A pair of charged conducting plates produces a uniform field of 12,000 N/C, directed to the right, between the plates. The separation of the plates is 40 mm. An electron is projected from plate A, directly toward plate B, with an initial velocity of vo = 2.0 × 10⁷ m/s, as shown in the figure. (e = 1.60 × 10^-19 C, ε0 = 8.85 × 10^-12 C²/N ∙ m², mel = 9.11 × 10^-31 kg) The velocity of the electron as it strikes plate B is closest to

Question

Quizplus · Accepted Answer

The force on the electron in the uniform field is given by F = qE where q is the charge of the electron and E is the electric field strength. Since the electron is negatively charged, it experiences a force opposite in direction to the electric field, i.e., to the left. The force on the electron is then F = -eE.
Using Newton's second law, we have F = ma where m is the mass of the electron and a is its acceleration.
Combining the above two equations, we get -eE = ma, which gives a = -eE/m
Substituting the given values, we get a = -(1.60 × 10^-19 C)(12000 N/C)/(9.11 × 10^-31 kg) = -2.11 × 10^14 m/s^2
The velocity of the electron as it strikes plate B can be found using the kinematic equation vf^2 = vi^2 + 2ax, where vf is the final velocity (which we need to find), vi is the initial velocity (given as 2.0 × 10^7 m/s), a is the acceleration (which we have calculated), and x is the distance the electron travels between the plates (which is the separation of the plates, given as 40 mm or 0.04 m).
Substituting the values, we get vf^2 = (2.0 × 10^7)^2 + 2(-2.11 × 10^14)(0.04) = 4.00 × 10^14
Taking the square root, we get vf = sqrt(4.00 × 10^14) = 2.0 × 10^7 m/s
Therefore, the closest answer is B) 1.5 × 10^7 m/s (rounded to two significant figures).

A Pair of Charged Conducting Plates Produces a Uniform Field