Animated Solution for Physics - Electrostatics: Two charged thin infinite plane sheets of uniform surface charge densities ${\sigma }_{+}$ and ${\sigma }_{-}$, where $|{\sigma }_{+}|>|{\sigma }_{-}|$, intersect at right angle. Which of the following best represents the electric field lines for this system?

The problem of finding the electric field lines for two intersecting infinite plane sheets is a beautiful exercise in vector superposition and symmetry. Let's break down the physics and the geometry behind this elegant setup.

Imagine two infinite plane sheets intersecting at right angles. We are given two surface charge densities: ${\sigma }_{+}$ on the horizontal sheet and ${\sigma }_{-}$ on the vertical sheet. The problem also states a crucial condition: $|{\sigma }_{+}|>|{\sigma }_{-}|$.

Before we jump into the resultant field, let's recall the behavior of a single infinite charged sheet. The electric field produced by an infinite thin plane sheet is uniform everywhere in space. Its magnitude is given by:

Notice that this formula does not depend on the distance from the sheet! The field is constant. Furthermore, the field points away from a positive charge and towards a negative charge.

Superposition

Let's pick a point in the first quadrant. The horizontal sheet (${\sigma }_{+}$) is positive, so it creates an electric field ${\vec{E}}_{+}$ that points vertically upwards, away from the sheet. The vertical sheet (${\sigma }_{-}$) is negative, so it creates an electric field ${\vec{E}}_{-}$ that points horizontally to the left, towards the sheet.

The net electric field at our chosen point is simply the vector sum of these two perpendicular components:

Because the individual fields ${\vec{E}}_{+}$ and ${\vec{E}}_{-}$ are constant everywhere in the first quadrant, their vector sum ${\vec{E}}_{net}$ must also be a constant vector. A constant vector field means the electric field lines will be straight, parallel lines. This immediately eliminates any options showing curved lines or circles!

Now, let's use the condition $|{\sigma }_{+}|>|{\sigma }_{-}|$. Since the magnitude of the electric field is directly proportional to the surface charge density, this inequality tells us that the vertical component is stronger than the horizontal component:

Geometrically, if you draw a right-angled triangle with these two vectors, the vertical side is longer than the horizontal side. Therefore, the resultant vector ${\vec{E}}_{net}$ will be steeper—it will make an angle greater than $45^{\circ }$ with the horizontal axis.

If we apply this exact same logic to all four quadrants, we find: - Quadrant 1: Field points Up-Left. - Quadrant 2: Field points Up-Right. - Quadrant 3: Field points Down-Right. - Quadrant 4: Field points Down-Left.

Connecting these straight, steep field lines across the boundaries creates a pattern of nested rhombuses that are stretched vertically. This perfectly matches the visual representation in the correct option.