Answer:
E = (2.29 i ^ - 0.917 j ^) 10⁶ N / C
E_{total} = 2,467 10⁶ N / A θ = -21.8
Explanation:
For this exercise we will use that the electric field is a vector quantity, so the total field is
E_total = E₁₃ + E₂₃
bold font vectors
. We can work with the components of the electric field in each axis
X- axis
E_ total x = E₁₃ₓ + E_{23x}
y-axis
E_{total y} = E_{13y} + E_{23y}
the expression for the electric field is
E = k q / r²
where r is the distance between the charge and the positive test charge
in this exercise
Let's find the field created by charge 1
q₁ = 10 μC = 10 10⁻⁶ C
x₁ = 5 cm = 0.05 m
x₃ = 21 cm = 0.21 m
E_{13x} = 9 10⁹ 10 10⁻⁶ / (0.21 -0.05)²
E_{13x} = 3.516 10⁶ N / C
y₁ = 6 cm = 0.06 cm
y₃ = -12 cm = -0.12 m
E_{13y} = 9 10⁹ 10 10⁻⁶ / (-0.12 - 0.06)²
E_{13y} = 2,777 10⁶ N / C
let's find the field produced by charge 2
q₂ = -27 μC = - 27 10⁻⁶ C
x₂ = -6 cm = -0.06 m
x₃ = 0.21 m
E_{23x} = 9 10⁹ 27 10⁻⁶ / (0.21 + 0.06)²
E_{23x} = 1.23 10⁶ N / A
y₂ = 10 cm = 0.10 m
y₃ = -0.12 m
E_{23y} = 9 10⁹ 27 10⁻⁶ / (-0.12 - 0.10)²
E_{23y} = 1.86 10⁶ N / C
Taking the components we can calculate the total electric field, we must use that charge of the same sign repel and attract the opposite sign, remember that the test charge is always considered positive.
E_{total x} = E_{13x} - E_{23x}
E_{total x} = (3.516 - 1.23) 10⁶
E_{total x} = 2.29 10⁶ N / A
E_{total y} = -E_{13y} + E_{23y}
E_{total y} = (-2.777 +1.86) 10⁶ N / A
E_{total y} = -0.917 10⁶ N / A
we can give the result in two ways
E = (2.29 i ^ - 0.917 j ^) 10⁶ N / C
or in the form of modulus and angle, let's use the Pythagorean theorem to find the modulus
E_{total} = √ (E_{total x}^2 + E_{total y}^ 2)
E_{total} = √ (2.29² + 0.917²) 10⁶
E_{total} = 2,467 10⁶ N / A
let's use trigonometry for the angle
tan θ = E_total and / E_totalx
θ = tan⁻¹ E_{total y} / E_{total x}
θ = tan⁻¹ (-0.917 / 2.29)
θ = -21.8
The negative sign indicates that the angle is measured with respect to the x-axis in a clockwise direction.
