Answer to Question #135630 in General Chemistry for Yaaseen Andor

Question

An electron is released from rest in a uniform electric field. The electron accelerates vertically upward, travelling 4.5 m in the first 3.00 µs after it is released. What are the magnitude and direction of the electric field?

Note: Show your free body diagram (FBD) and solutions accurately, systematically and orderly with the standard unit, and express it into prefix word if possible.

Answer 1

using equations of motion

$S=ut+\frac12at^2$

$4.5=0+0.5\times a \times(3\times10^{-6})^2\\4.5=0.5\times9\times10^{-12}\times a\\a=10^{12} m/s^2$

$a=eE/m \\10^{12}=\frac{1.6\times10^{-19}\times E}{9.1\times10^{-31}}$

$E=5.7 V/m$