# If the radius of the gaussian surface enclosing a charge is halved, how does the electric flux through the gaussian surface change?

According to question, electric flux (ϕ) due to a point charge enclosed by a spherical Gaussian surface is given by ϕ = E.A

$phi =frac{kq}{{{r}^{2}}}cdot 4pi {{r}^{2}}=kqcdot 4pi left( because E=frac{kq}{{{r}^{2}}},and,A=4pi {{r}^{2}} right)$

So, there is no effect of change in radius on the electric flux. Flux is function of q, here q is changed to q/2, flux becomes $phi =frac{kq}{{{r}^{2}}}cdot 4pi {{r}^{2}}=frac{kq}{2}cdot 4pi left( because E=frac{kq}{{{r}^{2}}},and,A=4pi {{r}^{2}} right)$