Question

proof gauss's theorm ?

Answer 1

Gauss’s Theorem and its Proof


GAUSS’S LAW


The surface integral of electrostatic field E produce by any source over any closed surface S enclosing a volume V in vacuum i.e. total electric flux over the closed surface S in vacuum is 1/epsilon times the total charge Q contained inside S.

                                          



The surface chosen to calculate the surface integral is called Gaussian surface.


PROOF OF GAUSSIAN THEOREM


The flux through area element ΔS is



The unit vector rˆ is along the radius vector from the centre to the area element. Now, since the normal to a sphere at every point is along the radius vector at that point, the area element ΔS and rˆ have the same direction. Therefore,




 Since the magnitude of a unit vector is 1. The total flux through the sphere is obtained by adding up flux through all the different area elements:




 Since each area element of the sphere is at the same distance r from the charge,




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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