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 \[ I(x) = \int \frac{dx}{(x-11)^{\frac{11}{13}} (x+15)^{\frac{15}{13}}} \] If \[ I(37) - I(24) = \frac{1}{4} \left( b^{\frac{1}{13}} - c^{\frac{1}{13}} \right) \] where \( b, c \in \mathbb{N} \), then \[ 3(b + c) \] is equal to:
For the thermal decomposition of \( N_2O_5(g) \) at constant volume, the following table can be formed, for the reaction mentioned below: \[ 2 N_2O_5(g) \rightarrow 2 N_2O_4(g) + O_2(g) \] Given: Rate constant for the reaction is \( 4.606 \times 10^{-2} \text{ s}^{-1} \).