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 speed of v₀ = 2.0 × 10⁷ m/s. What is the speed of the electron as it strikes plate B? (e = 1.6 × 10^-19 C, m_electron = 9.11 × 10^-31 kg) 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 electric field acceleration acting on the electron can be determined as follows:
F = ma = eE
a = eE/m = 1.6 × 10^-19 C × 12,000 N/C / 9.11 × 10^-31 kg = 2.11 × 10^16 m/s^2
where F is the electrostatic force, m is the mass of the electron, and e is the electron charge.

Next, we can use kinematic equations to determine the final speed of the electron. Since the electron is projected horizontally, we only need to consider the vertical motion. The vertical displacement of the electron can be determined as follows:
y = 1/2 at^2 = 1/2 (2.11 × 10^16 m/s^2) (t/2)^2 = 1.06 × 10^16 (t/2)^2
where t/2 is the time it takes for the electron to reach the other plate (since it starts from rest).

The vertical velocity of the electron can be determined as follows:
v = at/2 = (2.11 × 10^16 m/s^2) (t/2)

Finally, the speed of the electron can be determined using the Pythagorean theorem, since the horizontal velocity remains constant throughout the motion:
v^2 = (v0)^2 + (v_y)^2 = (2.0 × 10^7 m/s)^2 + [(2.11 × 10^16 m/s^2) (t/2)]^2

We can solve for t using the vertical displacement equation and the fact that the separation of the plates is 40 mm:
40 × 10^-3 m = 1.06 × 10^16 (t/2)^2
t/2 = sqrt[(40 × 10^-3 m) / (1.06 × 10^16)] = 1.84 × 10^-9 s

Substituting t into the speed equation, we get:
v = sqrt[(2.0 × 10^7 m/s)^2 + [(2.11 × 10^16 m/s^2) (1.84 × 10^-9 s/2)]^2] ≈ 1.5 × 10^7 m/s ≈ 1.5 × 10^17 cm/s ≈ 0.05c

Therefore, the best answer choice is B.

A Pair of Charged Conducting Plates Produces a Uniform Field