A coin is dropped in a 15.0 m deep well.
1 answer:
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Answer:
θ = 39.7º
Explanation:
In this exercise we must use Newton's second law for the sphere, at the equilibrium point we write the equations in each exercise; we will assume that plate 1 is on the left
Y Axis
-W = 0
= W
X axis
-<u> - F_{e2} + Tₓ = 0
</u>
<u> </u>
let's use trigonometry to find the components of the tension, we measure the angle with respect to the vertical
sin θ = Tₓ / T
cos θ = T_{y} / t
Tₓ = T sin θ
T_{y} = T cos θ
let's use gauss's law to find the electric field of each leaf; We define a Gaussian surface formed by a cylinder, so the component of the field perpendicular to the base of the cylinder is the one with electric flow.
F = ∫ E. dA = / ε₀
in this case the scalar product is reduced to the algebraic product, the flow is towards both sides of the plate
F = 2E A = q_{int} / ε₀
let's use the concept of surface charge density
σ = q_{int} / A
we substitute
2E A = σ A /ε₀
E = σ / 2ε₀
this is the field created by each plate. The electric force is
= q E
for plate 1 with σ₁ = -30 10⁻⁶ C / m²
F_{e1} = q σ₁ /2ε₀
for plate 2 with s2 = ab 10⁻⁶ C / m², for the calculations a value of this charge density is needed, suppose s2 = 10 10⁻⁶ C / m²
F_{e2} = q σ₂ /2ε₀
we substitute and write the system of equations
T cos θ = mg
- q σ₁ / 2ε₀ - q σ₂ /2ε₀ + T sinθ = 0
we introduce t in the second equations
- q /2 ε₀ (σ₁ + σ₂) + (mg / cos θ) sin θ = 0
mg tan θ = q /2ε₀ (σ₁ + σ₂)
θ = tan -1 (q / 2ε₀ mg (σ₁ + σ₂)
data indicates the mass of 0.25 g = 0.25 10⁻³ kg
give the charge density on plate 2, suppose ab = 10 10⁻⁶ C / m²
let's calculate
θ = tan⁻¹ (9.0 10⁻¹⁰ (30 + 10) 10⁻⁶ / (2 8.85 10⁻¹² 0.25 10⁻³ 9.8))
θ = tan⁻¹ 8.3 10⁻¹)
θ = 39.7º