Applications of Gauss Law

The Gauss law can be applied to solve many electrostatic problems, which involve unique symmetries like spherical, planar or cylindrical. Calculating electric fields in complex problems can be challenging and involves tricky integration. The Gauss law evaluates the electric field. There are many fields or problems where the Gauss law can be applied, and it can be applied in the following way:

• To make the evaluation of the electric field easy, choose a Gaussian surface. 
• To solve the problems efficiently, use symmetry. 
• The Gaussian surface does not need to coincide with the actual surface. It can be outside or inside the Gaussian surface.

Electric field due to parallel charged sheets

Take two infinite parallel sheets with opposite and equal charge density

 +𝝈 and -𝝈. The magnitude will be E=𝝈./2ɛ0 and is perpendicular to the sheet. 

Case 1. When the point P1 is between the sheets, the resultant field at P1 is E=E1+E2=𝝈2ɛ0+𝝈2ɛ0=𝝈ɛ0

Case 2. When the point P2 is outside the sheets, the electric field will be in the opposite direction and equal in magnitude. 

The resultant will be  E=E1–E2=𝝈2ɛ0–𝝈2ɛ0=0.

Electric field due to a uniformly charged spherical shell 

Case 1. Taking a point outside the shell

Taking a charged shell with:

R = Radius of the charged shell. 

P = a point outside the shell. 

R = a distance of the point from the centre.

O = centre. 

The flux crossing the Gaussian sphere will be ɸ=sĒ . ds=sE ds=E(4r2) .

Simplifying by Gauss Law E(4r2) =q/ɛ0 or E=14r2qr2.

Case 2. At a point on the surface of the shell

E=14r2qr2 because r = R.

Case 3. At a point inside the shell

Take a point P’ inside the shell at a distance r’ from the centre of the shell and a Gaussian surface with radius r’.

The flux crossing through the Gaussian sphere in an outward direction is ɸ=sĒ . ds=sEds=E(4r’2) .

According to Gauss law, E4r‘2=qɛ0=0.

Electric field due to infinite charged plate sheet

Take an infinite charged plate sheet with 

𝝈 = density of surface charge. 

P = a point outside the shell. 

r = distance of point P from the sheet. 

E = be the electric field at point P. 

Let us assume the Gaussian surface to be a cylinder of crossing the sheet and the sectional crossing area be A. Then the length will be 2r of the cylinder perpendicular to the sheet.

The electrical field is at a right angle to the end of the cpas and away from the plane.

Its magnitude is the same as P and the other P’. 

Hence, the total flux through the closed surface will be 

ɸ=[∮E.ds]p+[[∮E.ds]p’ =EA+EA=2EA.

If 𝝈 is the charge per unit area in the plane sheet, then the net positive charge q within the Gaussian surface is q=𝝈A.

Simplifying using Gauss Law 2EA=𝝈Aɛ0.

E=𝝈2ɛ0

Electric field caused by an infinitely long straight charged wire

Take a uniformly charged wire of an infinite length with a constant linear density 

P =  a point. 

r = distance from the wire. 

E = electric point at the point P. 

l = the length of the cylinder. 

r = radius. 

Take a small area ds on the Gaussian surface. 

Ē and ds are in the same direction. 

Then the electric flux with a curved surface will be 

ɸ= E ds .

ɸ=E(2rl).

The electric flux through the plane caps = 0. 

Total flux will be ɸ=E(2rl). 

Simplifying by Gaussian law  

E(2rl)=0 or E=2rl0.

Electric field outside the spherical shell

P = a point away from the shell. 

r = distance away from the spherical shell. 

r = radius of the Gaussian surface.

O = centre.  

According to Gauss law ɸ=q/ɛ0.

The charge inside the Gaussian surface will be 4R2 .

Simplifying by Gauss law 

E4R2=4R2/0

E=R20 r2

Putting the value of surface charge density as q/4R2.

Therefore, E=kqr2 r.

r is the radius vector.

Conclusion 

There are several cases where Gauss law can be used to solve problems quickly. When it comes to solving electrostatic problems that include unique shapes like cylindrical, spherical or planar symmetry, Gauss law can be used to solve these problems. 

We have discussed several applications such as Electric field outside the spherical shell, electric field caused by an infinitely long straight charged wire, Electric field due to uniformly charged spherical shell, Electric field due to parallel charged sheets, and Electric field due to infinite charged plate sheet.