A positive ion $A$ and a negative ion $B$ has charges $6.67 \times 10^{-19} \mathrm{C}$ and $9.6 \times 10^{-10} \mathrm{C}$, and masses $19.2 \times 10^{-27} \mathrm{~kg}$ and $9 \times 10^{-27} \mathrm{~kg}$ 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 $\mathrm{P} \times 10^{-13}$, where the value of P is _________.

(Take $\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \mathrm{Nm}^2 \mathrm{C}^{-1}$ and universal gravitational constant as $6.67 \times 10^{-11} \mathrm{Nm}^2 \mathrm{~kg}^{-2}$ )

<p>To find the ratio of the magnitudes of electrostatic force to gravitational force between ions $A$ and $B$, we use the following formulas for electrostatic force ($F_e$) and gravitational force ($F_g$):</p> <p>$ F_e = \frac{k \cdot q_1 \cdot q_2}{r^2} $</p> <p>$ F_g = \frac{G \cdot m_1 \cdot m_2}{r^2} $</p> <p>To find the ratio $\frac{F_e}{F_g}$, simplify as follows:</p> <p>$ \frac{F_e}{F_g} = \frac{k \cdot q_1 \cdot q_2}{G \cdot m_1 \cdot m_2} $</p> <p>Using the given values:</p> <p><p>$k = 9 \times 10^9 \, \text{Nm}^2 \text{C}^{-1}$</p></p> <p><p>$q_1 = 6.67 \times 10^{-19} \, \text{C}$</p></p> <p><p>$q_2 = 9.6 \times 10^{-10} \, \text{C}$</p></p> <p><p>$G = 6.67 \times 10^{-11} \, \text{Nm}^2 \text{kg}^{-2}$</p></p> <p><p>$m_1 = 19.2 \times 10^{-27} \, \text{kg}$</p></p> <p><p>$m_2 = 9 \times 10^{-27} \, \text{kg}$</p></p> <p>The formula becomes:</p> <p>$ \frac{F_e}{F_g} = \frac{9 \times 10^9 \times 6.67 \times 10^{-19} \times 9.6 \times 10^{-10}}{6.67 \times 10^{-11} \times 19.2 \times 10^{-27} \times 9 \times 10^{-27}} $</p> <p>Calculate the result:</p> <p>$ = \frac{10^{-20}}{2 \times 10^{-65}} $</p> <p>This calculation gives the ratio of the electrostatic force to the gravitational force without considering their separation distance $r$, as it cancels out. The value of $P$ mentioned in the prompt can be deduced from the simplified final expression.</p>