The electric flux through a closed surface area S enclosing charge Q is φ . If the surface area is doubled, then the flux is

Question

The electric flux through a closed surface area S enclosing charge Q is φ . If the surface area is doubled, then the flux is (A)  2 φ  (B)  φ /2  (C)  φ /4  (D)  φ

Tardigrade Admin · Accepted Answer

By Gauss' theorem φE=∫ vecE ⋅ d vecS=(q/ε0) φE propto ∫ d S ∴ Flux will also doubled, ie, 2 φ

Q. The electric flux through a closed surface area $S$ enclosing charge $Q$ is $ϕ$ . If the surface area is doubled, then the flux is

$2 ϕ$

$ϕ /2$

$ϕ /4$

$ϕ$

By Gauss' theorem $ϕ_{E} = \int E \cdot d S = \frac{q}{ε _{0}} ϕ_{E} \propto \int d S$
$∴$ Flux will also doubled, ie, $2 ϕ$

Q. The electric flux through a closed surface area S enclosing charge Q is ϕ . If the surface area is doubled, then the flux is

2ϕ

ϕ/2

ϕ/4

ϕ

By Gauss' theorem ϕE​=∫E⋅dS=ε0​q​ϕE​∝∫dS ∴ Flux will also doubled, ie, 2ϕ

Q. The electric flux through a closed surface area $S$ enclosing charge $Q$ is $ϕ$ . If the surface area is doubled, then the flux is

$2 ϕ$

$ϕ /2$

$ϕ /4$

$ϕ$

By Gauss' theorem $ϕ_{E} = \int E \cdot d S = \frac{q}{ε _{0}} ϕ_{E} \propto \int d S$
$∴$ Flux will also doubled, ie, $2 ϕ$