Two infinite planes each with uniform surface charge density $+ σ$ are kept in such a way that the angle between them is $30^{0}$ . The electric field in the region shown between them is given by

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$\frac{σ}{2 ε_{0}} [(1 + \sqrt{3}) \hat{y} - \frac{\hat{x}}{2}]$

$\frac{σ}{2 ε_{0}} [(1 - \frac{\sqrt{3}}{2}) \hat{y} - \frac{\hat{x}}{2}]$

$\frac{σ}{ε_{0}} [(1 + \frac{\sqrt{3}}{2}) \hat{y} + \frac{\hat{x}}{2}]$

$\frac{σ}{2 ε_{0}} [(1 + \sqrt{3}) \hat{y} + \frac{\hat{x}}{2}]$

answer is A.

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Detailed Solution

$\begin{array}{l} \vec{E} = {\vec{E}}_{d u e t o U p p e r p l a t e} + {\vec{E}}_{d u e t o L o w e r p l a t e} \\ \Rightarrow \vec{E} = \frac{σ}{2 ε_{0}} \cos 60 º (- \hat{x}) + \frac{σ}{2 ε_{0}} \sin 60 º (- \hat{y}) + \frac{σ}{2 ε_{0}} (\hat{y}) \\ \Rightarrow \vec{E} = \frac{σ}{2 ε_{0}} \cos 60 º (- \hat{x}) + [\frac{σ}{2 ε_{0}} - \frac{σ}{2 ε_{0}} \sin 60 º] (\hat{y}) \end{array}$