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Sagot :
Answer:
E = 10.10 10⁵ N / C, θ = 184.75º
Explanation:
The electric field created by a point charge is
E = k q / r²
where r is the distance from the load to the desired point
Since the electric field is a vector quantity, one of the simplest ways to calculate it is to find the components of the electric field on each axis and add them.
charge 1 q₁ = 4 10⁻⁶ C placed at position x₁ = 0.2 m
the distance is
r² = (x-x₀) ² + (y-y₀) ²
r² = (0 - 0.2) ² + (0.15 - 0) ²
r² = 0.0625
E₁ = 9 10⁹ 4 10⁻⁶ / 0.0625
E₁ = 5.76 10⁵ N / C
let's find the angle of this vector
tan θ = y / x
θ = tan⁻¹ y / x
θ = tan⁻¹ 0.15 / 0.2
θ = 36.9º
This angle is in the second quadrant since as the load is positive the electric field is salient, this angle measured from the positive side of the x axis is
θ’= 180- θ
θ’= 180- 36.9
θ’= 143.1º
we use trigonometry
cos 143.1 = E₁ₓ / E₁
sin 143.1 = E_{1y} / E₁
E₁ₓ = E₁ cos 143.1
E_{1y} = E₁ sin 143.1
E₁ₓ = 5.76 10⁵ cos 143.1 = -4.606 10⁵ N / C
E_{1y} = 5.76 10⁵ sin 143.1 = 3.458 10⁵ N / C
charge 2 q₂ = 5 10⁻⁶ at position x = -0.2 m
r²2 = 0.2² + 0.15²
r² = 0.0625
E₂ = 9.10⁹ 5 10⁻⁶/0.0625
E₂ = 7.20 10⁵ N / c
we look for the angles
tan θ = y / x
θ = tan⁻¹ 0.15 / 0.2
θ = 36.9º
in that case the charge is negative, therefore the electric field is directed to the charge and therefore the angle is in the third quadrant
θ'= 180 + θ
tea = 180 + 36.9 = 216.9º
the components of the electric field are
E₂ₓ = E₂ cos 216.9
E_{2y} = E₂ sin 216.9
E₂ₓ = 7.20 10⁵ cos 216.9 = -5.76 10⁵ N / C
E_{2y} = 7.20 10⁵ sin 216.9 = -4.32 10⁵ N / C
The components of the total electric field are
Eₓ = E₁ₓ + E₂ₓ
Eₓ = -4.606 10⁵ -5.76 10⁵
Eₓ = - 10,366 10⁵ N / C
E_y = E_{1y} + E_{2y}
E_y = 3.458 10⁵ -4.32 10⁵
E_y = -0.862 10⁵ N / C
We can give the result in two ways
1) E = Ex i ^ + Ey j ^
E = (- 10.366 i ^ -0.862 j ^) 10⁵ N / C
2) in the form of module and angle
let's use the Pythagorean theorem
E = [tex]\sqrt{E_x^2 + E_y^2 }[/tex]Ra Ex² + Ey²
E = [tex]\sqrt{ (1.3666^2 + 0.862^2)}[/tex] 10⁵
E = 10.10 10⁵ N / C
trigonometry
tan θ’= E_y / Eₓ
θ'= tan⁻¹ (0.862 / 10366)
θ’= 4.75º
this angle is in the third quadrant, therefore measured from the positive side of the x-axis is
θ = 180 + θ'
θ = 180 + 4.75
θ = 184.75º
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