Answer:
α = 5 10⁻³ rad / s²
Explanation:
For this exercise we can use Newton's second law for rotational movement, where the force is electric
τ = I α
Where the torque is
τ = F x r = F r sin θ
Strength is
F = q E
The moment of inertia of a small ball, which we approximate to a point is
I = m r²
We replace
2 (q E) r sin θ = 2m r² α
The number 2 is because the two forces create the same torque
α = q E sin θ / m r
Let's reduce the magnitudes to the SI system
m = 1.0g = 1.0 10⁻³ kg
L = 2.0 cm = 2.0 10⁻² m
q = 10 nc = 10 10⁻⁹ C
E = 1.0 10 N / C
r = L / 2
r = 1.0 10⁻² m
Let's calculate
α = 10 10⁻⁹ 1.0 10 sin 30 / 1.0 10⁻³ 1.0 10⁻²
α = 5 10⁻³ rad / s²