Question

The total electric flux coming out from a closed surface enclosing a stationary
unit positive charge in air is​

Answer 1

HELLO DEAR,

By appling Gauss' Theorem:-

The total electric flux linked with closed surface S is

Φ = ∫E.dS = q/ε0

Where ,q is the total charge enclosed by the closed Gaussian (imaginary) surface.

By using above theorem ,the total electric field passes through the closed shereical gaussion surface is,

∫E.dS= q/ε0

The electric field passes normal to the Gaussian surface (sphere) and uniform.

Therefore E.S= q/ε0

E= q/ε0 S

E= q/4πr^2 ε0

[S = surface area of sphere ]

[S= 4πr^2. ]

E= q/4πε0r^2 .

And TOTAL ELECTIC FLUX PASSES THROUGH THE SPHERE IS,

=> Φ = q/ε0

Because ,

Electric Flux:- Electric flux linked with any surface is proportional to the total number of electric field lines that normally passes through that surface. It is a Scalar quantity.

I HOPE IT'S HELP YOU DEAR,

THANKS.

Answer 2

To find:

The total electric flux coming out from a closed surface enclosing a stationary unit positive charge in air?

Calculation:

We will apply Gauss' Law to find out the total electric flux coming out of the the Gaussian surface enclosing the charge.

According to Gauss' Theorem:

• The net electric flux through the surface (enclosing a charge) is equal to the ratio of the charge enclosed and the permittivity of free space.

• Since the enclosed charge is a point charge, we can assume that a spherical Gaussian surface around the charge.

Let flux be denoted as $\phi$ , Electrostatic Field Intensity be E and elemental area be ds:

$\therefore \: \displaystyle\oint \vec{E} \: . \: \vec{ds} = \frac{q}{\epsilon_{0} }$

$\implies \: \phi = \dfrac{q}{\epsilon_{0} }$

So, final answer is:

$\boxed{ \bold{net \: flux\: = \phi = \dfrac{q}{\epsilon_{0} } }}$