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Q. The electric flux from a cube of edge $l$ is $\phi$. If an edge of the cube is made $2 l$ and the charge enclosed is halved, its value will be

Electric Charges and Fields

Solution:

$\phi=\frac{q}{\varepsilon_{0}}$ (Independent of dimensions)
New flux, $\phi^{\prime}=\frac{\phi}{2}$