MCQ
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
  • A
    $\frac{\sigma}{\varepsilon_{0}}\left[\left(1+\frac{\sqrt{3}}{2}\right) \hat{\mathrm{y}}+\frac{\hat{\mathrm{x}}}{2}\right]$
  • $\frac{\sigma}{2 \varepsilon_{0}}\left[\left(1-\frac{\sqrt{3}}{2}\right) \hat{\mathrm{y}}-\frac{\hat{\mathrm{x}}}{2}\right]$
  • C
    $\frac{\sigma}{2 \varepsilon_{0}}\left[(1+\sqrt{3}) \hat{\mathrm{y}}+\frac{\hat{\mathrm{x}}}{2}\right]$
  • D
    $\frac{\sigma}{2 \varepsilon_{0}}\left[(1+\sqrt{3}) \hat{\mathrm{y}}-\frac{\hat{\mathrm{x}}}{2}\right]$

Answer

Correct option: B.
$\frac{\sigma}{2 \varepsilon_{0}}\left[\left(1-\frac{\sqrt{3}}{2}\right) \hat{\mathrm{y}}-\frac{\hat{\mathrm{x}}}{2}\right]$
b
Electric field due to each sheet is uniform and equal to $\mathrm{E}=\frac{\sigma}{2 \varepsilon_{0}}$

Now net electric field between plates

$\overrightarrow{\mathrm{E}}_{\mathrm{net}} =\mathrm{E} \cos 60^{\circ}(-\hat{\mathrm{x}})+\left(\mathrm{E}-\mathrm{E} \sin 60^{\circ}\right)(\hat{\mathrm{y}})$

$=\frac{\sigma}{2 \varepsilon_{0}}\left[-\frac{\hat{\mathrm{x}}}{2}+\left(1-\frac{\sqrt{3}}{2}\right) \hat{\mathrm{y}}\right]$

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