Respuesta :
Answer:
B) x-component of the total force exerted on the third charge by the other two (Fn₃x)
Fn₃x =8,56*10⁻⁵ N (+x)
B) y-component of the total force exerted on the third charge by the other two (Fn₃y)
Fn₃y = 5.7*10⁻⁵ N (-y)
Explanation:
Theory of electrical forces
Because the particle q₃ is close to two other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.
Equivalences:
1nC= 10⁻⁹C
1cm = 10⁻²m
Data
q₁=4.96 nC = +4.96*10⁻⁹C
q₂=-1.99 nC = -1.99*10⁻⁹C
[tex]d_{1} =\sqrt{ 4.01^{2}+2.98^{2} } = 4.996 cm = 4.996*10^{-2} m=49.96*10^{-3} m[/tex]
d₂= 2.98 cm
Graphic attached
The directions of the individual forces exerted by q₁ and q₂ on q₃ are shown in the attached figure.
The force (F₁₃) of q₁ on q₃ is repulsive because the charges have equal signs.
the force F₂₃ of q₂ and q₃ is attractive because the charges have opposite signs.
Calculation of the forces exerted on the charge q₁ and q₂ on q₃
To calculate the magnitudes of the forces exerted by the charges q₁, q₂, on q₃ we apply Coulomb's law:
F₁₃=k*q₁*q₃/d₁² =9*10⁹*4.96*10⁻⁹*5.99*10⁻⁹/(49.96*10⁻³)²=10.7**10⁻⁵ N
F₂₃=k*q₂*q₃/d₂² =9*10⁹*1.99*10⁻⁹*5.99*10⁻⁹/(2.98*10⁻²)²=12.08*10⁻⁵ N
F₁₃x=F₁₃cosβ=10.7*10⁻⁵* (4.01/4.996)=8,56*10⁻⁵ N (+x)
F₁₃y=F₁₃sinβ= 10.7*10⁻⁵* (2.98/4.996)=6,38*10⁻⁵ N (+x)
F₂₃x= 0
F₂₃y= F₂₃=12.08*10⁻⁵ N (-y)
A) x-component of the total force exerted on the third charge by the other two (Fn₃x)
Fn₃x= F₁₃x + F₂₃x= 8,56*10⁻⁵ N + 0
Fn₃x =8,56*10⁻⁵ N (+x)
B) y-component of the total force exerted on the third charge by the other two (Fn₃y)
Fn₃y = F₁₃y + F₂₃y= 6,38*10⁻⁵ N - 12.08*10⁻⁵ N
Fn₃y = 5.7*10⁻⁵ N (-y)
The charges, 4.96 nC, -1.99 nC, and 5.99 nC, forming a triangle gives
the following approximate values of the force at the 5.99 nC charge.
- x-component is [tex]\underline{8.6 \times 10^{-5} \, \mathbf{\hat i} \ N}[/tex]
- y-component is [tex]\underline{-5.69 \times 10^{-5} \, \mathbf{\hat j} \ N}[/tex]
- The magnitude of the total force is 1.03 × 10⁻⁴ N
- The direction of the force is 33.5° clockwise from the horizontal x-axis
How can the force acting at the 5.99 nC charge be resolved?
The given charges are;
Q₁ = 4.96 nC, at point (0, 0)
Q₂ = -1.99 nC, at point (4.01, 0)
Q₃ = 5.99 nC at point (4.01, 2.98)
According to Coulomb's Law, we have;
[tex]F_{13} = \mathbf{\dfrac{9 \times 10^9 \times \left(4.96 \times 10^{-9} \right)\times \left(5.99\times 10^{-9} \right)}{4.01^2 + 2.98^2} } }\approx 1.07 \times 10^{-4}[/tex]
F₁₃ ≈ 1.07 × 10⁻⁴ N
The components of the force are;
[tex]cos\left(arctan \left(\dfrac{2.96}{4.01} \right) \right) \times 1.07 \times 10^{-4} \, \mathbf{\hat i} + sin\left(arctan \left(\dfrac{2.96}{4.01} \right) \right) \times 1.07 \times 10^{-4} \, \mathbf{\hat j}[/tex]
Which gives;
[tex]\vec{F_{13}} \approx \mathbf{8.6 \times 10^{-5} \, \mathbf{\hat i} + 6.39 \times 10^{-5} \, \mathbf{\hat j}}[/tex]
Therefore;
[tex]F_{23}= \dfrac{9 \times 10^9 \times \left((-1.99)\times 10^{-9} \right)\times \left(5.99\times 10^{-9} \right)}{2.98^2} } \approx\mathbf{ 1.208 \times 10^{-4}}[/tex]
Which gives;
[tex]\vec{F_{23}} \approx \mathbf{-1.208 \times 10^{-4} \, \mathbf{\hat j}}[/tex]
The components of the force at Q₃ is therefore;
- x-component = [tex]8.6 \times 10^{-5} \, \mathbf{\hat i} + 0 = \underline{8.6 \times 10^{-5} \, \mathbf{\hat i}}[/tex]
- y-component = [tex]6.39 \times 10^{-5} \, \mathbf{\hat j} + -1.208 \times 10^{-4} \, \mathbf{\hat j} \approx \underline{-5.69 \times 10^{-5} \, \mathbf{\hat j}}[/tex]
The magnitude of the total force is therefore;
|F| ≈ √((8.6×10⁻⁵)² + (-5.69 × 10⁻⁵)²) ≈ 1.03 × 10⁻⁴
- The magnitude of the total force, |F| ≈ 1.03 × 10⁻⁴ N
The direction of the total force is found as follows;
[tex]The \ direction, \ \theta \approx \mathbf{arctan \left(\dfrac{-5.69}{8.6} \right)} \approx -33.5 ^{\circ}\alpha[/tex]
- The force acta in a direction of approximately 33.5° clockwise from the horizontal x-axis
Learn more about Coulomb's Law on electric force here
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