Question

Please show all work and equations!

An electron of mass me 9.11 x 10-31 kg and a charge of ge-1.60 x 10-19 C is released, from rest, in a region of uniform electric field that points to the right with a magnitude of E 3.23 N/C. How long does it take for the electron to reach a speed of 2.64 x 10 m/s? Assume the experiment is performed in a vacuum and that you can ignore the effects of gravity and friction. Give you answer in seconds using the exponential form (.e., use 1.23e-4 to represent 123.4 microseconds rather than 0.0001234). 4 Enter answer here seconds

Answer 1

Given m_e = 9.1 times 10^-31 kg q_e = 1.6 times 10^-39 c v = 2.64 times 10^4 Vertical-bar E^rightarrow Vertical-bar = 3.23 N/c electric force F_e = q_e E = ma -1.6 times 10^-19 times 3.23 = 9.1 times 10^-31 times a a = -1.6 times 10^-13 times 3.23/9.1 times 10^-31 a = -5.168 times 10^-13/9.1 times 10^-31 a = -0.567 times 10^12 m/s^2 initial velocity (u) = 0 final velocity (v) = 2.64 times 10^4 v = u + at t = v/a t = 2.64 times 1064/0.567 times 10^12 t = 4.656 times 10^-8 s