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

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

J & K CETJ & K CET 2008Electric Charges and Fields

Solution:

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, $2 \phi$