Question

gauss law and it's application

Answer 1

Gauss Law states that, the flux of net Electric Field through a closed surface is equal to the net charge enclosed by the closed surface divided by permitivity of space.

Therefore, mathematically it can be written as

∫E.ds = Qint/ξ     (Integration is done over the entire surface.)

where Qint = Total charge enclosed by the close surface

and ξ = permitivity of space = ξ0ξr

Let us consider a closed surface as shown in figure below and apply Gauss Law.

The direction of Electric Field through the closed surface is shown in figure. The flux of net electric field through the closed surface

∫E.ds = (q1+q2+q4-q3) / ξ

It shall be noted here that while applying Gauss Law, the electric field considered through the surface shall be the net electric field due to both the charges inside and outside the closed surface. But the Qint shall only account for the charges enclosed by the surface, it shall not account for charges outside the surface.

Application of Gauss Law

Gauss Law is very useful for calculating the Electric Field at any point. Let us use this law to find the value of field at any point P from a uniformly charged infinite long rod. Let the surface charge density of the infinitely long rod be λ.

 

Well, for applying Gauss Law very first step should be to draw a closed surface passing through point P. This closed surface is obviously a cylinder. Let the length of cylinder is ‘l’ and radius is ‘r’.

As the line is uniformly charged and its length is infinite, therefore the electric field on its curved surface will be uniform and perpendicular to surface. Thus the flux of electric field through the closed surface

Ø = ∫E.ds

Since the direction of area vector ds is perpendicular to the surface ds and outside the volume therefore area vector ds and E are parallel. Therefore their dot product E.ds = Eds. Hence,

Ø = ∫Eds

Since E is constant on the surface, therefore we can take it outside the integration.

= E ∫ds

But ds is the total curved surface area which is equal to 2πrl.

= Ex(2πrl)

But as per Gauss Law, the flux of net electric field is equal to the total charge enclosed divided by permittivity. Hence,

Ex(2πrl) = Qint / ξ0

Now, we need to find the total charged enclosed by the surface. Since surface charge density of line is λand the length of line enclosed by the closed surface is ‘l’ therefore Qint = λl.

Therefore,

Ex(2πrl) = λl / ξ0

⇒E = λ / 2πξ0r

The above expression gives the value of electric field at a distance r from the uniformly charged infinitely long rod.

Answer 2

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Gauss law: Gauss law states that the total electric flux through a surface is equal to the total charge within that surface divided by permittivity of free space ( can be air or any other medium ).

To prove Gauss Theorem, we need to prove

Φ = q/ ε0

We know that for a closed surface 

∮ E→ . dA→ = Φ = q/ ε0 ............(1)

First we will calculate LHS of equation (1) and prove that it is equal to RHS

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Step 1

Consider a sphere with a point charge ‘q’ as its centre and radius as ‘r’. Since the charge is a point charge so the electric field will be radial in all directions.(see attached file)

Step 2

Take an infinitely small area on the surface.

So, the electric field at a distance ‘r’ over the Gaussian surface due to charge ‘q’ will be:  

Er =q/ 4πε0 r2

Step 3

Multiply both side of the equation with dS.

We get − 

∮ E→ . dA→ = ∮( q dS/4πε0 r2)

⇒( q/4πε0 r2) ∮dS ..........(2)

(Here; 4πε0 ,q, r are constants)

Now complete area of a sphere = ∮ dS = 4πr2

Putting the value in equation (2) we get −

∮ E→ . dA→ = Φ = q/ ε0 ..........(3)

We can see that ; equation (1) = equation (3)

Since LHS = RHS, Hence Gauss theorem is proved.

The total flux according to our knowledge is 

∮ E.dS

Since electric field E ∝ 1/r2. That means it follows inverse square law.

Suppose electric field does not follow inverse square law, instead it follows inverse cube law as in case of a dipole.

In that case E ∝ 1/r3

So, in case of a dipole E = q/4πε0 r3

On Multiply both side of the equation with dS.

We get − ∮ E.dS = ∮ (q/4πε0 r3)dS

⇒ (q/4πε0 r3 )∮ dS ............(2)

(Here; 4πε0, q, r are constants)

Now complete area of a sphere = ∮ dS = 4πr2

Putting the value in equation (2) we get −

⇒ ∮ Er.dS = q/ ε0 ≠ Φ ............(3)

The above equation is not satisfying Gauss Theorem.

So Gauss Theorem is applicable only when electric field follows inverse square law.

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Gauss’s law is used for calculation of electrical field for a symmetrical distribution of charges. For a highly symmetric configuration of electric charges such as cylindrical, or spherical distribution of charges, the Law can be used to obtain the electric field E without taking any hard integrals.

Some other famous applications include:
- calculation of the electric field close to a large plane sheet of charges
- Calculation of the electric filed inside a uniformly charged sphere
- …
For some applications, the inverse problem of finding the charge inside a volume given the field at the surface of the volume can also be done.




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