A charge Q is enclosed by a Gaussian surface of radius $R$. If the radius is doubled then the outward…
A charge Q is enclosed by a Gaussian surface of radius $R$. If the radius is doubled then the outward electric flux will
be reduced to half
be doubled
remain the same
increase four times
Solution
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.