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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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remain the same
We know from Gauss's law that the electric flux through any closed spherical surface is given by
$\phi=\frac{\mathrm{q}_{\text {enclosed. }}}{\varepsilon_0}$
So electric flux is independent of the radius of the closed surface and only depends on the charge enclosed.
Therefore if the radius of Gaussian spherical surface is doubled then there will be no effect on flux and the outward electric flux will remain the same.
$\phi=\frac{\mathrm{q}_{\text {enclosed. }}}{\varepsilon_0}$
So electric flux is independent of the radius of the closed surface and only depends on the charge enclosed.
Therefore if the radius of Gaussian spherical surface is doubled then there will be no effect on flux and the outward electric flux will remain the same.
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