A charge Q is enclosed by a Gaussian surface | vedclass

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(C) According to Gauss's law,the electric flux $\phi$ through any closed surface is given by the formula:$\phi = \frac{Q_{	ext{enclosed}}}{\varepsilo

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remain the same

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(C) According to Gauss's law,the electric flux $\phi$ through any closed surface is given by the formula:$\phi = \frac{Q_{	ext{enclosed}}}{\varepsilon_0}$Here,$Q_{	ext{enclosed}}$ is the total charge enclosed by the Gaussian surface and $\varepsilon_0$ is the permittivity of free space.From this expression,it is clear that the electric flux depends only on the magnitude of the charge enclosed within the surface.It does not depend on the shape or the size (radius $R$) of the Gaussian surface.Therefore,if the radius of the Gaussian surface is doubled,the enclosed charge $Q$ remains the same,and consequently,the outward electric flux will remain the same.

$A$ charge $Q$ is enclosed by a Gaussian surface of radius $R$. If the radius is doubled,then the outward electric flux will

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