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 $ 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: