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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?

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up vote 2 down vote accepted

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 = $q/6\epsilon_0$

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Thank you, do you mean in Q1 that the flux is still the same, but only field is change? Also she said the flux is decrease because A= 4*pi*r^2, and the new radius is 1/2r so the flux is decrease :) – user32104 Nov 2 '13 at 15:27
i made a new id to just answer your question because i am a begginer here too and dont have preveledge to answer.try to learn latex math on was my pleasure to answer your question as i know how it feels to being unanswered. – user32111 Nov 2 '13 at 15:37

I wanna attempt question 3, inside the cube the electric field is equal to zero since the charge is enclosed inside the cube and the electric flux remain the same because the electric field inside the cube affected the expression and make it to be equal to zero.

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