Question

Gauss's law ?
And where it is applicable ?

Answer 1

also known as Gauss's flux theorem, is a law relating the distribution of electric charge to the resulting electric field and it is applicable in electrostatics

Answer 2

$<b>____________♦♦♥♦♦____________$

$\huge\mathfrak\purple{\:\:\:\:\:\:\:\:hello\:\:\:mate}$

$<b>____________♦♦♥♦♦____________$
$\tt{\pink{\huge{here\:is\:your\:answer}}}$
$<b>____________♦♦♥♦♦____________$


$\huge\mathfrak\purple{\:\:\:\:\:\:\:\:gauss\:law}$
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

$\huge\mathfrak\purple{\:\:\:\:\:\:\:\: proof}$

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.(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.

$\huge\mathfrak\purple{\:\:\:\:\:\:\:\: applications}$

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.




$<b>____________♦♦☺♦♦____________$

$<marquee behaviour-move bigcolour-pink><h1>☺☺ThankYou✌✌</h1></marquee>$

$<b>________▶▶▶⭐⭐⭐◀◀◀________$