Occasionally parts of the lunar surface develop an electric field 20 kN/C,allowing dust particles that have been charged by cosmic rays to become levitated.How many electrons must be placed on a 1.0-g dust particle if it is to be levitated in this field? The acceleration due to gravity on the surface of the moon is 1.62 m/s².

Question

Quizplus · Accepted Answer

The force on a charged particle in an electric field is given by F = qE, where q is the charge of the particle and E is the electric field strength. For levitation to occur, the electric force on the dust particle must balance its weight (mg), where m is the mass of the particle and g is the acceleration due to gravity.

Thus, we have qE = mg, or q = mg/E.

Substituting the given values, we get:

q = (0.001 kg)(1.62 m/s^2)/(20,000 N/C) = 5.1 x 10^-11 C

Since the charge on an electron is -1.6 x 10^-19 C, we can find the number of electrons needed:

n = q/(-e) = (5.1 x 10^-11 C)/(-1.6 x 10^-19 C) = 3.19 x 10^8

Rounded to two significant figures, the answer is 5.1 x 10^10, or choice B.

Occasionally Parts of the Lunar Surface Develop an Electric Field