Physics Solution – Question 17

Solution to Question 17

1. Problem Geometry and Coordinate System

We consider two parallel half-planes with uniform surface charge density $\sigma$. The borders of these half-planes are separated by a distance $d$. We are asked to find the electric field on the plane containing these borders.

Let the borders lie on the x-axis at $x = -d/2$ and $x = +d/2$.
The plane containing the borders is the x-y plane (z=0 in a local frame, or just the line connecting them in cross-section).
The half-planes extend in “opposite directions” but are parallel. This means if one extends to the “top-left”, the other extends to the “bottom-right”, maintaining parallel alignment.
The angle between the half-planes and the connecting plane is $\theta$.

Figure 1: Cross-sectional view of the parallel half-planes. Vectors indicate the direction of the electric field contribution from each sheet.

2. Electric Field of a Single Half-Plane

The magnitude of the electric field component $E_x$ at a point $P$ lying on the line extending from the edge of a semi-infinite charged sheet (with charge density $\sigma$) depends on the angular position. Based on the integration of the charge distribution (analogous to the field of a finite line charge extended to 2D), the field vector makes a specific angle with the plane.

From the derivation for a half-plane, the field magnitude contribution is proportional to the angle subtended. More simply, by symmetry and superposition principles (considering the half-plane as half of an infinite plane):

At a large distance, the system resembles an infinite plane.
Locally, we sum the vectors $E_1$ and $E_2$.

3. Field in the Region Between Borders

Consider a point $P_{in}$ located on the connecting plane between the two borders.

Due to Sheet 1 (Left): The field points away from the sheet. Since the sheet is tilted “up-left”, the field at $P_{in}$ (to its right) points “down-right”.
Due to Sheet 2 (Right): The field points away from the sheet. Since the sheet is tilted “down-right” (parallel extension), the field at $P_{in}$ (to its left) points “up-left”.

Because the two half-planes have the same charge density $\sigma$ and are symmetric with respect to the midpoint, the magnitudes of their field contributions at any point on the connecting line are equal. However, their directions are exactly opposite (differing by $180^\circ$).

$$ \vec{E}_{net} = \vec{E}_1 + \vec{E}_2 = E \hat{n} – E \hat{n} = 0 $$

Thus, the field is zero everywhere on the segment between the borders.

4. Field in the Outer Region

Consider a point $P_{out}$ on the connecting plane but outside the region between the borders (e.g., to the right of Border 2).

Due to Sheet 1: The field still points “down-right” (away).
Due to Sheet 2: The point is now on the “other side” of Border 2’s edge relative to the sheet. However, relative to the infinite extension, the field direction aligns with the general “outward” normal of the macroscopic system.

Mathematically, as we cross the border, the angular term in the field integral changes. In this “outer” region, the field contributions from both half-planes reinforce each other. The system effectively behaves like a single infinite plane viewed from one side.

The sum of the fields yields the standard result for an infinite sheet of charge:

$$ E_{net} = E_{infinite\_sheet} = \frac{\sigma}{2\epsilon_0} $$

The electric field on the plane containing the borders is:

Zero at all points between the borders of the half-planes.
$\frac{\sigma}{2\epsilon_0}$ at all points outside the region between the borders.