Assignment 9: Field Energy and Dielectrics

Due: 8:00am on Wednesday, February 8, 2012

Note: To understand how points are awarded, read your instructor's Grading Policy.

Energy in Capacitors and Electric Fields

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 , where is the charge of the capacitor and 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, , and the formula for the capacitance of a

parallel-plate capacitor,

, where is the area of each of the plates and is the plate separation. As usual, is the permittivity of free

space.

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

Part A

Find the energy of the capacitor in terms of and by using the definition of capacitance and the formula for

the energy in a capacitor.

Express your answer in terms of and .

ANSWER:

=

Correct

Part B

Find the energy of the capacitor in terms of and by using the definition of capacitance and the formula for

the energy in a capacitor.

Express your answer in terms of and .

ANSWER:

=

Correct

All three of these formulas are equivalent:

.

Depending on the problem, one or another may be more convenient to use. However, any one of them would give

you the correct answer. Note that these formulas work for any type of capacitor.

Part C

A parallel-plate capacitor is connected to a battery. The energy of the capacitor is . 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 . Find the ratio .

Hint C.1 Determine what remains constant

Hint not displayed

Hint C.2 Identify which formula to use

Hint not displayed

ANSWER:

=

0.5

Correct

Part D

A parallel-plate capacitor is connected to a battery. The energy of the capacitor is . 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 . Find the ratio .

Hint D.1 Determine what remains constant

Hint not displayed

Hint D.2 Identify which formula to use

Hint not displayed

ANSWER:

=

2

Correct

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 and plate separation , and it is charged to voltage . Use the formulas from

the problem introduction to obtain the formula for the energy of the capacitor.

Express your answer in terms of , , , and appropriate constants.

ANSWER:

=

Correct

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 , 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

,

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

Part F

A parallel-plate capacitor has area and plate separation , and it is charged so that the electric field inside is .

Use the formulas from the problem introduction to find the energy of the capacitor.

Hint F.1 How to approach the problem

Hint not displayed

Express your answer in terms of , , , and appropriate constants.

ANSWER:

=

Correct

As mentioned before, we can think of the energy of the capacitor as the energy of the electric field inside the

capacitor.

Part G

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

inside the capacitor is .

Hint G.1 How to approach the problem

Hint not displayed

Hint G.2 Volume between the plates

Hint not displayed

Express your answer in terms of and appropriate constants.

ANSWER:

=

Correct

Note that the answer for does not contain any reference to the geometry of the capacitor: and do not appear in

the formula. In fact, the formula

describes the energy density in any electrostatic field, whether created by a capacitor or any other source.

Force between Capacitor Plates

Consider a parallel-plate capacitor with plates of area and with separation .

Part A

Find , the magnitude of the force each plate experiences due to the other plate as a function of , the

potential drop across the capacitor.

Hint A.1 How to approach the problem

Hint not displayed

Hint A.2 Method 1: Use the stored energy to find the force

Hint not displayed

Hint A.3 Method 2: Use the product of the charge and the field to find the force

Hint not displayed

Express your answer in terms of given quantities and .

ANSWER:

= Correct

Exercise 24.34

A capacitor is formed from two concentric spherical conducting shells separated by vacuum. The inner sphere has

radius 10.5 , and the outer sphere has radius 16.5 . A potential difference of 110 is applied to the capacitor.

Part A

What is the energy density at = 10.6 , just outside the inner sphere?

ANSWER:

=

3.54×10−5

Correct

Part B

What is the energy density at = 16.4 , just inside the outer sphere?

ANSWER:

=

6.17×10−6

Correct

Part C

For a parallel-plate capacitor the energy density is uniform in the region between the plates, except near the edges

of the plates. Is this also true for a spherical capacitor?

ANSWER:

Yes

No

Correct

Exercise 24.37

Two parallel plates have equal and opposite charges. When the space between the plates is evacuated, the electric

field is = 3.50×105 . When the space is filled with dielectric, the electric field is = 2.30×10

5 .

Part A

What is the charge density on each surface of the dielectric?

ANSWER:

= 1.06×10

−6

Correct

Part B

What is the dielectric constant?

ANSWER:

=

1.52

Correct

part C of problem 24.42 is ungraded essay. Submit something but do not request the answer or give up

Exercise 24.42

A constant potential difference of 12 is maintained between the terminals of a 0.25- parallel-plate air capacitor.

Part A

A sheet of Mylar is inserted between the plates of the capacitor, completely filling the space between the plates.

When this is done, how much additional charge flows onto the positive plate of the capacitor? (Dielectric

constant of Mylar is 3.1 )

Express your answer using two significant figures.

ANSWER:

=

6.3×10−6

Correct

Part B

What is the total induced charge on either face of the Mylar sheet?

Express your answer using two significant figures.

ANSWER:

= 6.3×10

−6

Correct

Part C

What effect does the Mylar sheet have on the electric field between the plates? Explain how you can reconcile

this with the increase in charge on the plates, which acts to increase the electric field.

Essay answers are limited to about 500 words (3800 characters maximum, including spaces).

ANSWER: My Answer: increases electric field

Capacitor with Partial Dielectric

Consider a parallel-plate capacitor that is partially filled with a dielectric

of dielectric constant . The dielectric has the same same height as the separation of the plates of the capacitor but

fills a fraction of the area of the capacitor. The capacitance of the capacitor when the dielectric is completely

removed is .

Part A

What is the capacitance of this capacitor as a function of ?

Hint A.1 Modeling the partly filled capacitor

Hint not displayed

Hint A.2 Find the capacitance of the air-filled portion

Hint not displayed

Hint A.3 Find the capacitance of the dielectric-filled portion

Hint not displayed

Hint A.4 Special cases

Hint not displayed

Express in terms of , , and .

ANSWER:

= Correct

± Energy of a Capacitor in the Presence of a Dielectric

A dielectric-filled parallel-plate capacitor has plate area = 25.0 , plate separation = 8.00 and dielectric

constant = 4.00. The capacitor is connected to a battery that creates a constant voltage = 7.50 . Throughout

the problem, use = 8.85×10−12

.

Part A

Find the energy of the dielectric-filled capacitor.

Hint A.1 Energy of a charged capacitor in terms of its capacitance and voltage

Hint not displayed

Hint A.2 Capacitance of a dielectric-filled capacitor

Hint not displayed

Hint A.3 Find the capacitance

Hint not displayed

Express your answer numerically in joules.

ANSWER:

=

3.11×10−10

Correct

Part B

The dielectric plate is now slowly pulled out of the capacitor, which remains connected to the battery. Find the

energy of the capacitor at the moment when the capacitor is half-filled with the dielectric.

Hint B.1 What quantity remains constant?

Hint not displayed

Hint B.2 Modeling the capacitor

Hint not displayed

Hint B.3 Finding the energy

Hint not displayed

Express your answer numerically in joules.

ANSWER:

=

1.94×10−10

Correct

Part C

The capacitor is now disconnected from the battery, and the dielectric plate is slowly removed the rest of the way

out of the capacitor. Find the new energy of the capacitor, .

Hint C.1 What quantity remains constant?

Hint not displayed

Hint C.2 Energy of a charged capacitor in terms of its capacitance and charge

Hint not displayed

Hint C.3 Calculate the charge in the capacitor

Hint not displayed

Express your answer numerically in joules.

ANSWER: = 4.86×10

−10

Correct

Comparing the expressions for and , one can see that ; in other words, the energy of the capacitor

increases as the plate is being pulled out.

Part D

In the process of removing the remaining portion of the dielectric from the disconnected capacitor, how much

work is done by the external agent acting on the dielectric?

Hint D.1 Conservation of energy

Hint not displayed

Express your answer numerically in joules.

ANSWER:

=

2.92×10−10

Correct

Since for every dielectric , the work done in the last process is positive; in other words, an external agent

must apply a force to pull the plate out; the capacitor would exert a net force that would "resist" the pullout.