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kykrilka [37]
2 years ago
7

A parallel plate capacitor is charged by connecting it to a battery. After the steady state is reached, the electric energy stor

ed in the capacitor is UE. Which of the following makes a correct claim about the amount of work needed to charge the capacitor and gives evidence to support this claim?
A) No work is needed to charge the capacitor, because the charges will naturally move to their equilibrium position.
B) The amount of work needed to charge the capacitor is UE, because in all physical situations, the work done is equal to the final potential energy of the system.
C) The amount of work needed to charge the capacitor is UE, because when integrating the equation W = integral qdV with the correct limits yield the equation for the energy stored on a capacitor, UE = 1/2qV.
D) The amount of work needed to charge the capacitor is 2UE, because it takes UE of work to remove the charges from the battery and UE to place the charges on the capacitor (E).
E) The amount of work needed to charge the capacitor is 2UE, because for a capacitor UE = 1/2qV while for the battery W = qV.
Physics
1 answer:
Vladimir79 [104]2 years ago
3 0

Answer: C. The amount of work needed to charge the capacitor is UE, because when integrating the equation W = integral qdV with the correct limits yield the equation for the energy stored on a capacitor, UE = 1/2qV.

Explanation:

The claim about the amount of work that is needed to charge the capacitor and give evidence to support this claim is option C "The amount of work needed to charge the capacitor is UE, because when integrating the equation W = integral qdV with the correct limits yield the equation for the energy stored on a capacitor, UE = 1/2qV".

Option C is the correct answer because when we a capacitor is being charged, the amount of work that's being stored as a potential energy.

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A rock climber hangs freely from a nylon rope that is 15 m long and has a diameter of 8.3 mm. If the rope stretches 5.1 cm, what
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Answer:

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Explanation:

Change in length

        \Delta L=\frac{PL}{AE}

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6 0
3 years ago
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3 years ago
Read 2 more answers
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