Question

find the magnitude of an electron's electric field if the field itself is 50.0 cm away from it

Answer 1

The magnitude of the electric field, created by a point charge $q$ at distance $r$ is $E = k\frac{q}{r^2}$, where $k = \frac{1}{4 \pi \varepsilon_0} = 8.85 \cdot 10^{-12} N m^2 C^{-2}$.

Hence, for $q = e = 1.6 \cdot 10^{-19}C$ and $r = 50 cm = 0.5 m$, the magnitude of electric field is $E = 5.7 \cdot 10^{-30} \frac{V}{m}$.