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 v_o = 2.0 × 10⁷m/s, as shown in the figure. (e= 1.60 × 10^-19C, ε₀ = 8.85 × 10^-12 C²/N ∙ m², m_el = 9.11 × 10^-31 kg) The velocity of the electron as it strikes plate B is closest to A) 1.2 × 10⁷ m/s. B) 1.5 × 10⁷ m/s. C) 1.8 × 10⁷ m/s. D) 2.1 × 10⁷ m/s. E) 2.4 × 10⁷ m/s.

Question

Quizplus · Accepted Answer

The force acting on the electron can be found using the formula F = qE, where q is the charge of the electron and E is the electric field strength between the plates. Plugging in the values, we get:

F = (1.60 × 10^-19 C)(12,000 N/C) = 1.92 × 10^-15 N

The acceleration of the electron can then be found using the formula a = F/m, where m is the mass of the electron. Plugging in the values, we get:

a = (1.92 × 10^-15 N)/(9.11 × 10^-31 kg) = 2.11 × 10^15 m/s^2

Using the kinematic equation v^2 = u^2 + 2as, where u is the initial velocity of the electron (2.0 × 10^7 m/s), s is the distance the electron travels (the separation between the plates, 40 mm or 0.04 m), and v is the final velocity of the electron (the velocity as it strikes plate B), we can solve for v. Plugging in the values, we get:

v = sqrt((2.0 × 10^7 m/s)^2 + 2(2.11 × 10^15 m/s^2)(0.04 m)) = 1.5 × 10^7 m/s

Therefore, the closest choice to the velocity of the electron as it strikes plate B is 1.5 × 10^7 m/s, which corresponds to choice B.

A Pair of Charged Conducting Plates Produces a Uniform Field