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Question

(B) The electric field due to an infinite plane sheet with surface charge density $\sigma$ is $E = \frac{\sigma}{2\varepsilon_{0}}$,directed away from

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) The electric field due to an infinite plane sheet with surface charge density $\sigma$ is $E = \frac{\sigma}{2\varepsilon_{0}}$,directed away from the sheet.Let the horizontal sheet be along the $x$-axis. Its electric field is $\vec{E}_{1} = \frac{\sigma}{2\varepsilon_{0}} \hat{y}$.The second sheet is at an angle of $30^{\circ}$ with the $x$-axis. Its normal makes an angle of $30^{\circ} + 90^{\circ} = 120^{\circ}$ with the positive $x$-axis.The electric field due to the second sheet is $\vec{E}_{2} = \frac{\sigma}{2\varepsilon_{0}} (\cos 120^{\circ} \hat{x} + \sin 120^{\circ} \hat{y}) = \frac{\sigma}{2\varepsilon_{0}} (-\frac{1}{2} \hat{x} + \frac{\sqrt{3}}{2} \hat{y})$.The net electric field is $\vec{E}_{net} = \vec{E}_{1} + \vec{E}_{2} = \frac{\sigma}{2\varepsilon_{0}} \hat{y} + \frac{\sigma}{2\varepsilon_{0}} (-\frac{1}{2} \hat{x} + \frac{\sqrt{3}}{2} \hat{y})$.$\vec{E}_{net} = \frac{\sigma}{2\varepsilon_{0}} [-\frac{1}{2} \hat{x} + (1 + \frac{\sqrt{3}}{2}) \hat{y}]$.

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

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