1. Write the forces

For two particles separated by distance $r$:

• Electrostatic force magnitude: $F_e = \frac{1}{4\pi\varepsilon_0}\frac{|q_1 q_2|}{r^2}$

• Gravitational force magnitude: $F_g = G\frac{m_1 m_2}{r^2}$

So their ratio is $\frac{F_e}{F_g} = \frac{\frac{1}{4\pi\varepsilon_0}|q_1q_2|}{Gm_1m_2}$

Notice that $r^2$ cancels, so the ratio is independent of distance.

2. Substitute the given values

Given: $q_1 = 6.67\times 10^{-19}\,\text{C}, \qquad q_2 = 9.6\times 10^{-10}\,\text{C}$ $m_1 = 19.2\times 10^{-27}\,\text{kg}, \qquad m_2 = 9\times 10^{-27}\,\text{kg}$ $\frac{1}{4\pi\varepsilon_0}=9\times 10^9, \qquad G=6.67\times 10^{-11}$

Thus $\frac{F_e}{F_g}=\frac{(9\times 10^9)(6.67\times 10^{-19})(9.6\times 10^{-10})}{(6.67\times 10^{-11})(19.2\times 10^{-27})(9\times 10^{-27})}$

3. Simplify carefully

First, cancel common factors $6.67$ and $9$ from numerator and denominator: $\frac{F_e}{F_g}=\frac{10^9\cdot 9.6\times 10^{-19}\times 10^{-10}}{10^{-11}\cdot 19.2\times 10^{-27}\times 10^{-27}}$

Now, $\frac{9.6}{19.2}=\frac{1}{2}$

So $\frac{F_e}{F_g}=\frac{1}{2}\times 10^{9-19-10+11+27+27}$

Add exponents: $9-19-10+11+27+27 = 45$

Hence $\frac{F_e}{F_g}=\frac{1}{2}\times 10^{45} = 5\times 10^{44}$

4. Match with the given form

The question states the ratio is $P\times 10^{-13}$.

So, $P\times 10^{-13} = 5\times 10^{44}$

Therefore, $P = 5\times 10^{57}$

5. Interpretation

Since this is an integer-type question, $P$ is expected to be an integer. But here, $P = 5\times 10^{57}$ which is not a usual finite integer entry format for such exams.

This strongly indicates a misprint in the question statement, most likely in the power of $10$ attached to one of the charges, or in the form $P\times 10^{-13}$.

If the intended form had been $P\times 10^{44}$, then $P=5$.

So the given data and asked format are inconsistent, which justifies the stored answer being BONUS.