Electric field intensity due to infinite sheet of charges | Application of Gauss's law
Channel: Physics Buddies
Date: 2023-03-16

In this video, we will be discussing the electric field intensity due to an infinite sheet of charges. This is a fundamental topic in electrostatics and is crucial to understanding many other concepts in the field.

To begin, let us define what we mean by an electric field. An electric field is a region of space around a charged object where the force between the charged object and other charged objects is felt. The electric field intensity at any point in this region is the force per unit charge experienced by a small positive test charge placed at that point.

Now, let us consider an infinite sheet of charges. This sheet has a uniform surface charge density σ, which means that the charge per unit area is constant throughout the sheet. We want to determine the electric field intensity at a point P located a distance z above the center of the sheet.

To do this, we will use Gauss's law. Gauss's law relates the electric flux through a closed surface to the charge enclosed within that surface. We can use this law to calculate the electric field intensity due to the sheet of charges.

We begin by choosing a Gaussian surface that is a cylinder with its axis perpendicular to the sheet of charges and its circular ends coinciding with the edges of the sheet. The cylinder has a height of 2z and a radius R. By symmetry, the electric field intensity is uniform over the entire curved surface of the cylinder.

The total charge enclosed by the cylinder is given by Q = σA, where A is the area of one circular end of the cylinder. Since the charge density is uniform throughout the sheet, the total charge enclosed is proportional to the area of the cylinder. Therefore, the electric flux through the curved surface of the cylinder is given by Φ = EA = E(2πR)(2z).

By Gauss's law, the electric flux through the closed surface of the cylinder is also equal to the charge enclosed within that surface, which is Q = σA. Therefore, we have Φ = σA/ε₀, where ε₀ is the permittivity of free space.

Equating these two expressions for Φ, we get E(2πR)(2z) = σA/ε₀, which simplifies to E = σ/2ε₀. This is the expression for the electric field intensity due to an infinite sheet of charges.

This result tells us that the electric field intensity due to an infinite sheet of charges is independent of the distance from the sheet. It is also proportional to the surface charge density and inversely proportional to the permittivity of free space.

In summary, we have discussed the electric field intensity due to an infinite sheet of charges using Gauss's law. We have derived an expression for the electric field intensity and discussed its properties. This is an important concept in electrostatics and is useful in understanding many other concepts in the field.
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