Animated Solution for Physics - Electrostatics: Two infinite planes each with uniform surface charged 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

Imagine you are standing in a vast, empty space, and suddenly, two infinite, flat sheets of charge appear. Both sheets carry a uniform positive surface charge density, denoted by $\sigma$. One sheet lies perfectly flat along the horizontal axis. The other sheet is tilted, slicing through the horizontal sheet at a crisp $30^{\circ }$ angle.

Our mission is to find the net electric field in the specific wedge-shaped region trapped between these two infinite planes. To do this, we must rely on the principle of superposition, which tells us that the total electric field at any point is simply the vector sum of the electric fields produced by each plate individually.

Before we dive into the vectors, let's recall a beautiful result from Gauss's Law. The electric field generated by a single, infinite plane sheet of charge is remarkably simple. It doesn't depend on how far away you are from the sheet! The field is uniform everywhere, and its magnitude is given by:

Because our plates have a positive charge density ($+\sigma$), the electric field vectors will always point directly away from the plates, perpendicular to their surfaces.

Let's break the problem down and look at the horizontal plate (let's call it Plate 2) first. The region we are interested in lies entirely above this horizontal plate.

Since the electric field must point away from the positive charge, the field from Plate 2, which we'll call ${\vec{E}}_2$, will point vertically upwards. In vector notation, this is purely in the positive y-direction:

Now comes the slightly tricky part: the inclined plate (Plate 1). The region we are analyzing is situated below this tilted plate. Therefore, the electric field ${\vec{E}}_1$ must point away from it, directing downwards and to the left.

To add ${\vec{E}}_1$ to ${\vec{E}}_2$, we need to resolve ${\vec{E}}_1$ into its x and y components. Let's look closely at the geometry. Plate 1 is inclined at $30^{\circ }$ to the horizontal. The electric field vector is perpendicular to the plate. A line perpendicular to a $30^{\circ }$ incline will make an angle of $90^{\circ }-30^{\circ }=60^{\circ }$ with the vertical, or equivalently, $60^{\circ }$ with the horizontal axis.

Since ${\vec{E}}_1$ points downwards and to the left (into the third quadrant relative to a local origin), both its x and y components will be negative. We can write this as:

Substituting the standard trigonometric values ($\cos 60^{\circ }=\frac{1}{2}$ and $\sin 60^{\circ }=\frac{\sqrt{3}}{2}$), we get:

We have our two field vectors. Now, we simply add them together to find the net electric field, ${\vec{E}}_{net}$:

Let's group the y-components together to tidy up the expression:

Finally, we substitute the magnitude of the electric field, $E=\frac{\sigma }{2{\epsilon }_0}$, back into our equation:

This perfectly matches option (b). The elegance of this problem lies entirely in carefully visualizing the geometry and rigorously applying vector resolution!