Without the dielectric, the charge stored is given by:
\[ Q = \frac{A \epsilon_0 V}{d} \]
With the dielectric inserted:
\[ Q' = \frac{A \epsilon_0 V}{d - t + \frac{t}{K}} \]
Given:
\[ Q' = 1.25 Q \implies \frac{A \epsilon_0 V}{d - t + \frac{t}{K}} = 1.25 \cdot \frac{A \epsilon_0 V}{d} \]
Canceling common terms:
\[ \frac{d}{d - t + \frac{t}{K}} = 1.25 \]
Substituting \( d = 5 \, \text{mm} \) and \( t = 2 \, \text{mm} \):
\[ 1.25 \left( 5 - 2 + \frac{2}{K} \right) = 5 \]
Simplifying:
\[ 1.25 \left( 3 + \frac{2}{K} \right) = 5 \]
\[ 3 + \frac{2}{K} = 4 \]
\[ \frac{2}{K} = 1 \implies K = 2 \]
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: