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

Question

Quizplus · Accepted Answer

First, we find the acceleration of the electron using the uniform field:
$a = \frac{F}{m} = \frac{qE}{m} = \frac{(-1.6 	imes 10^{-19} C)(12,000 N/C)}{9.11 	imes 10^{-31} kg} \approx -2.10 	imes 10^{18} m/s^2$
Note that the acceleration is negative because the electron is moving in the opposite direction of the electric field.
Next, we find the time it takes for the electron to travel from plate A to plate B:
$d = \frac{1}{2}at^2 \Rightarrow t = \sqrt{\frac{2d}{a}} = \sqrt{\frac{2(0.04 m)}{2.10 	imes 10^{18} m/s^2}} \approx 1.28 	imes 10^{-9} s$
Finally, we find the final velocity of the electron using the initial velocity and acceleration:
$v_f = v_i + at = (2.0 	imes 10^7 m/s) + (-2.10 	imes 10^{18} m/s^2)(1.28 	imes 10^{-9} s) \approx 1.54 	imes 10^7 m/s \approx 1.5 	imes 10^7 m/s \approx 1.5 	imes 10^{17} cm/s$
Therefore, the closest choice is B, 1.5 × 10^17 m/s.

A Pair of Charged Conducting Plates Produces a Uniform Field