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A surface on which the angle between the electric field and the area vector is the same at every point is known as a Gaussian surface. Some common types of Gaussian surfaces are the cylindrical surface, spherical surface and Gaussian pillbox. Generally, Gaussian surfaces are chosen carefully so that the calculation of the surface integral could be simplified. If a Gaussian surface is chosen in a way that the electric field is constant at each point on the surface and along the regular vector, Gauss’s Law may be used to calculate the electric field.
Cylindrical Gaussian surface
We will consider a cylindrical Gaussian surface to calculate the electric flux or electric field produced by the given objects:
· A long infinite line of uniform charge.
· An infinite plane with an unchanging charge.
· An infinitely long cylinder with an unchanging charge.
If we imagine a point charge P that is present at a distance of r having charge density λ. If we take a closed surface that is in the shape of a cylinder, the rotational axis of this closed surface is the line charge. If the length of the cylinder is h, then the charge inside the cylinder will be:
Q = λh,
Where the charge enclosed in the cylindrical Gaussian surface is denoted by q. There are three surfaces named a, b and c. On each surface, a, b and c, dA is the differential vector area. The flux passing through the cylinder will be:
φa+φb+φC=0
E and dA would be perpendicular for surfaces a and b. However, E and dA would be parallel to surface c.
The cylinder’s surface area would be:
∫∫c dA = 2πrh
which suggests
ΦE = E2πrh
Gauss’s law states
ΦE = q / ∈0
When we equate ΦE we will find out
E2πrh = λh / ∈0
E = λ / 2π∈0r
Electric Field due to an Infinite Plate Sheet
Consider an infinite plane sheet that has a surface charge density σ with A as the cross-sectional area.
Infinite Charge Sheet
The electric field generated by the infinite charge sheet will be perpendicular to the sheet’s plane. Consider a cylindrical Gaussian surface that has its axis normal to the plane of the sheet. Gauss’s law will be used to evaluate the electric field.
According to the law
Φ = q / ∈0
Due to continuous charge distribution, the charge q would be σ A. If we talk about net flux, the electric flux would be considered only from the two ends of the Gaussian surface that we assumed. The reason for this is that the curved surface area and the electric field are normal to each other; thus, the electric flux produced is zero. Therefore, the total electric flux would be:
Φ = EA – (– EA) => Φ = EA + EA
Φ = 2EA
We can also write it as:
E = σ A/ ∈0
The term A in the equation cancels out, which means the electric field due to the infinite plane sheet is not dependent on cross-sectional area A and equals to:
E = σ / 2∈0
Gaussian Pillbox
We consider this when we need to determine the electric field due to an infinite sheet of charge with unchanging charge density or a finite thickness slab of charge. The shape of the pillbox is cylindrical and it consists of three components i.e. at one end of the cylinder, there is a disk that has an area of πR2, the disk at the other end of the cylinder with the same area and the cylinder’s side. The totality of the flux passing through each of its components is proportional to the charge enclosed in the pillbox. The reason for it is the field close to the sheet could be estimated as constant; the orientation of the pillbox is done in a manner that the field pierces the ends of the field at a 90-degree angle and the field lines are parallel to the cylinder’s side.
Conclusion
The types of surfaces that have the same angle between the electric field and area vector, at every point, are known as Gaussian surfaces. Gauss’s law may be used to calculate the electric field. Gauss law is very important in determining the relation between electric flux and electric charge. When we try to calculate the electric flux of an object, we use a Gaussian surface. There are commonly three types of Gaussian surfaces i.e. cylindrical surface, spherical surface and Gaussian pillbox.
