Question

What is the magnitude of the electric force between a proton and an electron when they are at a distance of 3.57 angstrom from each other?

Answer 1

Answer:

$1.812\times 10^{-9} N$ is the magnitude of the electric force between a proton and an electron.

Explanation:

Coulomb's law is given as ;

$F=K\times \frac{q_1\times q_2}{r^2}$

$q_1,q_2$ = Charges on both charges

r = distance between the charges

K = Coulomb constant =$9\times 10^{9} N m^2/C^2$

We have ;

Charge of proton =$q_1=1.602\times 10^{-19} C$

Charge of electron =$q_2=-1.602\times 10^{-19} C$

$r=3.57 \AA=3.57\times 10^{-10} m$

Force between the proton and electron at r distance will be :

$F=9\times 10^{9} N m^2/C^2\times \frac{1.602\times 10^{-19} C\times 1.602\times 10^{-19} C}{(3.57\times 10^{-10} m)^2}$

$F=-1.812\times 10^{-9} N$

(negative sign means that attractive force)

$1.812\times 10^{-9} N$ is the magnitude of the electric force between a proton and an electron.