Question

An electron is traveling at 1.9x10 m/s when it enters an electric field that slows it. The field has a mag nitude of 6200 N/C. How far does the electron penetrate the field before it stops.

Answer 1

The energy-work theorem says

$$\Delta E=W\\ \frac{mv^2}{2}=Fd$$

Hence

$$d=\frac{mv^2}{2F}=\frac{9.1*10^{-31}*(1.9*10)^2}{2*6200}=2.6*10^{-32}\;\rm m$$