Question

A closed surface S is constructed around a conducting wire connected to a battery and a switch (figure 30-Q6). As the switch is closed, the free electrons in the wire start moving along the wire. In any time interval, the number of electrons entering the closed surface S is equal to the number of electrons leaving it. On closing the switch, the flux of the electric field through the closed surface
(a) is increased
(b) is decreased
(c) remains unchanged
(d) remains zero
Figure

Answer 1

Answer:

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Answer 2

On closing the switch, the electric field’s flux via the closed surface (c) remains unchanged and (d) remains zero.

Explanation:

• In the closed surface, initially, there is no charge. Initially, the flux is zero as the wire has neutrality. Now, if the battery is connected and there is a flow of current through it, the flux remains zero.
• This is because of the reason that the surface accepts certain no. of electrons that is equal to the no. of leaving electrons. Thus, the net charge is zero.
• Gauss’ law says that electric flux via a shut surface is equal to 1/ϵ0 times the charge bounded by the surface.

                       $\Phi=\int E \cdot d s=\frac{q}{\epsilon_{0}}$

• Where = electric flux, ds = area element, E = electric field, ϵ0 = electric permittivity of vacuum and q = charge enclosed.
• Here, the total charge enclosed by the surface (Q) is zero.  
• Therefore, the electric field’s flux is zero and it is always unchanged.