Q. The electric flux through a closed surface area S enclosing charge $\mathcal{Q\ }$is$\ \phi$. If the surface area is doubled, then the flux is
A) 2$\phi$
B) $\phi/2$
C) $\phi/4$
D) $\phi$
Explanation
(a) By Gauss’s theorem
$\phi_{E} = \int_{}^{}{\mathbf{E}.d\mathbf{S} = \frac{q}{\varepsilon_{0}}.}$
$\phi_{E} \propto \int_{}^{}{ds}$
$\mathbf{\therefore}$ Flux will also doubled, $ie,\ 2\phi.$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.