Question

Imagine a Carbon atom which has lost four of its electrons. What is strength of the electric field due this atom at a distance of 16 nm?

Answer 1

If the atom of carbon has lost 4 electrons, it has now an excess of charge equal to +4e:
$Q=+4e$
where $e=1.6 \cdot 10^{-19}C$

Its charge is concentrated in the nucleus, so we can treat it as a single-point charge, whose electric field is given by:
$E=k \frac{Q}{r^2}$
where k is the Coulomb's constant and r is the distance from the charge. In our problem,
$r=16 nm=16 \cdot 10^{-9} m$
therefore the electric field at this distance is
$E=k \frac{Q}{r^2}=(8.99 \cdot 10^{9} Nm^2C^{-2}) \frac{4(1.6 \cdot 10^{-19}C)}{(16\cdot 10^{-9}m)^2}= 2.25 \cdot 10^7 N/C$