Question

Estimate the magnitude of the electric field due to the proton in a hydrogen atom at a distance of
5.29×10 to the power of −11 m, the expected position of the electron in the atom.

Answer 1

Electric field due to a point charge is given as

$E = \frac{kq}{r^2}$

here we know that

$q = 1.6 \times 10^{-19} C$

also the distance is given as

$r = 5.29 \times 10^{-11} m$

now we will have

$E = \frac{(9\times 10^9)(1.6 \times 10^{-19})}{(5.29 \times 10^{-11})^2}$

so we will have

$E = 5.14 \times 10^{11} N/C$

so above is the electric field due to proton