Respuesta :
Answer:
Eₓ = -4,187 N / C, E_y = 6,937 N / C
Explanation:
To solve this exercise we will calculate the electric field at the desired point (0, 0) and then add it vectorially
E = [tex]k \frac{q}{r^2}[/tex]
charge 1
the value of q₁ = - 2,600 nC = -2,600 10⁻⁹ C
r₁ = [tex]\sqrt{ (x_1-x_o)^2 + (y_1-y_o)^2}[/tex]
r₁ = [tex]\sqrt{0 +(1.04-0)^2}[/tex]RA 0 + (1.04 - 0) 2
r₁ = 1.04 m
we substitute
E₁ = k q1 / r12
E₁ = 9 10⁹ 2.6 10⁻⁹ / 1.04²
E₁ = 21.635 N / C
This electric field is directed towards the negative charge and is in the direction of the y axis.
E₁ = 21.635 j ^ N / C
charge 2
the value of q₂ = 3,600 nC = 3,600 10-9 C
r₂ = Ra (1.4 -0) 2 + (0.4 -0) 2
r₂ = 1.4560 m
we substitute
E₂ = k q₂ / r₂²
E₂ = 9 10⁹ 3,600 10⁻⁹ / 1.4560²
E₂ = 15.283 N / C
This field leaves the charge since the charge is positive, let's find the angle
tan θ = y / x
θ = tan⁻¹ y / x
θ = tan⁻¹ 0.400 / 1.400
θ = 15.9º
let's decompose the electric field
sin 15.9 = E_{2y} / E2
cos 15.9 = E₂ₓ / E2
E_{2y} = E2 sin 15.9
E₂ₓ = E2 cos 15.9
E_{2y} = 15.283 sin 15.9 = 4.187 N / C
E₂ₓ = 15.283 cos 15.9 = 14.698 N / C
the field lines leave the positive charge and are directed to the left, therefore
E_{2y} = - 4.187 N / C
E₂ₙ = -14.698N / C
the total field on each axis is
Eₓ = E_{1x} + E_{2x}
Eₓ = 0 -4,187
Eₓ = -4,187 N / C
E_y = 21.635 - 14.698
E_y = 6,937 N / C
