Suppose a point charge is located at the center of a spherical surface. The electric field at the surface of the sphere and the total flux through the sphere are determined.
1).What happens to the flux and the magnitude of The electric field if the radius of the sphere is halved?
Our teacher said the flux decreases and the filed increases.
But here:
A spherical gaussian surface surrounds a point charge $q$. Describe what happens to the: flux through the surface if
2) The radius of the sphere is doubled
Our teacher said the electric flux will not change
3) The shape of the surface is changed to that of a cube
Also the electric flux will not change
So why the electric flux changed in question 1 but in question 2 did not change?
And when does the electric flux and electric field change?
Best Answer
From gauss law we have $$\Phi_E=\frac{q_i}{\epsilon_0}$$ $q_i$ represents total change inside a closed surface and is independent of surface area and radius of that closed surface.
to the other end remember $$\Phi_E=\oint E.dA$$ where $$|E|=\frac{q}{4\pi\epsilon_0r^2}.$$
$\Phi_E$ is of course constant in your example (all questions including Q.1.)
and the reason the electric flux changed in question 1 is you are misunderstood. Field will increase if $r$ decrease because $E\propto \frac{1}{r^2}$.
Q.3's answer : (taken freom H.C verma concepts of physics chapter 30 exercise 5): Since the charge is placed at the centre of the cube. Hence the flux passing through each side = $q/6\epsilon_0$