JEE Mains · Physics · STD 12 - 1. Electric charges and fields
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 \(\frac{\sigma}{\varepsilon_{0}}\left[\left(1+\frac{\sqrt{3}}{2}\right) \hat{\mathrm{y}}+\frac{\hat{\mathrm{x}}}{2}\right]\)
- B \(\frac{\sigma}{2 \varepsilon_{0}}\left[\left(1-\frac{\sqrt{3}}{2}\right) \hat{\mathrm{y}}-\frac{\hat{\mathrm{x}}}{2}\right]\)
- 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 & Solution
Correct Answer
(B) \(\frac{\sigma}{2 \varepsilon_{0}}\left[\left(1-\frac{\sqrt{3}}{2}\right) \hat{\mathrm{y}}-\frac{\hat{\mathrm{x}}}{2}\right]\)
Step-by-step Solution
Detailed explanation
Electric field due to each sheet is uniform and equal to \(\mathrm{E}=\frac{\sigma}{2 \varepsilon_{0}}\) Now net electric field between plates…
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