Question:

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

Updated On: Jul 29, 2022
  • $ 2\,\phi $
  • $ \phi /2 $
  • $ \phi /4 $
  • $ \phi $
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The Correct Option is A

Solution and Explanation

By Gauss' theorem $\phi_{E}=\int \vec{E} \cdot d \vec{S}=\frac{q}{\varepsilon_{0}} \phi_{E} \propto \int d S$ $\therefore$ Flux will also doubled, ie, $2 \phi$
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Concepts Used:

Gauss Law

Gauss law states that the total amount of electric flux passing through any closed surface is directly proportional to the enclosed electric charge.

Gauss Law:

According to the Gauss law, the total flux linked with a closed surface is 1/ε0 times the charge enclosed by the closed surface.

For example, a point charge q is placed inside a cube of edge ‘a’. Now as per Gauss law, the flux through each face of the cube is q/6ε0.

Gauss Law Formula:

As per the Gauss theorem, the total charge enclosed in a closed surface is proportional to the total flux enclosed by the surface. Therefore, if ϕ is total flux and ϵ0 is electric constant, the total electric charge Q enclosed by the surface is;

Q = ϕ ϵ0

The Gauss law formula is expressed by;

ϕ = Q/ϵ0

Where,

Q = total charge within the given surface,

ε0 = the electric constant.