A grain of dust weighing 8 mg is at rest in a vertical uniform electric field.

A grain of dust weighing 8 mg is at rest in a vertical uniform electric field. Determine the value of the field strength if the charge of the dust particle is 2 μC.

Data: m (mass of a dust grain at rest) = 8 mg (8 * 10-6 kg); q (charge of the considered dust grain) = 2 μC (2 * 10-6 C).

Constants: g (acceleration due to gravity) ≈ 10 m / s2.

Since the grain of dust under consideration is at rest, the force of the action of the electric field is equal to the force of gravity and the sought-for field strength can be expressed from the equality: E * q = m * g, whence E = m * g / q.

Let’s calculate: E = 8 * 10-6 * 10/2 * 10-6 = 40 V / m.

Answer: The sought field strength is 40 V / m.