A parallel plate capacitor with plate separation d is connected to a battery. Th
ID: 1779785 • Letter: A
Question
A parallel plate capacitor with plate separation d is connected to a battery. The capacitor is fully charged to Q Coulombs and a voltage of V. (C is the capacitance and U is the stored energy.) Answer the following questions regarding the capacitor charged by a battery. For each statement below, select True or False.
After being disconnected from the battery, increasing d increases V. A: True B: False
With the capacitor connected to the battery, inserting a dielectric with will increase U. A: True B: False
After being disconnected from the battery, increasing d increases U. A: True B: False
After being disconnected from the battery, decreasing d increases C. A: True B: False
With the capacitor connected to the battery, inserting a dielectric with will increase C. A: True B: False
After being disconnected from the battery, inserting a dielectric with will decrease V. A: True B: False
Explanation / Answer
After being disconnected from the battery, increasing d increases V:
True
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With the capacitor connected to the battery, inserting a dielectric with will increase U:
True. U = CV²/2 so energy will increase to U because capacitance will increase to C (V remains constant).
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After being disconnected from the battery, increasing d increases U :
True. U = Q²/2C and C = A/d. C increases linearly with decrease in d, and U decreases linearly with increase in C, therefore U increases linearly with increase in d.
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After being disconnected from the battery, decreasing d increases C :
True. using the above relation.
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With the capacitor connected to the battery, inserting a dielectric with will increase C:
True. Capacitance will increase to C.
_________________________
After being disconnected from the battery, inserting a dielectric with will decrease V:
True. V will decrease. The dielectric sets up a back EMF which partially cancels V.
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