A small sphere of mass 0.25 g that carries a charge of 9.0 × 10−10 C.
Two parallel vertical infinite charged sheets of
charge densities σ1= -30 × 10−6 C/m2 and σ2= ab ×
10−6 C/m2 respectively. The sphere is attached to
one end of a very thin silk string 5.0 cm long. The
other end of the string is attached to the 2nd sheet as
shown in the figure. At equilibrium, the string will
make an angle (ϴ) with the vertical. Calculate the
angle that the string makes with the vertical?

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

         - - 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 = / 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

second mirror gathers light from a faraway object and reflects it to another mirror. - a.