mp em ass 8: capacitance

Assignment 8: Capacitance

Due: 8:00am on Monday, February 6, 2012

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

[Switch to Standard Assignment View]

Introduction to Capacitance

Learning Goal: To understand the meaning of capacitance and ways of calculating capacitance

When a positive charge is placed on a conductor that is insulated from ground, an electric field emanates from

the conductor to ground, and the conductor will have a nonzero potential relative to ground. If more charge is

placed on the conductor, this voltage will increase proportionately. The ratio of charge to voltage is called the

capacitance of this conductor: .

Capacitance is one of the central concepts in electrostatics, and specially constructed devices called capacitors are

essential elements of electronic circuits. In a capacitor, a second conducting surface is placed near the first (they

are often called electrodes), and the relevant voltage is the voltage between these two electrodes.

This tutorial is designed to help you understand capacitance by assisting you in calculating the capacitance of a

parallel-plate capacitor, which consists of two plates each of area separated by a small distance with air or

vacuum in between. In figuring out the capacitance of this configuration of conductors, it is important to keep in

mind that the voltage difference is the line integral of the electric field between the plates.

Part A

What property of objects is best measured by their capacitance?

ANSWER:

ability to conduct electric current

ability to distort an external electrostatic field

ability to store charge

Correct

Capacitance is a measure of the ability of a system of two conductors to store electric charge and energy.

Capacitance is defined as . This ratio remains constant as long as the system retains its geometry and the

amount of dielectric does not change. Capacitors are special devices designed to combine a large capacitance

with a small size. However, any pair of conductors separated by a dielectric (or vacuum) has some capacitance.

Even an isolated electrode has a small capacitance. That is, if a charge is placed on it, its potential with

respect to ground will change, and the ratio is its capacitance .

Part B

Assume that charge is placed on the top plate, and is placed on the bottom plate. What is the magnitude

of the electric field between the plates?

Hint B.1 How do you find the magnitude of the electric field?

Hint not displayed

Hint B.2 What is the electric flux integral due to the electric field?

Hint not displayed

Hint B.3 Find the enclosed charge

Hint not displayed

Hint B.4 Recall Gauss's law

Hint not displayed

Express in terms of and other quantities given in the introduction, in addition to and any other constants

needed.

ANSWER:

= Correct

Part C

What is the voltage between the plates of the capacitor?

Hint C.1 The electric field is the derivative of the potential

Hint not displayed

Express in terms of the quantities given in the introduction and any required physical constants.

ANSWER:

= Correct

Part D

Now find the capacitance of the parallel-plate capacitor.

Express in terms of quantities given in the introduction and constants like .

ANSWER:

= Correct

You have derived the general formula for the capacitance of a parallel-plate capacitor with plate area and plate

separation . It is worth remembering.

Part E

Consider an air-filled charged capacitor. How can its capacitance be increased?

Hint E.1 What does capacitance depend on?

Hint not displayed

ANSWER:

Increase the charge on the capacitor.

Decrease the charge on the capacitor.

Increase the spacing between the plates of the capacitor.

Decrease the spacing between the plates of the capacitor.

Increase the length of the wires leading to the capacitor plates.

Correct

Part F

Consider a charged parallel-plate capacitor. Which combination of changes would quadruple its capacitance?

ANSWER:

Double the charge and double the plate area.

Double the charge and double the plate separation.

Halve the charge and double the plate separation.

Halve the charge and double the plate area.

Halve the plate separation; double the plate area.

Double the plate separation; halve the plate area.

Correct

A Capacitor with Thick Plates

Consider a capacitor that consists of two metal plates of nonzero thickness separated by a positive distance .

When such a capacitor is connected across the terminals of a battery with emf , two things happen

simultaneously: Charge flows from one plate to another, and a potential difference appears between the two plates.

In this problem, you will determine the amount and distribution of charge on each metallic plate for an ideal

(infinite) parallel-plate capacitor. There is competition between the repulsive forces from like charges on a plate

and the attractive forces from the opposite charge on the other plate. You will determine which force "wins."

Let , , , and be the respective surface charge densities on surfaces A, B, C, and D. Take all the charge

densities to be positive for now. Some of them are in fact negative, and this will be revealed at the end of the

calculation.

Part A

If we assume that the metallic plates are perfect conductors, the electric field in their interiors must vanish. Given

that the electric field due to a charged sheet with surface charge is given by

,

and that it points away from the plane of the sheet, how can the condition that the electric field in plate I vanishes

be written?

Hint A.1 How to approach the problem

Hint not displayed

ANSWER:

Correct

Part B

Similarly, how can the condition that the electric field in plate II vanishes be written?

Hint B.1 How to approach the problem

Hint not displayed

ANSWER:

Correct

Part C

There is an electric field in the region between the two plates. The magnitude of this electric field is . This

imposes another condition on the charge densities on the surfaces of the plates. How can this condition be

expressed?

Hint C.1 How to approach the problem

Hint not displayed

ANSWER:

Correct

Part D

Finally, how can the condition that the total charge of the system is conserved be expressed?

Hint D.1 How to approach the problem

Hint not displayed

ANSWER:

Correct

You now have four equations in four variables. If you solve the four equations simultaneously you obtain

,

,

which is the familiar expression for the surface charge density on a parallel-plate capacitor. However, you have

now also established that all the charge resides on the insides of the capacitor plates.

Capacitors in Parallel

Learning Goal: To understand how to calculate capacitance, voltage, and charge for a parallel combination of

capacitors.

Frequently, several capacitors are connected together to form a collection of capacitors. We may be interested in

determining the overall capacitance of such a collection. The simplest configuration to analyze involves capacitors

connected in series or in parallel. More complicated setups can often (though not always!) be treated by combining

the rules for these two cases. Consider the example of a parallel combination of capacitors: Three capacitors are

connected to each other and to a battery as shown in the figure.

The individual capacitances are , , and , and the battery's voltage is .

Part A

If the potential of plate 1 is , then, in equilibrium, what are the potentials of plates 3 and 6? Assume that the

negative terminal of the battery is at zero potential.

Hint A.1 Electrostatic equilibrium

Hint not displayed

ANSWER:

and

and

and

and

Correct

Part B

If the charge of the first capacitor (the one with capacitance ) is , then what are the charges of the second and

third capacitors?

Hint B.1 Definition of capacitance

Hint not displayed

Hint B.2 Voltages across the capacitors

Hint not displayed

ANSWER:

and

and

and

and

Correct

Part C

Suppose we consider the system of the three capacitors as a single "equivalent" capacitor. Given the charges of

the three individual capacitors calculated in the previous part, find the total charge for this equivalent

capacitor.

Express your answer in terms of and .

ANSWER:

=

Correct

Part D

Using the value of , find the equivalent capacitance for this combination of capacitors.

Hint D.1 Using the definition of capacitance

Hint not displayed

Express your answer in terms of .

ANSWER:

=

Correct

The formula for combining three capacitors in parallel is

.

How do you think this formula may be generalized to capacitors?

Capacitors in Series

Learning Goal: To understand how to calculate capacitance, voltage, and charge for a combination of capacitors

connected in series.

Consider the combination of capacitors shown in the figure.

Three capacitors are connected to each other in series, and then to the battery. The values of the capacitances are

, , and , and the applied voltage is . Initially, all of the capacitors are completely discharged; after the

battery is connected, the charge on plate 1 is .

Part A

What are the charges on plates 3 and 6?

Hint A.1 Charges on capacitors connected in series

Hint not displayed

Hint A.2 The charges on a capacitor's plates

Hint not displayed

ANSWER:

and

and

and

and

0 and

0 and

Correct

Part B

If the voltage across the first capacitor (the one with capacitance ) is , then what are the voltages across the

second and third capacitors?

Hint B.1 Definition of capacitance

Hint not displayed

Hint B.2 Charges on the capacitors

Hint not displayed

ANSWER:

and

and

and

0 and

Correct

Part C

Find the voltage across the first capacitor.

Hint C.1 How to analyze voltages

Hint not displayed


ANSWER:

= Correct

Part D

Find the charge on the first capacitor.

Express your answer in terms of and .

ANSWER:

= Correct

Part E

Using the value of just calculated, find the equivalent capacitance for this combination of capacitors in

series.

Hint E.1 Using the definition of capacitance

Hint not displayed


ANSWER:

= Correct

The formula for combining three capacitors in series is

.

How do you think this formula may be generalized to capacitors?

Exercise 24.22

The Figure below shows a system of four capacitors, where the

potential difference across is 50.0 .

Part A

Find the equivalent capacitance of this system between and .

ANSWER:

= 3.47

Correct

Part B

How much charge is stored by this combination of capacitors?

ANSWER:

= 174

Correct

Part C

How much charge is stored in the 10.0- capacitor?

ANSWER:

= 174

Correct

Part D

How much charge is stored in the 9.0- capacitor?

ANSWER:

= 174

Correct

Problem 24.60

The capacitors in the Figure are initially uncharged and are

connected, as in the diagram, with switch open. The applied potential difference is = + 210 .

Part A

What is the potential difference ?

Express your answer using two significant figures.

ANSWER:

= 70

Correct

Part B

What is the potential difference after switch is closed?

ANSWER:

= 105

Correct

Part C


ANSWER:

= 105

Correct

Part D


ANSWER:

= 105

Correct

Part E


ANSWER:

= 105

Correct

Part F

How much charge flowed through the switch when it was closed?

ANSWER:

= 315

Correct

± Capacitance and Electric Field of a Spherical Capacitor

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

sphere has a radius of = 12.5 , and the outer sphere has a radius of = 14.9 . A potential difference of

120 is applied to the capacitor.

Part A

What is the capacitance of the capacitor?

Hint A.1 How to approach the problem

Hint not displayed

Hint A.2 Which Gaussian surface to use

Hint not displayed

Hint A.3 Evaluate the integral in Gauss's law

Hint not displayed

Hint A.4 Find an expression for the potential difference

Hint not displayed

Hint A.5 Definiton of capacitance

Hint not displayed

Use = 8.85×10−12

for the permittivity of free space.

ANSWER:

= 8.63×10

−11

Correct

Part B

What is the magnitude of the electric field at radius 12.6 , just outside the inner sphere?

Hint B.1 How to approach the problem

Hint not displayed

Hint B.2 Calculate the charge on the inner sphere

Hint not displayed

ANSWER:

= 5870

Correct

Part C

What is the magnitude of at 14.6 , just inside the outer sphere?

Hint C.1 How to approach the problem

Hint not displayed

ANSWER:

= 4370

Correct

In a parallel-plate capacitor, the field between the plates is uniform. In a spherical capacitor, however, the field is

a function of the radial distance from the center, so the field is not uniform between the spheres.

In problem 24.48 the square plates have side lengths of 16 cm.

Problem 24.48

A parallel-plate air capacitor is made by using two plates 16 square, spaced 3.7 apart. It is connected to a

12- battery.

Part A

What is the capacitance?


ANSWER:

= 6.1×10

−11

Correct

Part B

What is the charge on each plate?


ANSWER:

= 7.4×10

−10

Correct

Part C

What is the electric field between the plates?


ANSWER:

= 3200

Correct

Part D

What is the energy stored in the capacitor?


ANSWER:

= 4.4×10

−9

Correct

Part E

The battery is disconnected and then the plates are pulled apart to a separation of 7.4 .

What is the capacitance in this case?


ANSWER:

= 3.1×10

−11

Correct

Part F

What is the charge on each plate in this case?


ANSWER:

= 7.4×10

−10

Correct

Part G

What is the electric field between the plates in this case?


ANSWER:

= 3200

Correct

Part H

What is the energy stored in the capacitor?


ANSWER:

= 8.8×10

−9

Correct

mp em ass 8: capacitance

Documents