An electron with a speed of 2.0 × 10⁶ m/s moves into a uniform electric field of 550 N/C that is parallel to the electron's motion.How long does it take to bring the electron to rest? (me = 9.11 × 10−³¹ kg,e = 1.6 × 10−¹⁹ C) A)2.1E-8 s B)6.4E-9 s C)1.6E-6 s D)1.6E-6 s. E)3.1E-7 s

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Quizplus · Accepted Answer

The time it takes to bring the electron to rest can be calculated using the equation $t = \frac{v}{a}$, where $v$ is the initial velocity of the electron and $a$ is the acceleration caused by the electric field. The force on the electron due to the electric field is $F = eE$, where $e$ is the charge of the electron and $E$ is the electric field strength. The acceleration $a$ is then $F/m_e = eE/m_e$. Substituting the given values: $a = (1.6 \times 10^{-19} \, \text{C}) \times (550 \, \text{N/C}) / (9.11 \times 10^{-31} \, \text{kg})$, and then calculating $t = v/a$ with $v = 2.0 \times 10^6 \, \text{m/s}$, we find the time to bring the electron to rest.

An Electron with a Speed of 2