Answer to Question #133079 in Electricity and Magnetism for Herlyn Nelmida

Question

An electron is released from rest in a uniform electric field. The electron accelerates
vertically upward, travelling 4.50 m in the first 3.00 micro second after it is released. What
are the magnitude and direction of the electric field? [Note: Neglect the gravitational force]

Answer 1

First let us determine the acceleration of the electron. It starts from the rest, so the distance is $s = \dfrac{at^2}{2}.$ Therefore, the acceleration is

$a=\dfrac{2s} {t^2} =\dfrac{2\cdot4.5} {(3.00\cdot10^{-6}) ^2} = 10^{12}\mathrm{m/s}^2$

Next, let us determine the force acting on the electron $F=m_ea = 9.1\cdot10^{-31}\cdot 10^{12} = 9.1\cdot 10^{-19}\mathrm{N}.$

The electric field is $E=\dfrac{F} {q_e} =\dfrac{9.1\cdot10^{-19}} {1.6\cdot10^{-19}} =5.7\mathrm{N/C}.$

The acceleration is directed upwards, so the force is also directed upwards. The charge is negative, so the electric field is directed downwards.