Question

The electric flux from a cube of edge $l$ is $\phi$ in an enclosed charge. If the edge of the cube is made $\frac{2}{3} l$ and the charge enclosed in the cube is doubles, then the elctric flux value will be

• A. $4 \phi$
• B. $2 \phi$
• C. $\frac{\phi}{2}$
• D. $\phi$

Answer 1

Answer: (B)

According to Gauss's law, the electric flux through a closed surface is Electric flux, $\phi=\frac{Q}{\varepsilon_{0}} \ldots$ (i)
Electric flux through a closed surface does not depend on the dimensions of closed surface, it depends only the amount of charge enclosed by the closed surface.
According to question,
$Q'=2 Q$
$\Rightarrow \phi'=\frac{Q'}{\varepsilon_{0}}=\frac{2 Q}{\varepsilon_{0}}=2 \phi[$ using Eq. (i) $]$