A parallel-plate capacitor carries a constant charge Q (i.e., + Q on one plate,

Question

A parallel-plate capacitor carries a constant charge Q (i.e., +Q on one plate, and -Q on the other). When a dielectric is inserted between the plates, what happens to the voltage difference between the plates, and the potential energy stored in the capacitor?

The voltage and potential energy both increase.
The voltage decreases, and the potential energy increases.
The voltage increases, and the potential energy decreases.
The voltage and potential energy both decrease.
None of these choices.

A parallel-plate capacitor is attached to a battery that maintains a constant voltage difference V across the plates. When a dielectric is inserted between the plates, what happens to the charge on the plates, and the potential energy stored in the capacitor?

The charge and potential energy both increase.
The charge increases, and the potential energy decreases.
None of these choices.
The charge and potential energy both decrease.
The charge decreases, and the potential energy increases.

Explanation / Answer

1) The voltage and potential energy both decrease.

when dielctric is introduced,
because, V' = V/k

Q' = Q

U' = 0.5*Q'*V'

= 0.5*Q*(V/K)

= 0.5*Q*V/k

= U/K

2) The charge and potential energy both increase.

when dielctric is introduced,
here, V' = V

C' = k*C

Q' = C'*V'

= K*C*V

= K*Q

U' = 0.5*Q'*V'

= 0.5*K*Q*V

= k*U

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