The potential at the surface of a conducting sphere is given by:
\( V = \frac{Kq}{r} \).
Since the two spheres are connected, their potentials will be the same:
\[ \frac{Kq_1}{a} = \frac{Kq_2}{b}. \]
Simplifying:
\[ \frac{q_1}{q_2} = \frac{a}{b}. \]
Final Answer: \( \frac{a}{b} \).
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: