Respuesta :
Answer:
F = 0.111015 N
Explanation:
For this exercise the force is given by Coulomb's law
F = k q₁q₂ / r₂₁²
we calculate the electric force of the other two particles on the charge q1
Charges q₁ and q₂
the distance between them is
r₁₂ = y₁ -y₂
r₁₂ = 0.30 + 0.30
r₁₂ = 0.60 m
let's calculate
F₁₂ = 9 10⁹ 2 10⁻⁶ 2 10⁻⁶ / 0.60 2
F₁₂ = 1 10⁻¹ N
directed towards the positive side of the y-axis
Charges 1 and 3
Let's find the distance using the Pythagorean Theorem
r₁₃ = RA [(0.40-0) 2 + (0-0.30) 2]
r₁₃ = 0.50 m
F₁₃ = 9 10⁹ 2 10⁻⁶ 4 10⁻⁶ / 0.50²
F₁₃ = 1.697 10⁻² N
The direction of this force is on the line that joins the two charges (1 and 3), let's use trigonometry to find the components of this force
tan θ = y / x
θ = tan⁻¹ y / x
θ = tan⁻¹ 0.3 / 0.4
tea = 36.87º
The angle from the positive side of the x-axis is
θ ’= 180 - θ
θ ’= 180 - 36.87
θ ’= 143.13º
sin143.13 = F_13y / F₁₃
F_13y = F₁₃ sin 143.13
F{13y} = 1.697 10⁻² sin 143.13
F_13y = 1.0183 10⁻² N
cos 143.13 = F_13x / F₁₃
F₁₃ₓ = F₁₃ cos 143.13
F₁₃ₓ = 1.697 10⁻² cos 143.13
F₁₃ₓ = -1.357 10-2 N
Now we can find the components of the resultant force
Fx = F13x + F12x
Fx = -1,357 10-2 +0
Fx = -1.357 10-2 N
Fy = F13y + F12y
Fy = 1.0183 10-2 + 1 10-1
Fy = 0.110183 N
We use the Pythagorean theorem to find the modulus
F = Ra (Fx2 + Fy2)
F = RA [(1.357 10-2) 2 + 0.110183 2]
F = 0.111015 N
Let's use trigonometry for the angles
tan tea = Fy / Fx
tea = tan-1 (0.110183 / -0.01357)
tea = 1,448 rad
to find the angle about the positive side of the + x axis
tea '= pi - 1,448
Tea = 1.6936 rad
The magnitude of the net force on q1 exerted by the other two charges is 0.357 N.
The direction of the net force on q1 is 50⁰.
The given parameters;
- q1 = 2.0 μC located at (0, 0.3) m
- q2 = 2.0 μC located at (0, -0.3) m
- q3 = 4.0 μC located at (0.4, 0) m
The force on q1 due to q2 occurs only in y-direction and can be calculated using Coulomb's law as shown below;
[tex]F_1_2 = \frac{kq^2}{r^2}j = \frac{(9\times 10^9) \times (2\times 10^{-6})^2 }{(0.3 +0.3)^2} j \\\\\F_{12} = (0.1)j[/tex]
The force on q1 due to q3 occurs both in x-direction and y-direction, and it is calculated as follows;
[tex]distance \ between \ q1 \ and \ q3\ , r_{13} = \sqrt{0.3^2 \ + \ 0.4^2} = 0.5 \ m\\\\F_{13} = \frac{kq^2}{r_{13}^2} (\frac{0.4i}{0.5} \ + \ \frac{0.3j}{0.5} )\\\\F_{13} = \frac{kq^2}{r_{13}^2} (0.8i + 0.6j)\\\\F_{13} = \frac{9\times 10^9 \times (2\times 10^{-6}) \times (4\times 10^{-6})}{0.5^2} (0.8i + 0.6j)\\\\F_{13} = 0.288(0.8i + 0.6j)\\\\F_{13} = 0.23i + 0.173j[/tex]
The net force is calculated as follows;
[tex]F_{net} = F_{12} \ + \ F_{13}\\\\F_{net} = (0.1j) \ + \ (0.23i + 0.173j)\\\\F_{net} = (0.23i + 0.273j)[/tex]
The magnitude of the net force on q1 is calculated as follows;
[tex]|F| = \sqrt{(0.23^2) \ + \ (0.273^2)} \\\\|F| = 0.357 \ N[/tex]
The direction of the net force on q1 is calculated as follows;
[tex]tan(\theta )= \frac{F_y}{F_x} \\\\\theta = tan^{-1} (\frac{0.273}{0.23} )\\\\\theta = 50^0[/tex]
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