Question

A small sphere of

Answer 1

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

       $T_{y}$ -W = 0

       $T_{y}$ = W

X axis

         -$F_{e1}$ - F_{e2} + Tₓ = 0

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 = $q_{int}$ / 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

        $F_{e}$ = 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