Question

Consider two infinite sheets forming the x-z plane and y-z plane having charge density $+\sigma$ and $-\sqrt{3}\sigma$ respectively. Find the angle (in degree) which the electric field makes at a point in the first quadrant with the positive x-axis.

Answer 1

150

Answer 2

The electric field due to an infinite plane sheet of charge with uniform surface charge density $\sigma$ is given by $\vec{E} = \frac{\sigma}{2\epsilon_0} \hat{n}$, where $\hat{n}$ is the unit vector normal to the sheet, pointing away from the sheet if $\sigma > 0$ and towards the sheet if $\sigma < 0$.

The first sheet is in the x-z plane, which is the plane $y=0$. It has a charge density $\sigma_1 = +\sigma$. For a point in the first quadrant, $y > 0$. The electric field due to this sheet points away from the sheet, i.e., in the positive y-direction. The electric field $\vec{E}_1$ is: $\vec{E}_1 = \frac{\sigma_1}{2\epsilon_0} \hat{j} = \frac{+\sigma}{2\epsilon_0} \hat{j}$.

The second sheet is in the y-z plane, which is the plane $x=0$. It has a charge density $\sigma_2 = -\sqrt{3}\sigma$. For a point in the first quadrant, $x > 0$. The electric field due to this sheet points towards the sheet since the charge density is negative. The direction towards the sheet $x=0$ from a point with $x > 0$ is in the negative x-direction. The electric field $\vec{E}_2$ is: $\vec{E}_2 = \frac{|\sigma_2|}{2\epsilon_0} (-\hat{i}) = \frac{|-\sqrt{3}\sigma|}{2\epsilon_0} (-\hat{i}) = \frac{\sqrt{3}\sigma}{2\epsilon_0} (-\hat{i}) = -\frac{\sqrt{3}\sigma}{2\epsilon_0} \hat{i}$.

The net electric field at the point in the first quadrant is the vector sum of $\vec{E}_1$ and $\vec{E}_2$: $\vec{E} = \vec{E}_1 + \vec{E}_2 = \frac{\sigma}{2\epsilon_0} \hat{j} - \frac{\sqrt{3}\sigma}{2\epsilon_0} \hat{i} = -\frac{\sqrt{3}\sigma}{2\epsilon_0} \hat{i} + \frac{\sigma}{2\epsilon_0} \hat{j}$.

Let $E_x$ and $E_y$ be the components of the net electric field along the x and y axes, respectively. $E_x = -\frac{\sqrt{3}\sigma}{2\epsilon_0}$ $E_y = \frac{\sigma}{2\epsilon_0}$

We want to find the angle $\theta$ which the electric field vector $\vec{E}$ makes with the positive x-axis. This angle is given by $\tan \theta = \frac{E_y}{E_x}$. $\tan \theta = \frac{\frac{\sigma}{2\epsilon_0}}{-\frac{\sqrt{3}\sigma}{2\epsilon_0}} = -\frac{1}{\sqrt{3}}$.

Since $E_x < 0$ and $E_y > 0$, the electric field vector lies in the second quadrant. The angle $\theta$ in the second quadrant such that $\tan \theta = -\frac{1}{\sqrt{3}}$ is $150^\circ$. This can be found by considering the reference angle $\alpha = \arctan \left| -\frac{1}{\sqrt{3}} \right| = \arctan \left( \frac{1}{\sqrt{3}} \right) = 30^\circ$. Since the vector is in the second quadrant, the angle with the positive x-axis is $180^\circ - \alpha = 180^\circ - 30^\circ = 150^\circ$.

The angle the electric field makes with the positive x-axis is $150^\circ$.