1.14.2. Field due to a uniformly charged infinite plane sheet

Let $s$ be the uniform surface charge density of an infinite plane sheet. We take the x-axis normal to the given plane. By symmetry, the electric field will not depend on y and z coordinates and its direction at every point must be parallel to the x-direction.

The electric field created by an infinite plane sheet with uniform charge density is examined. The symmetry of this system allows us to conclude that the electric field only varies with respect to the perpendicular distance from the sheet. Importantly, this means that the field is uniform and points either towards or away from the sheet, depending on the sign of the charge density.

Imagine a large, flat surface like a trampoline that is uniformly pressed down at every spot. No matter where you are above the trampoline, the 'pressure' felt (akin to the electric field) is uniform as long as you are at the same height above this surface.

We can take the Gaussian surface to be a rectangular parallelepiped of cross-sectional area $A$. As seen from the figure, only the two faces 1 and 2 will contribute to the flux; electric field lines are parallel to the other faces and they, therefore, do not contribute to the total flux.

In this chunk, we define a Gaussian surface to calculate the electric field. We use a closed surface that intersects the charged plane. The electric flux through the surface is calculated based on the fact that only the faces of the Gaussian surface that are perpendicular to the electric field contribute to the flux. The other faces, which are parallel to the electric field, contribute nothing to the total flux.

Think of holding a net in a flowing river. If you place the net flat against the water, it will collect lots of water (representing the flux), but if you turn it on its side so that it runs parallel to the flow, it will collect very little as it isn't facing the water directly. This example helps visualize how flux works through different surfaces.

The charge enclosed by the closed surface is $sA$. Therefore by Gauss’s law,
$2EA = \frac{sA}{\epsilon_0}$
or, $E = \frac{s}{2\epsilon_0}$

In this part, we apply Gauss's law to relate the net electric flux through our Gaussian surface to the charge enclosed. By rearranging the law, we can solve for the electric field $E$. The result shows that the electric field around an infinite plane sheet of charge is constant and does not depend on the distance from the sheet.

Consider a flat fan that blows air uniformly in all directions. No matter how close or far you stand from the fan's surface, the wind feels the same force pushing against you as long as it is in front of you, illustrating how the electric field is uniform around the charged plane.

Vectorically,
$E = \frac{s}{2\epsilon_0} \hat{n}$
where $\hat{n}$ is a unit vector normal to the plane and going away from it. E is directed away from the plate if s is positive and toward the plate if s is negative.

The direction of the electric field is defined based on the sign of the surface charge density. If the sheet has positive charge density, the electric field exists in a direction moving away from the sheet. Conversely, if it has negative charge density, the electric field moves towards the sheet. This directional dependency is crucial for understanding how charges interact.

Imagine how a magnet behaves when you bring another magnet close. If the magnets are of opposite charge (like a positive and a negative surface), they attract. This is similar to how an electric field directs towards a negatively charged sheet, showcasing attraction.

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