Question

A point charge is placed at each corner of square with side leanth a. The charges all have same magnitude q. My question, What is the its magnitude in terms of q and a?

Answer 1

Answer:

 E = k q / a²   (1.3535) (- i ^ + j ^)

  E = k q / a²  1.914  ,      θ’= 135

Explanation:

For this exercise we will use Newton's second law where we must add as vectors

        E_total = E₁₂ i ^ + E₁₄ j ^ + E₁₃

Let's look for the value of each term

On the x axis

       E₁₂ = k q / a²

On the y axis

       E₁₄ = k q / a²

For the charge in the opposite corner we look for the distance

        d = √ (a² + a²) = a √2

let's look for the field

      E₁₃ = k q / d²

      E₁₃ = k q / 2a²

let's use trigonometry to find the two components of this field

       cos 45 = E₁₃ₓ / E₁₃

       E₁₃ₓ = E₁₃ cos 45

       

       sin 45 = E_{13y} / E₁₃

       E_{13y} = E₁₃ sin 45

       E₁₃ₓ = k q / 2a²  cos 45

       E_{13y} = k q / 2a²  sin 45

let's find each component of the electric field

X axis

      Eₓ = -E₁₂ - E₁₃ₓ

      Eₓ = - k q / a² - k q / 2a² cos 45

      Eₓ = - k q / a² (1 + cos 45/2)

      cos 45 = sin 45 = 0.707

      Eₓ = - k q / a²   (1 + 0.707 / 2)

      Eₓ = - k q / a²    (1.3535)

Y axis  

      E_y = E₁₄ + E_{13y}

       E_y = k q / a² + k q / 2a²     sin 45

       E_y = k q / a² (1 + sin 45/2)

       E_y = k q / a²       (1.3535)

we can give the results in two ways

       E = k q / a²   (1.3535) (- i ^ + j ^)

In modulus and angle form, let's use Pythagoras' theorem for the angle

       E = √ (Eₓ² + E_y²)

        E = k q / a²    1.3535 √2

        E = k q / a²     1.914

we use trigonometry for the angle

        tan θ = E_y / Eₓ

         θ = tan⁻¹  (E_y / Eₓ)

         θ = tan⁻¹ (1 / -1)

         θ = 45

in the third quadrant, if we measure the angle of the positive side of the x-axis

           θ‘= 90 + 45

           θ’= 135