The electric field at \( P \) is the vector sum of the fields due to the three charged sheets.
Using Gauss’s Law:
\[ \vec{E} = \frac{\sigma}{2\epsilon_0} \]
For the point \( P \):
\[ \vec{E}_P = \left( \frac{\sigma}{2\epsilon_0} + \frac{2\sigma}{2\epsilon_0} + \frac{\sigma}{2\epsilon_0} \right) (-\hat{i}) \]
Simplify:
\[ \vec{E}_P = \frac{4\sigma}{2\epsilon_0} (-\hat{i}) = \frac{2\sigma}{\epsilon_0} (-\hat{i}) \]
Thus, \( x = 2 \).
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 \( 4.606 \times 10^{-2} \text{ s}^{-1} \).