Answer
(a)magnitude of force on electron will be $ 5.76\times10^{-11}N$
(b) The net force on the system will be zero.
Work Step by Step
Distance between electron and proton $r=2.0nm=2.0\times10^{-9}m$
magnitude of charge on proton $q_{p}=1.6\times10^{-19}C$
magnitude of charge on electron $q_{e}=1.6\times10^{-19}C$
from coulombs law magnitude of force on electron will be
$F_{e}=k\frac{q_{e}\times q_{p}}{r^{2}}$ where $k= 9.00\times10^{9}N.m^{2}/C^{2}$
$F_{e}=9.00\times10^{9}N.m^{2}/C^{2}\times \frac{1.6\times10^{-19}C\times 1.6\times10^{-19}C}{(2.0\times10^{-9}m)^{2}}=5.76\times10^{-11}N$
(b) The net force on the system will be zero.
The proton exerts a attractive force of $ 5.76\times10^{-11}N$ on electron.
the electron also exerts same amount of attractive force on proton .
since both the forces are equal and opposite in nature the net force will be zero.
