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A parallel plate capacitor is charged and then is disconnected from the battery. By what factor does the stored energy change when the plate separation is doubled?

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Amal Jeeva
Amal JeevaLv10
1 Dec 2020

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Related textbook solutions

Physics: Principles with Applications
7th Edition, 2013
Giancoli
ISBN: 9780321625922

Related questions

The parallel plates in a capacitor, with a plate area of 5.80 cm2 and an air-filled separation of 4.90 mm, are charged by a 8.10 V battery. They are then disconnected from the battery and pulled apart (without discharge) to a separation of 6.70 mm. Neglecting fringing, find (a) the potential difference (in V) between the plates, (b) the initial stored energy, (c) the final stored energy, and (d) the work required to separate the plates.

Below is the question (its a True and False question). Ashortexplanation as to why the answer is what it is would behelpful. Iwill rate!
A parallel plate capacitor with plate separation d isconnectedto a battery. The capacitor is fully charged to QCoulombs and avoltage of V. (C is the capacitance and U is thestored energy.)Answer the following questions regarding thecapacitor charged by abattery. For each statement below, selectTrue or False.

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