Two very large parallel metal plates, separated by 0.20 m, are connected across a 12-V source of potential. An electron is released from rest at a location 0.10 m from the negative plate. When the electron arrives at a distance 0.050 m from the positive plate, how much kinetic energy has the electron gained? (e = 1.60 × 10^-19 C)

Question

Quizplus · Accepted Answer

First, we need to calculate the electric field between the two plates. The electric field is given by E = V/d, where V is the potential difference between the plates and d is the distance between them. Plugging in the given values, we get E = 12/0.20 = 60 N/C.

The force on the electron is F = e*E, where e is the charge of the electron. Plugging in the given values, we get F = (1.60 × 10^-19 C)(60 N/C) = 9.60 × 10^-18 N.

The work done by the electric field on the electron as it moves from 0.10 m to 0.050 m is W = F*d, where d is the distance moved. Plugging in the given values, we get W = (9.60 × 10^-18 N)(0.050 m - 0.10 m) = -4.80 × 10^-19 J. The negative sign indicates that the work done by the electric field is negative (i.e., the electric field does negative work on the electron).

The change in kinetic energy of the electron is equal to the work done on it by the electric field, since there are no other forces doing work on the electron. So the kinetic energy gained by the electron is equal to |-4.80 × 10^-19 J| = 4.80 × 10^-19 J.

Therefore, the best choice is B, which is 4.8 × 10^-19 J.

Two Very Large Parallel Metal Plates, Separated by 0