Anucleus 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:

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

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

the force required to move the cart is divided by three or 33.33 n.

where's the following? you gotta add it to know what the choices are

explanation: