Question

Two infinite planes each with uniform surface charge density $+\sigma C/m^2$ are kept in such a way that the angle between them is $30°$. The electric field in the region shown between them is given by

• A. $\frac{\sigma }{2{\epsilon }_0}\left[\left(1-\frac{\sqrt{3}}{2}\right)\hat{y}-\frac{\hat{x}}{2}\right]$
• B. $\frac{\sigma }{2{\epsilon }_0}\left[\left(1+\frac{\sqrt{3}}{2}\right)\hat{y}+\frac{\hat{x}}{2}\right]$
• C. $\frac{\sigma }{2{\epsilon }_0}\left[\left(1+\sqrt{3}\right)\hat{y}+\frac{\hat{x}}{2}\right]$
• D. $\frac{\sigma }{2{\epsilon }_0}\left[\left(1-\sqrt{3}\right)\hat{y}-\frac{\hat{x}}{2}\right]$

Answer 1

The correct option is A

$\frac{\sigma }{2{\epsilon }_0}\left[\left(1-\frac{\sqrt{3}}{2}\right)\hat{y}-\frac{\hat{x}}{2}\right]$

Step 1: Given data

Given, the uniform surface charge density of the the infinite sheets is $\sigma$.

Angle between the two planes, $\theta =30°$.

Step 2: Formulas used

Electric field due to infinite sheet is given as,
$\overrightarrow{E}=\frac{\sigma }{2{\epsilon }_0}\hat{n}$
where ${\epsilon }_0$ is the permittivity of free space and $\hat{n}$ is the unit vector along the normal of the plane and in the direction of electric field.

Step 3: Calculation

Let $\overrightarrow{E_1}$ be the electric field due to $S_1$ and $\overrightarrow{E_2}$ be the electric field due to $S_2$

From the figure, the unit normal vector of $\overrightarrow{E_1}$ is,
$$\begin{gathered} -\cos 60°\hat{x}-\sin 60°\hat{y} \\ =-\frac{1}{2}\hat{x}-\frac{\sqrt{3}}{2}\hat{y} \end{gathered}$$

Thus,
$\overrightarrow{E_1}=\frac{\sigma }{2{\epsilon }_0}\left(-\frac{1}{2}\hat{x}-\frac{\sqrt{3}}{2}\hat{y}\right)$

We have,
$\overrightarrow{E_2}=\frac{\sigma }{2{\epsilon }_0}\hat{y}$

Adding them both,

$\begin{array}{rcl}\overrightarrow{E_1}+\hspace{0.17em}\overrightarrow{E_2}& =& \frac{\sigma }{2{\epsilon }_0}\left(-\frac{1}{2}\hat{x}-\frac{\sqrt{3}}{2}\hat{y}\right)+\frac{\sigma }{2{\epsilon }_0}\hat{y}\\ & =& \frac{\sigma }{2{\epsilon }_0}\left(-\frac{1}{2}\hat{x}-\frac{\sqrt{3}}{2}\hat{y}+\hat{y}\right)\\ & =& \frac{\sigma }{2{\epsilon }_0}\left(-\frac{1}{2}\hat{x}-\frac{\sqrt{3}}{2}\hat{y}+\hat{y}\right)\\ & =& \frac{\sigma }{2{\epsilon }_0}\left[\left(1-\frac{\sqrt{3}}{2}\right)\hat{y}-\frac{\hat{x}}{2}\right]\end{array}$

Therefore, the electric field in the region shown between them is $\frac{\sigma }{2{\epsilon }_0}\left[\left(1-\sqrt{3}\right)\hat{y}-\frac{\hat{x}}{2}\right]$.

Hence, option A is correct.