Gauss’s Law : The First Of Four Maxwell’s Equations

In this article, I will discuss about Gauss’s Law.It is one of the Maxwell’s four equations of electromagnetism.

This is the second part of a series. Read the first part here : Gauss’s Law Part 1 : Electric Flux

Gauss’s Law states that :

The net electric flux through any hypothetical closed surface is equal to (1/ε0) times the net electric charge within that closed surface

If you don’t know about electric flux, then read the previous part of the series . I explained electric flux in the previous part.

A closed surface is a surface bound from all sides, which has no open part. For example : surface of a sphere, balloon, box etc.

Let, we have an closed surface. There is a point charge inside the closed surface.

The picture above looks like a 2d shape, but assume it as a 3D irregular closed surface. Now , we have to calculate the net electric flux through the closed surface.

The electric flux φ through a surface with area A in an uniform electric field E is :

As the surface is an irregular shape, the distance of the surface from the point charge q is different ins each point, and so, E changes everywhere on the surface.

So, we need to divide the surface into smaller pieces. For each smaller piece dA , we will calculate flux dφ=E*dA*cos θ. Then we have to integrate through the surface.

So, the net electric flux φ will be integration of dφ through the whole surface.

φ = ∮E*dA*cos θ

The “∮” symbol is used for closed integration. Because our surface is closed, we used ∮ symbol instead of “∫” symbol.

φ = ∮E*dA*cos θ………..(1)

The integration in (1) is not as easy as it looks. Both E and θ change over each point on the surface. E depends on the distance r from the +q charge. As the distance from the charge varies in each point of the surface, E changes over each pint of the surface. θ also changes because the direction of E and dA change for each piece of area. And we don’t know how these values change for an irregular shaped surface. So, the integration in (1) is very difficult to solve. In fact there is no way you can solve it mathematically. You can solve it experimentally with the help of a computer , by computing each dφ and then summing up. But we need a mathematical solution .So, we need to adopt another strategy.

Let us assume a hypothetical sphere inside the surface. And let , the charge +q is in the center of that sphere.

We know, that the electric field lines from a positive point charge spread outwards from the charge.

So all the field lines will spread out from the +q charge.

All the field lines that will cross the surface of the sphere , will also cross the outer irregular surface. In the previous part of this series we assumed electric flux as the number of electric field lines that cross through a surface. If we use the fluid flow analogy (we used it in the previous part) we can assume an water source in place of the charge and water is spreading out from the source just like the electric field lines are spreading out. We can say that the amount of water that will cross the outer irregular surface will also cross the inner spherical surface. So, if we calculate the flux for the sphere , we will also get the flux for the outer surface. Because according to the fluid flow analogy, the total amount of water that will cross the outer surface , must cross the inner spherical surface.

The net electric flux on the sphere = The net electric flux on the outer surface

Calculating electric flux for the sphere is easy because it is not an irregular shape. E is constant throughout the surface because the distance r from center is constant through all points on the sphere.

So, let us calculate the integration φ = ∮E*dA*cos θ for the sphere.

As you see from the picture , the directions of dA and E are the same. Because radius is always perpendicular to the surface of the sphere and the direction of E is along the radius , so E is also perpendicular to the surface. And the direction of an area dA is always perpendicular to the surface (see the previous part). So, dA and E have same direction . So, the angle between them is 0 degree. So, θ = 0

Now, ∮dA is the area of the surface of the sphere. So, ∮dA = 4πr²

So, φ = E * 4πr² ………(1)

So, the total flux through the sphere is ,

This is the Gauss’s Law . The net electric flux φ =the net charge q * (1/ε0).

General Form Of Gauss’s Law

We described flux as the number of electric field lines that cross through the surface. So, if +q charge is put inside a closed surface, the number of electric field lines that will cross the surface is φ. Now, what will happen when more than 1 point charge is put inside a closed surface? What will be the flux through the surface then? Let , two point charges +q₁ and +q₂ are put inside a closed surface .

Now, we have to calculate the total flux through the surface for these two charge. First , let us see what will happen if we put these two charges individually inside the closed surface.

If we put these two charges individually , for the charge q₁ , the number of electric lines that will cross the surface is φ1 =q1/ε₀ . And for the charge q2 , it is φ2 =q2/ε₀ . If we put both the charges inside the closed surface , then all of the electric field lines generated from both the charge q1 and q2 will cross the surface . Because no electric filed line will be lost or destroyed.

So, the total number of electric field lines that will cross the closed surface is ,

Σ q is the net charge inside the closed surface.

This is the general form of the Gauss’s Law. Now let us see a few more examples. If a negative charge is put inside a closed surface , then Σ q is negative. So, the net flux will be negative.

For a negative charge, all electric field lines point towards the negative charge. So, all the field lines go into the surface. In previous examples , when the charge was positive , the field lines went out of the surface, if this is considered as positive flux , then the field lines going into the surface will be considered as negative flux. So, in this case the net flux is ,

φ = — q /ε₀

If two opposite point charges but same magnitude are put inside a closed surface, then

Σ q = +q — q = 0

So, φ = 0/ε₀

or, φ = 0

As, you can see from the picture , any electric field lines that goes out of the surface , must come inside the surface . Because all electric field lines goes from positive charge to negative charge. So, the net flux will be 0.

If there is no charge inside the closed surface , but there is a point charge outside the closed surface , then the total charge inside the surface is Σ q = 0. the net flux φ = 0

As you can see from the picture, the total number of field lines that goes into the surface equals the total number of field lines that goes out of the surface. So, the net flux φ = 0

Applications Of Gauss’s Law

Gauss’s law can be used to measure the electric field of distributed charges like electric fields due to a uniformly charged spherical shell, cylinder , plate etc.

Electric Field Of Charged Sphere

In this section, we will use Gauss’s law to measure electric field of a uniformly charged spherical shell . Gauss’s law states that :

The net electric flux through any hypothetical closed surface is equal to (1/ε0) times the net electric charge within that closed surface

According to Gauss’s law, if the net charge inside a Gaussian surface is Σq, then the net electric flux through the surface , φ = Σq/ε₀

Electric Field Of Charged Hollow Sphere

Let us assume a hollow sphere with radius r , made with a conductor. The conducting hollow sphere is positively charged with +q coulomb charges.

If the sphere has equal density all over its surface , then +q charge will be equally distributed all over the surface. So, the entire system is a symmetric system. From this symmetricity , we can say that the direction of the electric field will be radially outwards or inwards. Otherwise , the symmetricity will be lost. As the charges are positive , the sphere will repulse any positive point charge near it . So, the direction will be radially outwards. Also we can conclude that the magnitude of the electric field will be equal to the equidistant distances from the center because of the symmetricity.

Now, let us assume a hypothetical sphere with radius R and the same center as the charged sphere.

The outer spherical surface is our Gaussian Surface. The net charge inside the Gaussian surface , Σq = +q . According to Gauss’s Law, the total electric flux through the Gaussian surface ,

φ = q/ε₀ . . . . . (1)

Mathematically the flux is the surface integration of electric field through the Gaussian surface.

φ = ∮E . dA [dot product of E and dA]

or, φ = ∮E*dA*cos θ . . . . . (2)

From equation (1) and (2) ,

q/ε₀ = ∮E*dA*cos θ

E is constant through the surface . Because , all points on the surface are in same distance from the center. Previously in this article , we said that according to symmetricity, E will be constant in all equidistant places from the center. So, E can be brought out from the integration sign.

q/ε₀ = E ∮dA*cos θ …………..(3)

θ is always ⁰⁰ . As both the direction of dA and E are the same(radially outwards). So, the angle between them is 0.

So, equation (3) becomes ,

q/ε₀ = E ∮dA * cos ⁰⁰

q/ε₀ = E ∮dA * 1

q/ε₀ = E ∮dA

Now, ∮dA is the surface area of the outer sphere . So, ∮dA = 4πR²

q/ε₀ = E * 4πR²

So,

This expression is the same as that of a point charge. It is as if the entire charge is concentrated at the center of the sphere.

Electric Field On The Surface Of The Sphere (R = r)

On the surface of the conductor , where R = r , the electric field is :

E = (1/4πε₀) * (q/r²)

Electric Field Inside Hollow Sphere (R < r)

If we assume any hypothetical sphere inside the charged sphere, there will be no net charge inside the Gaussian surface . So, Σq = 0 .

So, the net flux φ = 0.

So, ∮E*dA*cos θ = 0

Or, E ∮dA*cos θ = 0

Or, E = 0

So, the electric field inside a hollow sphere is zero.

Electric Field Of Charged Solid Sphere

If the sphere is not hollow , instead it is a solid one , then the entire charge will be distributed on the surface of the solid sphere. Because, in electrostatic condition , there is no electric field inside a conductor. (This topic is explained here : Electric Field Inside A Conductor). As there is no electric field inside a conductor , if we assume any hypothetical surface inside a conductor , the net flux φ will be zero. Thus, the net charge inside a conductor Σq = 0.

Thus , if +q charge is given to a solid sphere, it will be distributed equally throughout the surface of the sphere . There will be no charge inside the sphere. So the electric fields will be the same as the hollow sphere.

Electric Field Of Charged Plate

Let we have a charged plane of infinite length and width. And let positive charges are equally distributed throughout the surface.

Though the plane in the picture doesn’t have infinite length and width , let us assume this as an infinite plane.

Let the charge density on the surface is λ coulomb/meter² . So, in 1m² area on the plane, there are λ coulomb charges.

The plane is symmetric. From the symmetricity of the system , we can say that the direction of electric field is perpendicular to the plane . If electric field is not perpendicular , then rotating the plane will break the symmetricity.

If the electric fields are perpendicular , the symmetricity will be preserved.

Also from the symmetricity , we can say that the magnitude of the electric field will be the same on equidistant distances from the plane.

Let us assume a hypothetical cylinder with height h and base area A. The outer surface of the cylinder is our Gaussian surface.

Now, we have to calculate flux through the Gaussian surface. The cylinder has 3 surfaces . Upper and lower bases and one curved surface.

Let , φ1 = flux through upper base

φ2 = flux through lower base

φ3 = flux through curved surface

So, φ1 = ∮E*dA*cos ⁰⁰ ………..[Direction between E and dA is ⁰⁰]

or, φ1 = ∮E*dA

or, φ1 = E ∮dA ……………[Because E is constant]

∮dA is the surface area of bases = A . So, the equation becomes :

φ1 = E * A

For lower base , the equations are the same . So, φ2 = E * A

For the curved surface the angle between E and dA is 9⁰⁰. So, cos 9⁰⁰ = 0.

So, φ3 = ∮E*dA*cos 9⁰⁰

or, φ3 = ∮E*dA* 0

or φ3 = 0

So, the net flux,

φ = φ1 + φ2 + φ3

or, φ = E*A + E*A + 0

or, φ = 2*E*A ……………(1)

According to Gauss’s Law , φ = Σ q /ε₀

Here, Σ q = total charge on the plane inside the cylinder. So, Σ q = λ*A

So, φ = (λ*A) /ε₀ ……………..(2)

From equation (1) and (2) ,

(λ*A) /ε₀ = 2*E*A

or, λ /ε₀ = 2*E

or, E = λ /2ε₀

So, for a infinite plane with charge density λ , the electric field ,

Notice that, r is not present in the equation . So, E does not change over distance from the plate.

Electric Field Between Two Charged Plates

Two Parallel Plane with Same Charge

Let , we have two parallel infinite plate each positively charged with charge density λ.

Now we want to calculate the electric field of these two parallel plate combined. In the previous section we learnt about their individual electric field is E = λ /2ε₀

In the middle of the two plate , both electric fields are opposite to each other . So, the cancel each other and the net electric field inside is zero.

On left and right side, both electric fields are in the same direction. So, their vector sum = E = λ /ε₀

This is the case in parallel plate capacitor. Now what will happen if the two plates have opposite charge ? The question is left for the reader.