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 electron experiences a constant electric force due to the uniform electric field between the plates. This force accelerates the electron to the right, and we can find the acceleration using the electric force equation:

F = qE

where F is the electric force, q is the charge of the electron, and E is the electric field. Plugging in the given values, we have:

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

Next, we can use the kinematic equation:

v_f^2 = v_i^2 + 2ax

where v_f is the final velocity, v_i is the initial velocity (given), a is the acceleration (from above), and x is the distance between the plates (given). Solving for v_f, we get:

v_f = sqrt[(2ax) + v_i^2] = sqrt[(2)(1.60 x 10^-19 C)(12,000 N/C)(0.040 m) + (2.00 x 10^7 m/s)^2]

v_f = 1.49 x 10^7 m/s ≈ 1.5 x 10^7 m/s

Therefore, the electron's velocity as it strikes plate B is closest to 1.5 x 10^7 m/s, or answer choice B.

A Pair of Charged Conducting Plates Produces a Uniform Field