question_answer

Two infinite planes each with uniform surface charge density \[+\sigma \] are kept in such a way that the angle between them is \[30{}^\circ \]. The electric field in the region shown between them is given by [JEE MAIN Held on 07-01-2020 Morning]

A) \[\frac{\sigma }{2{{\varepsilon }_{0}}}\left[ \left( 1+\sqrt{3} \right)\hat{y}+\frac{{\hat{x}}}{2} \right]\]

B) \[\frac{\sigma }{2{{\varepsilon }_{0}}}\left[ \left( 1+\sqrt{3} \right)\hat{y}-\frac{{\hat{x}}}{2} \right]\]

C) \[\frac{\sigma }{2{{\varepsilon }_{0}}}\left[ \left( 1-\frac{\sqrt{3}}{2} \right)\hat{y}-\frac{{\hat{x}}}{2} \right]\]

D) \[\frac{\sigma }{{{\varepsilon }_{0}}}\left[ \left( 1+\frac{\sqrt{3}}{2} \right)\hat{y}+\frac{{\hat{x}}}{2} \right]\]

Correct Answer: C

Solution :

[c] \[{{\vec{E}}_{1}}=\frac{\sigma }{2{{\varepsilon }_{0}}}\hat{y}\] \[{{\vec{E}}_{2}}=\frac{\sigma }{2{{\varepsilon }_{0}}}(-cos60{}^\circ \hat{x}-\sin 60{}^\circ \hat{y})\] \[=\frac{\sigma }{2{{\varepsilon }_{0}}}\left( -\frac{1}{2}\hat{x}-\frac{\sqrt{3}}{2}\hat{y} \right)\] \[\therefore \overrightarrow{{{E}_{P}}}=\overrightarrow{{{E}_{1}}}+\overrightarrow{{{E}_{2}}}=\frac{\sigma }{2{{\varepsilon }_{0}}}\left[ -\frac{1}{2}\hat{x}+\left( 1-\frac{\sqrt{3}}{2} \right)\hat{y} \right]\]