Answer
$ratio=4.16\times10^{42}$
Work Step by Step
The ratio between electrical and gravitational force is:
$ratio=\frac{F_e}{F_g}$ ............eq(1)
As $F_e=\frac{Kq_1q_2}{r^2}$ (in this case, $q_1=q_2=q$)
so $F_e=\frac{Kq^2}{r^2}$ .........eq(2)
Also, we know that
$F_g=\frac{Gm_1m_2}{r^2}$ (We know that $m_1=m_2=m$)
Hence:
$F_g=\frac{Gm^2}{r^2}$ ............eq(3)
Putting values from eq(2) and eq(3) in eq(1), we get:
$ratio=\frac{\frac{Kq^2}{r^2}}{ \frac{Gm^2}{r^2}}$
or $ratio=\frac{Kq^2}{r^2}\times\frac{r^2}{Gm^2}$
$ratio=\frac{Kq^2}{Gm^2}$
Plugging in the known values, we obtain:
$ratio=\frac{(9\times10^9)(1.60\times10^{-19})^2}{(6.67\times10^{-11})(9.11\times10^{-31})^2}$
$ratio=4.16\times10^{42}$
