Cuesta College, San Luis Obispo, CA
A potential difference of 4.2×103 V is applied to a 54 µF capacitor salvaged from a cardiac defibrillator, and the electric potential energy stored is used to explode a potato.[*] If a higher potential difference is applied to this capacitor, it will store more:
(B) electric potential energy.
(C) (Both of the above choices.)
(D) (Neither of the above choices.)
Correct answer (highlight to unhide): (C)
The capacitance is given by:
C = A/(4·π·k·d),
since the area A of each plate, and the distance d between the plates remain constant, the capacitance C remains constant regardless of how much potential difference is applied to it.
The relation between the capacitance, the charge stored, and the potential difference applied is:
C = Q/∆V.
Because the capacitance C remains constant, then increasing the potential difference ∆V will also increase the amount of charge Q stored by the capacitor.
The amount of electric potential energy stored by the capacitor is given by:
EPE = (1/2)·Q·(∆V),
As the amount of charge Q increases with the increase in potential difference ∆V, then the amount of electric potential energy stored by the capacitor must increase.
Sections 30882, 30883
Exam code: quiz04T4nK
(A) : 5 students
(B) : 14 students
(C) : 15 students
(D) : 2 students
Success level: 42%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.39