Question

electric flux through a closed surface enclosing charge q is q upon E0 . What will be the electric flux through a closed surface if it's size is doubled​

Answer 1

Given:

Electric flux through a closed surface enclosing charge q is q upon E_(0).

To find:

Electric flux when size of surface is doubled?

Calculation:

General Equation of Gauss' Law is :

• Electric Flux is equal to ratio of enclosed charge and Permittivity of Vacuum.

$E_{0} = \dfrac{ q_{enclosed} }{ \epsilon_{0} }$

$\implies E_{0} = \dfrac{ q }{ \epsilon_{0} }$

Now, even when the size of Gaussian Surface is doubled, still the charge enclosed remains same (i.e. q).

So, flux will still be E_(0).

Answer 2

Explanation:

Given:

Electric flux through a closed surface enclosing charge q is q upon E_(0).

To find:

Electric flux when size of surface is doubled?

Calculation:

General Equation of Gauss' Law is :

Electric Flux is equal to ratio of enclosed charge and Permittivity of Vacuum.

E_{0} = \dfrac{ q_{enclosed} }{ \epsilon_{0} } E

0

=

ϵ

0

q

enclosed

\implies E_{0} = \dfrac{ q }{ \epsilon_{0} } ⟹E

0

=

ϵ

0

q

Now, even when the size of Gaussian Surface is doubled, still the charge enclosed remains same (i.e. q).

So, flux will still be E_(0).