Question

Derivation Proof of gauss theorem??

Answer 1

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Ans = In physics, Gauss's law, also known as Gauss's fluxtheorem, is a law relating the distribution of electric charge to the resulting electric field. The surface under consideration may be a closed one enclosing a volume such as a spherical surface. ... Gauss's law can be used to derive Coulomb's law, and vice versa.


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

Proof

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.

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 = 14πε0 qr2

Step 3

Multiply both side of the equation with dS.

We get − ∮ E→ . dA→ = ∮ 14πε0 qr2 dS

⇒ 14πε0 qr2 ∮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 = 14πε0 qr3

On Multiply both side of the equation with dS.

We get − ∮ E.dS = ∮ 14πε0 qr3dS

⇒ 14πε0 qr3 ∮ 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.

thanks
#prareek001

Answer 2

Answer:

$\setlength{\unitlength}{20mm}\begin{picture}(1,1)\put(0,0){\vector(1,0){1}}\put(-0.1,-0.1){{\small \bold{+q}}}\put(1,0){{\circle*{0.13}}}\put(1,-0.2){{\red{ds}}}\put(0.5,0.1){{r}}\end{picture}\\\\\sf{Consider\:a\:charge\:+q\:on\:a\:closed\:surface (sphere)\:of\:radius\:r.}\\\\\sf{Electrical\:flux\:on\:a\:small\:area\:ds,}\\\\\sf{d\phi=E.ds}\\\\\sf{d\phi=Edcos\theta\:\implies d\phi=E.ds\:[cos\theta=cos0^{\circ}=1]}\\\\\sf{On\:Integrating\:both\:sides,}$

$\sf{\displaystyle \oint d\phi=\displaystyle \oint E.ds}\\\\\\\implies \sf{\phi=\displaystyle E\oint ds}\\\\\\\implies \sf{\phi=\dfrac{1}{4\pi \epsilon_0}\times\dfrac{q}{r^2}\displaystyle \oint ds}\:...........\sf{\bigg(\displaystyle \oint ds = 4\pi r^2\:ie, area\:of\:sphere\bigg)}$

$\sf{\phi =\dfrac{1}{4\pi \epsilon_0}\times \dfrac{q}{r^2}\times 4\pi r^2}\\\\\\\implies \boxed{\sf{\bold{\phi=\dfrac{1}{\epsilon_0}.q}}}$