Answer
a). $5.9\times10^{5}m/s$ , down
b). Lose potential energy
Work Step by Step
a). Change in potential energy = $qEd=-1.6\times10^{-19}\times1000\times10^{-3}=-1.6\times10^{-19}J$
This change is converted to kinetic energy, as the total energy is constant.
So, K.E. = $1.6\times10^{-19}J$
Therefore, $\frac{1}{2}m_{e}v_{e}^{2}=1.6\times10^{-19}$
Also, $m_{e}=9.1\times10^{-31}$
From above equation, $v_{e}=5.9\times 10^{5}m/s$
Since the electron is pointing upward, the electron is moving downwards.
b). The electron lose potential energy, which is converted to kinetic energy.
