Learning Goal: To be able to calculate the energy of a charged capacitor and to

Question

Learning Goal:

To be able to calculate the energy of a charged capacitor and to understand the concept of energy associated with an electric field.

The energy of a charged capacitor is given by U=QV/2, where Q is the charge of the capacitor and V is the potential difference across the capacitor. The energy of a charged capacitor can be described as the energy associated with the electric field created inside the capacitor.

In this problem, you will derive two more formulas for the energy of a charged capacitor; you will then use a parallel-plate capacitor as a vehicle for obtaining the formula for the energy density associated with an electric field. It will be useful to recall the definition of capacitance, C=Q/V, and the formula for the capacitance of a parallel-plate capacitor,

C=?0A/d, where A is the area of each of the plates and d is the plate separation. As usual, ?0 is the permittivity of free space.

First, consider a capacitor of capacitance C that has a charge Q and potential difference V.

Part A

Find the energy U of the capacitor in terms of C and Q by using the definition of capacitance and the formula for the energy in a capacitor.

Express your answer in terms of C and Q.

ANSWER:

Part B

This question will be shown after you complete previous question(s).

Part C

A parallel-plate capacitor is connected to a battery. The energy of the capacitor is U0. The capacitor remains connected to the battery while the plates are slowly pulled apart until the plate separation doubles. The new energy of the capacitor is U. Find the ratio U/U0.

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ANSWER:

Part D

A parallel-plate capacitor is connected to a battery. The energy of the capacitor is U0. The capacitor is then disconnected from the battery and the plates are slowly pulled apart until the plate separation doubles. The new energy of the capacitor is U. Find the ratio U/U0.

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ANSWER:

In this part of the problem, you will express the energy of various types of capacitors in terms of their geometry and voltage.

Part E

A parallel-plate capacitor has area A and plate separation d, and it is charged to voltage V. Use the formulas from the problem introduction to obtain the formula for the energy U of the capacitor.

Express your answer in terms of A, d, V, and appropriate constants.

ANSWER:

Let us now recall that the energy of a capacitor can be thought of as the energy of the electric field inside the capacitor. The energy of the electric field is usually described in terms of energy density u, the energy per unit volume.

A parallel-plate capacitor is a convenient device for obtaining the formula for the energy density of an electric field, since the electric field inside it is nearly uniform. The formula for energy density can then be written as

u=UV,

where U is the energy of the capacitor and V is the volume of the capacitor (not its voltage).

Part F

A parallel-plate capacitor has area A and plate separation d, and it is charged so that the electric field inside is E. Use the formulas from the problem introduction to find the energy U of the capacitor.

Express your answer in terms of A, d, E, and appropriate constants.

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ANSWER:

Part G

Find the energy density u of the electric field in a parallel-plate capacitor. The magnitude of the electric field inside the capacitor is E.

Express your answer in terms of E and appropriate constants.

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ANSWER:

Explanation / Answer

A.

B.

C.

D.

E.

F.

G. u = (?_0*E^2)/2

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