An electron with a speed of 2.0 x 10⁶ m/s moves into a uniform electric field of 500 N/C that is parallel to the electron's motion.How long does it take to bring the electron to rest? (m_e= 9.11 x 10^-31 kg,e = 1.6 x 10^-19 C)

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Quizplus · Accepted Answer

The force on the electron is F = eE = 1.6 x 10^-19 C * 500 N/C = 8 x 10^-17 N. The acceleration is a = F/m = 8 x 10^-17 N / 9.11 x 10^-31 kg = 8.78 x 10^13 m/s^2. The time to stop is t = v/a = (2.0 x 10^6 m/s) / (8.78 x 10^13 m/s^2) = 2.3 x 10^-8 s.

An Electron with a Speed of 2