A positive ion $A$ and a negative ion $B$ have charges $6.67\times10^{-19}\ \text{C}$ and $9.6\times10^{-10}\ \text{C}$, and masses $19.2\times10^{-27}\ \text{kg}$ and $9\times10^{-27}\ \text{kg}$ respectively. At an instant, the ions are separated by a certain distance $r$. At that instant the ratio of the magnitudes of electrostatic force to gravitational force is $P\times10^{45}$, where the value of $10P$ is ______. (Take $\dfrac{1}{4\pi\varepsilon_0} = 9\times10^9\ \text{N m}^2\text{C}^{-2}$ and universal gravitational constant as $6.67\times10^{-11}\ \text{N m}^2\text{kg}^{-2}$.) Assume that charge may not be an integral multiple of electrons.
Numerical value type. Enter your answer.
Answer: 5
Both forces vary as $1/r^2$, so $r$ cancels:
$$\frac{F_e}{F_g} = \frac{kq_1q_2}{Gm_1m_2} = \frac{9\times10^9\times6.67\times10^{-19}\times9.6\times10^{-10}}{6.67\times10^{-11}\times19.2\times10^{-27}\times9\times10^{-27}}$$
Cancel $9$ and $6.67$, and use $\dfrac{9.6}{19.2} = \dfrac{1}{2}$:
$$\frac{F_e}{F_g} = \frac{1}{2}\times\frac{10^{9}\times10^{-19}\times10^{-10}}{10^{-11}\times10^{-27}\times10^{-27}} = \frac{1}{2}\times\frac{10^{-20}}{10^{-65}} = 0.5\times10^{45}$$
$P = 0.5$, so $10P = 5$.
Solution by Sreeraj P, M.Sc Physics