ  = 298⁰

Explanation:

To find the electric force we must use Coulomb's law

           F = k q₁ q₂ / r₁₂²

Where k is the Coulomb constant that is worth 8.99 10⁹ N m² / C², q is the value of the charges and r the distance between them

In our case the total force is

         F = F₁₂ + F₃₂ + F₄₂

In the attached we can see the location of the charges. Distance

         r₁₂ = L

         r₃₂ = L

         r₄₂ = √ L² + L² = L √ 2

Let us explicitly write the equation, remember that force is a vector equation,

X axis

              Fₓ = F₁₂ +

Y Axis  

              = - F₃₂ -

With we have a square the angle of F₄₂ is 45⁰, using trigonometry we can find its components

        cos 45 = / F₄₂

        sin 45 =  / F₄₂

           = F₄₂ cos 45

           = F₄₂ sin45

Let's replace

        Fₓ = k Q 2Q / L² + k 4Q 2Q / 2L² cos 45

        = - k 3Q 2Q / L² - k 4Q 2Q / 2L²  sin  45

        Fₓ = k Q² (2 + 4 cos 45) = k Q² 4.82

         = - k Q² (6 + 4 sin 45) = k Q² 8.82

We can give strength in two ways

        F = k Q² / L² (4.82 i ^ - 8.82 j ^ N

In the form of magnitude and angle

      F = Ra Fₓ² +   ²

      F = kQ² / L² RA (4.82² + 8.82²)

      F = k Q² / L² 10.05 N

      θ = tan⁻¹    / Fₓ

      θ = tan⁻¹ (-8.82 / 4.82)

      θ = -61.34 = 298⁰