The electric flux from a cube of edge l is ph | vedclass

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(B) According to Gauss's law,the electric flux $\phi$ through a closed surface is given by $\phi = \frac{Q}{\varepsilon_0}$,where $Q$ is the total cha

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$2 \phi$

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(B) According to Gauss's law,the electric flux $\phi$ through a closed surface is given by $\phi = \frac{Q}{\varepsilon_0}$,where $Q$ is the total charge enclosed by the surface and $\varepsilon_0$ is the permittivity of free space.Gauss's law states that the electric flux through a closed surface is independent of the size or shape of the surface; it depends only on the net charge enclosed within it.Given that the initial flux is $\phi = \frac{Q}{\varepsilon_0}$.If the edge length is changed to $\frac{2}{3} l$,the flux remains unaffected by this change in geometry.If the enclosed charge is doubled,the new charge becomes $Q' = 2Q$.The new electric flux $\phi'$ is given by $\phi' = \frac{Q'}{\varepsilon_0} = \frac{2Q}{\varepsilon_0} = 2 \left( \frac{Q}{\varepsilon_0} \right) = 2 \phi$.Therefore,the new electric flux will be $2 \phi$.

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

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