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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 fro

Vedclass · Accepted Answer

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

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 lie along the $x$-axis. Its electric field is $\vec{E}_{1} = \frac{\sigma}{2\varepsilon_{0}} \hat{y}$.The second sheet is inclined at an angle of $30^{\circ}$ to 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 $\vec{E}_{2} = \frac{\sigma}{2\varepsilon_{0}} (\cos 120^{\circ} \hat{x} + \sin 120^{\circ} \hat{y}) = \frac{\sigma}{2\varepsilon_{0}} \left( -\frac{1}{2} \hat{x} + \frac{\sqrt{3}}{2} \hat{y} \right)$.The net electric field is $\vec{E}_{net} = \vec{E}_{1} + \vec{E}_{2} = \frac{\sigma}{2\varepsilon_{0}} \hat{y} + \frac{\sigma}{2\varepsilon_{0}} \left( -\frac{1}{2} \hat{x} + \frac{\sqrt{3}}{2} \hat{y} \right)$.Thus, $\vec{E}_{net} = \frac{\sigma}{2\varepsilon_{0}} \left[ -\frac{1}{2} \hat{x} + \left( 1 + \frac{\sqrt{3}}{2} \right) \hat{y} \right]$. This matches option $(B)$.

Two infinite planes, each with a uniform surface charge density $+\sigma$, are kept such that the angle between them is $30^{\circ}$. The electric field in the region between them is given by:

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