The coulombian force (\(F_c\)) is given by:
\[ F_c = \frac{k Q_1 Q_2}{r^2} = \frac{9 \times 10^9 \times 1.6 \times 10^{-19} \times 1.6 \times 10^{-19}}{r^2} \]
The gravitational force (\(F_g\)) is given by:
\[ F_g = \frac{G m_1 m_2}{r^2} = \frac{6.67 \times 10^{-11} \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-27}}{r^2} \]
The ratio of the forces is:
\[ \frac{F_c}{F_g} \approx 0.23 \times 10^{40} \approx 2.3 \times 10^{39} \]
Therefore, the answer is approximately \(10^{39}\).
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is: