Question

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A constant electric current passes through a wire whose cross-section changes from A₁ to A₂. If A₁ < A_{2, what happens to the current density and to the drift veocity of the moving charges as they go from region} A₁ to region A₂?
A) Current density increases and drift velocity decreases.
B) Current density decreases and drift velocity increases.
C) Both current density and drift velocity increase.
D) Both current density and drift velocity decrease.
E) Current density increases, and drift velocity stays the same as it doesn't depends on the cross-sectional area.
First question, I thought answer is C

2.) An air-FIlled parallel-plate capacitor is connected to a battery and allowed to charge up, and then disconnected from the battery. A slab of dielectric material is placed between the plates of the capacitor. After this is done, we find that:
A) the voltage across the capacitor has decreased.
B) the charge on the capacitor has decreased.
C) the charge on the capacitor has increased.
D) the voltage across the capacitor has increased.
E) the energy stored in the capacitor has increased.
I thought Second question answer was E.

explain please, only answer when you positive sure knows the answer, thanks!

Answer 1

1)

Current density is defined as the amount of current flowing per unit cross sectional area of the wire. Given that the cross sectional area of wire has increased from to , the current density decreases.

Now, relation between current and drift velocity is given by . Here, is cross sectional area and is drift velocity. For a constant current, if cross sectional area increases then the drift velocity must be decreased to keep current constant.

Thus, both current density and velocity decreases. Thus correct option is (D).

(2)

Initially when the air filled parallel plate capacitor is charged, there exists an electric field between the plates of capacitor from positive plate to negative plate. Now, when an dielectric is placed between the plates, it polarizes and negative charges accumulates on the side near positive plate and positive charges accumulates near the negative plate of capacitor. Thus there exists an additional field due to dielectric which opposes the field of capacitor. Thus the net electric field across the capacitor decreases and thus voltage across the capacitor decreases.

Thus correct option is (A)