Question

Two 1.0 g balls are connected by a 2.0-cm-long insulating rod of negligible mass. One ball has a charge of +10 nC, the other a 4 charge of - 10 nC. The rod is held in a 1.0 * 10 N/C uniform electric field at an angle of 30° with respect to the field, then released. What is its initial angular acceleration?

Answer 1

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