Answer
$distance =1.2\times10^{-7}m$
Work Step by Step
Charge on the proton $q=+e=1.6\times10^{-19}C$
given magnitude of electric field $E=1.0\times10^{5}N/C$
suppose above electric field is at a distance $r$
magnitude of the electric field due to a point charge is
$E= \frac{kq}{r^2}$
so $r=\sqrt \frac{kq}{E}$
putting $q=+e=1.6\times10^{-19}C$, $E=1.0\times10^{5}N/C$, $k=9.0\times10^{9}N.m^2/C^2$
$r=\sqrt \frac{9.0\times10^{9}N.m^2/C^2\times 1.6\times10^{-19}C}{1.0\times10^{5}N/C}$
$r=\sqrt (1.44\times10^{-14}m^2)=1.2\times10^{-7}m$
