Applications of Gauss Law to Find Electric Field

Now, let us see some of the application of gauss law to find electric field:

Electric Field due to a Line Charge

Suppose a line charge having linear charge density λ is given in the form of a thin charged rod.

To find the electric field intensity at point P along a wire, a cylindrical Gaussian surface is selected. This choice is made to apply Gauss’s law for finding the electric field, E, at point P.

The electric flux passing through the end surfaces of the cylindrical Gaussian surface is ,that is Φ1 = 0.

And , the electric flux passing through the curved surface of the cylindrical Gaussian surface is given as:

Surface area of curved part is given as: S = 2πrl

Total charged enclosed by gaussian surface is q = λ × l

Surface area and charge of gaussian surface (Equation-5)

The electric flux through the curved surface of the cylindrical Gaussian surface is given as:

Φ = E. cos θ.S = E × 1 × 2πrl

Total electric flux is given as:

Φ = Φ1 + Φ2

Φ = 0 + E.cos θ.S

Φ = E × 2πrl

From Gauss’s law, we know that,

Φ = q/ε₀ = λl/ε₀

2πrl × E = λl/ε₀

Electric Field Due to Ring

Now, let us take a look into the electric field due to Ring

For, Electric Field Intensity at Any Point on the Axis of a Uniformly Charged Ring, let us consider a wire forming a circular ring with negligible thickness and a radius of R, carrying a uniform charge +q distributed evenly around its circumference. Our aim is to calculate the electric field intensity at any point P along the axis of the loop, positioned at a distance x from the ring’s center, marked as O.

Let AB be the length of element dl.

The charge on the element AB is,

dq = (q × dl)/2πR

Electric field intensity at P due to charge element AB is,

|dE| = k dq/(CP)²

where, k = constant

|dE| = k dq/(R² + x²)

Now, resolve the electric field intensity dE into two rectangular components, that is

dE sinθ along the y-axis and dE cosθ along the x-axis.

And for diametrically opposite elements of the charged ring, the perpendicular components of the electric field intensity will nullify each other, resulting in,

∫dE sinθ = 0.

Whereas components along the axis of the charged ring will undergo integration. That is, ∫dE cosθ.

Hence, the resultant electric field intensity E at P is | E | = ∫dE cosθ

In △OPC, cos θ = OP/CP = x/√(R² + X² ) and |dE| = k dq/(R² + x²)

Therefore, |E| = ∫k × x × dq/(R² + X² )(√(R² + X² )

|E| = k × q × x/(R² + X² )^3/2

The direction of E is along the positive x-axis of the loop.

Electric Field Due to a Uniformly Charged Sphere

Now, let us talk about the electric field due to a uniformly charged sphere.

Electric Field Outside the Shell

To find out the electric field intensity at a point P outside the spherical shell when OP = r.

The Gaussian surface was taken as a sphere having radius r, while the electric field intensity will remain the same at every point seen on the Gaussian surface.

Thus, Gauss theorem becomes,

∮ = ∮

or it can be also be given as,

E∮dS = q/ε₀

E × (4πr²) = q/ε₀

Therefore, the electric field become,

E =

From the above equation we can say that the electric field outside the shell is similar to the electric field due to a point charge. Thus outside the sphere, the electric field behaves as though it is due to a point charge (carrying all the charge of the shell) at the Centre of the shell.

Electric Field Inside the Shell

If the point P lies inside the spherical shell, then the Gaussian surface is a surface of a sphere having radius r. Since no charge is present inside the spherical shell, the Gaussian surface encloses no charge. Hence, q = 0.

To derive the value you can put the value of q = 0 in the formula E = q/ε₀ which will give the result zero.

Electric Field at the Surface of the Shell

At the surface of the shell, r = R

E = q/(4πR²ε₀)

If σ is the surface charge density then

q = (4πR²ε₀).σ

Since, we have E = q/(4πR²ε₀)

Putting value of q = (4πR²ε₀).σ we get

E = σ/ε₀

Electric field is a fundamental concept in physics, defining the influence that electric charges exert on their surroundings. This field has both direction and magnitude. It guides the movement of charged entities, impacting everything from the spark of static electricity to the functionality of electronic devices Understanding electric fields will help you to understand how charge particles interact with each other and the surroundings and guide various natural and technological phenomena. In this article, we will learn in detail about electric field, its formula, calculation of electric field for ring, straight wire and continuous charge distribution.