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
is the buoyant force
is the density of the liquid
is the acceleration due to gravity