SAT Subject Physics Practice Question 81: Answer and Explanation

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Question: 81

8. A parallel-plate capacitor is charged to a potential difference of ∆V; this results in a charge of +Q on one plate and a charge of –Q on the other. The capacitor is disconnected from the charging source, and a dielectric is then inserted. What happens to the potential difference and the stored electrical potential energy?

A. The potential difference decreases, and the stored electrical potential energy decreases.
B. The potential difference decreases, and the stored electrical potential energy increases.
C. The potential difference increases, and the stored electrical potential energy decreases.
D. The potential difference increases, and the stored electrical potential energy increases.
E. The potential difference decreases, and the stored electrical potential energy remains unchanged.

Correct Answer: A

Explanation:

8 A Since Q cannot change and C is increased (because of the dielectric), ∆V = Q/C must decrease. Also, since UE = , an increase in C with no change in Q implies a decrease in UE.

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