Question

A nucleus of the boron-11 isotope consists of five protons and six neutrons. A particular ionized atom of boron-11, whose mass is 1.83 × 10^-26 kg, lacks 4 electrons from its neutral state. Find the magnitude and direction of the electric field that will levitate this ion, exactly balancing its weight. Take g = 9.81 m/s². Magnitude:

= ____________ N/C

Direction

Answer 1

The force from an electric field is F = qE   Since the ion is missing four electrons, it has a poisitve charge of...

(4)(1.6 X 10^-19) = 6.4 X 10^-19 C

To levitate, the electric force must match the weight of the ion

so qE = mg

E = mg/q

E = (1.83 X 10^-26)(9.81)/(6.4 X 10^-19)

E = 2.81 X 10^-7 N/C

Since the charge of the is positive, the Field will have to point upwards.