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 3 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:

Answer 1

Answer:

E = 3.74 X 10^{-7} N/C

upward direction

Explanation:

the force from an electric field is F = qE  

as in the given question, three electron is missing in the ion, so, it has a positive charge of-

(3)(1.6 X 10-19) = 4.8 X 10^{-19} C

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

so qE = mg

$E = \frac{mg}{q}$$E = \frac{(1.83 X 10-26)(9.81)}{(4.8 X 10^{-19})}$

E = 3.74 X 10^{-7} N/C

Since the charge is positive therefore the Field will  point in  upwards direction.