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 (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 (Q)/(r^2)=(8.99 \cdot 10^(9) Nm^2C^(-2)) (4(1.6 \cdot 10^(-19)C))/((16\cdot 10^(-9)m)^2)= 2.25 \cdot 10^7 N/C`