Respuesta :
Answer:
A) F_{A} = 36.26 N , θ = 204.4º
B) F_{B} = 43.45 N , θ = 349º
C) F_{C} = 25.16 N , θ = 112.52º
Explanation:
The force is a vector quantity, so we must take into account the direction of the force, for the calculation of the electric force we will use Coulomb's law
F₁₂ = k q₁ q₂ / r₁₂²
we will use numerical indexes 1 for A, 2 for B and 3 for C
a) Force on charge A
X axis
Fₓ = F₁₂
Fₓ = k q₁ q₂ / r₁₂²
Fₓ = 8.99 10⁹ 3 10⁻⁶ 6 10⁻⁶ / 0.07 2
Fₓ = 3,302 10¹ N
Since the two charges are of the same sign, the force is repulsive, so that it is directed towards the negative side of the x-axis.
Y Axis
[tex]F_{y}[/tex] = F₁₃
F_{y} = k q₁q₃ / r₁₃²
F_{y} = 8.99 10⁹ 3 10⁻⁶ 2 10⁻⁶ / 0.06 2
F_{y} = 1,498 10¹ N
Repulsive force in the direction of the negative part of the y axis
To find the magnitude we use the Pythagorean Theorem
[tex]F_{A}[/tex] = √ (Fₓ² + [tex]F_{y}[/tex]²)
F_{A}= √ (33.02 2 + 14.98 2)
F_{A} = 36.26 N
For the direction let's use trigonometry
tan θ = F_{y} / Fₓ
θ = tan⁻¹ F_{y} / Fₓ
θ = tan⁻¹ (-14.98 / -33.02)
θ = 24.4º in the third quadrant
Measured from the positive side of the x axis
θ = 180 + 24.4
θ = 204.4º
b) Force on charge B
X axis
Fₓ = F₂₁ + F₂₃ₓ
F₂₁ = k q₁ q₂ / r₁₂²
F₂₁ = 8.99 10⁹ 3 10⁺⁶ 6 10⁻⁶ / 0.07²
F₂₁ = - F₁₂ = 33.02 N
Directed towards the positive side of the x axis
F₂₃ = k q₃q₂ / r₃₂²
R₃₂² = (0.07² +0.06²) = 0.0085
F₂₃ = 8.99 10⁹ 2 10⁻⁶ 6 10⁻⁶ / 0.0085
F₂₃ = 1,269 10¹ N
For the direction of this repulsive force we use trigonometry
Tan θ = y / x
θ = tan⁻¹ 0.06 / 0.07
θ = 40.6º
As the force is repulsive therefore the angle is
θ = 360 - 40.6
θ = 319.4º
Now we can find the components of this force
F₂₃ₓ = F₂₃ cos 319.4
F₂₃ₓ = 12.69 cos 319.4
F₂₃ₓ = 9.635 N
[tex]F_{23y}[/tex] = F₂₃ sin319.4
F_{23y} = 12.69 sin319.4
F_{23y} = -8.258 N
The force on this charge is
X axis
Fₓ = F₂₁ + F₂₃ₓ
Fₓ = 33.02 + 9.635
Fₓ = 42,655 N
Y Axis
F_{y} = F_{23y}
F_{y} = -8.258 N
The Magnitude is
[tex]F_{B}[/tex] = √ (Fₓ² + F_{y}²)
F_{B} = √ (42,655 2 + 8,258 2)
F_{B} = 43.45 N
The address is
Tan θ = F_{y} / Fₓ
θ = tan⁺¹ (-8.258 / 42.655)
θ = -10.96
Measured from the positive side of the x axis is
θ = 360 -10.96
θ = 349º
c) The force on the charge C
X axis
Fₓ = F₃₂ₓ
Fₓ = -F₂₃ₓ = -9.635 N
Y Axis
F_{y} = F₃₁ + [tex]F_{32y}[/tex]
F_{y} = 14.98 + 8.258
F_{y}= 23,238 N
The magnitude is
[tex]F_{C}[/tex] = √ (Fₓ² + F_{y}²)
F_{C} = √ (9,635 2+ 23,238 2)
F_{C} = 25.16 N
The address is
tan θ = F_{y} / Fₓ
θ = tan-1 (-23.238 / 9.635)
θ = -67.48
Measured from the x axis
θ = 180 -67.48
θ = 112.52º
