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Physics 202, Lecture 7Today’s Topics

About Exam 1

Capacitance (Ch. 26-II) Review Energy storage in capacitors Dielectric materials, electric dipoles Dielectrics and Capacitance

Next lecture: More information about Exam 1. Current and Resistance (Ch. 27-I)

About the First Midterm Exam The exam will be on Monday Oct 1 at 5:30-7:00pm.Rooms: 2650 and 3650 Humanities

McBurney students: contact me for special arrangements. If conflicts with exam time due to an evening course:

• Stay tuned for special arrangements.• Be sure to have informed me via email by Friday at the latest.

An 8½ x 11” double-sided formula sheet is allowed. Has to be self prepared: hand written or printed, but no Xerox.

Bring a calculator: but do not use programming functionality. Absolutely no communication functionality will be allowed.

Format: word problems. More about this Thursday.

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First Midterm Exam Chapters Covered:

Chapter 23: Electric Fields Chapter 24: Gauss’s Law Chapter 25: Electric Potential Chapter 26.1-26.3: Capacitance

Exceptions: you will not be responsible for the material coveredin sections 24.5, 25.7, 25.8.

Office hours: I will be available in my office (5215 CH)on Friday afternoon, from 1-2:30, 3:30-5:30. TA’s are alsoavailable in the helproom as always.

Review sessions: Sunday, Sept 30, 1-3 PM, Rennebohm (this room) Saturday, Sept 29, 7 PM, place TBA

Capacitors: Summary

• Definition:

• Capacitance depends on geometry:

d

A

- - - - -+ + + +

Parallel Plates

a

b L

r

+Q

-Q

Cylindrical

a

b

+Q-Q

Spherical

C has units of “Farads” or F (1F = 1C/V) ε o has units of F/m

C !Q

"V

C =!oA

d

C = 4!"o

ab

b # a

C =2!"

oL

lnb

a

#$%

&'(

3

Capacitors in Series and Parallel

Parallel:

Series:

ΔV

C2

C1

ΔV

Q=Q1+Q2

Cp ΔV=Q

ΔV1 = ΔV2 = ΔV

ΔV1 ΔV2ΔV=ΔV1+ΔV2

Q

ΔV

C1 C2CSΔV=Q

Q1 = Q2 = Q

Cp = C1 + C2

Cs=

1

C1

+1

C2

!

"#$

%&

'1

ΔV

Example: Ch. 26 #27, 28 (board)

Energy of a Capacitor• How much energy is stored in a charged capacitor?

– Calculate the work provided (usually by a battery) to charge acapacitor to +/- Q:

Incremental work dW needed to add charge dq to capacitor at voltage V:- +

• In terms of the voltage V:

• The total work W to charge to Q is then given by:

Two ways to write W

dW = V (q)idq =q

C

!"#

$%&idq

W =1

Cqdq =

0

Q

!1

2

Q2

C

W =1

2CV

2

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Capacitor Variables

• In terms of the voltage V:

• The total work to charge capacitor to Q equals the energy Ustored in the capacitor:

You can do one of two things to a capacitor :

• hook it up to a battery specify V and Q follows

• put some charge on it specify Q and V follows

U =1

2CV

2

U =1

Cqdq =

0

Q

!1

2

Q2

C

Q = CV

V =Q

C

Example (I)

d

A

- - - - -+ + + +• Suppose the capacitor shown here is charged

to Q. The battery is then disconnected.

• Now suppose the plates are pulled further apart to a finalseparation d1.

• How do the quantities Q, C, E, V, U change?

• How much do these quantities change?.. See board.

• Q:• C:• E:• V:• U:

remains the same.. no way for charge to leave.

increases.. add energy to system by separating

decreases.. capacitance depends on geometry

increases.. since C ↓, but Q remains same (or d ↑ but E the same)remains the same... depends only on charge density

Answers: C1=d

d1

C V1=d1

dV

U1=

d1

dU

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• Suppose the battery (V) is keptattached to the capacitor.

• Again pull the plates apart from d to d1.

• Now what changes?

• C:• V:• Q:• E:• U:

decreases (capacitance depends only on geometry)must stay the same - the battery forces it to be Vmust decrease, Q=CV charge flows off the plate

d

A

- - - - -+ + + +V

• How much do these quantities change?.. See board.

Answers:

U1=

d

d1

U

C1=

d

d1

C

E1=

d

d1

E

must decrease ( , )

E =!

E0

E =V

D

must decrease ( ) U =

1

2CV

2

Example (II)

Where is the Energy stored?• Claim: energy is stored in the electric field itself.

• The electric field is given by:

⇒• The energy density u in the field is given by:

Units:

• Consider the example of a constant field generated by a parallelplate capacitor:

++++++++ +++++++

- - - - - - - - - - - - - -- Q

+Q

u =U

volume=

U

Ad=

1

2!

0E

2

E =!

"0

=Q

"0A

U =1

2!

0E

2Ad

U =1

2

Q2

C=

1

2

Q2

( A!0

/ d)

J

m3

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Question A parallel plate capacitor is holding a charge q. The

capacitor is not connected to a battery.If plates are pulled apart, what happens to the stored energy?

IncreasesDecreasesStays the same

- - - - - - -

+ + + + + + + +q

-qd

pull

pull

Arguments: volume increases, E same work done to it when

plates are being pulled apart

Dielectrics• Empirical observation:

Inserting a non-conducting material (dielectric) between theplates of a capacitor changes the VALUE of the capacitance.

• Definition:The dielectric constant of a material is the ratio of thecapacitance when filled with the dielectric to that without it:

κ values are always > 1 (e.g., glass = 5.6; water = 80)INCREASE the capacitance of a capacitorThey permit more energy to be stored on a given capacitor:

! =C

C0

! =C

C0

!U =CV

2

2="C

0V

2

2="U

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Dielectric Materials Dielectrics are electric insulators:

Charges are not freely movable, but can still have smalldisplacements in an external electric field

Atomic view: composed of permanent (or inducible)electric dipoles

- -

++

Permanent dipole

O--

H+ H+

P

+ - + -

+ - + -

Inducible dipole E>0

E=0

Electric Dipole in External E Field Electric dipole moment p.

Electric dipole moment in constant E field

+

-

d-q

+q

!F! = 0

!" =!p #!E

U = $!pi

!E

Net Force

Net Torque

Potential energy

!p = q

!d

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Quick Quiz 1 Which of the following configurations has the

highest potential energy?

Points at 45o

Points towards the E field

Points against the E field

Points normal to the E field

E

E

E

E

Quick Quiz 2 An electric dipole moment initially points at 45o with

respect to the x axis. When an external E field in thepositive x direction is applied, what will happen?

No change

Points towards the E field

Points against the E field

Points normal to the E field

E

E

E

E

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Dielectrics In External Field Alignment of permanent dipoles in external field

Induced field by non-permanent dipoles

Note: induced field always opposite to the external field E0

Applyingexternal E field

Zeroexternal field Equilibrium

Insert Dielectrics In Between Conductor Plates

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+++++++++++++++

Parallel Plate Example

• Deposit a charge Q on parallel platesfilled with vacuum (air)—capacitance C0

• Disconnect from battery• The potential difference is V0 = Q / C0.

Now insert material with dielectric constant κ .Charge Q remains constantCapacitance increases C = κ C0

Voltage decreases from V0 to:

Q+++++++++++++++

- - - - - - - - - - - - - - -

E0V0

Electric field decreases also:

- - - - - - - - - - - - - - -

Q

V E

+- +

-

+- +

-

+-

+-+-

Note: The field onlydecreases when thecharge is held constant!

V =Q

C=

Q

!C0

=V

!

E =V

d=V0

d!=E0

!

Example: Ch. 26 #47 (board)

Dielectric Constant For Various Materials

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Use of Dielectric Material Non-conducting dielectric material can be inserted

in between conductor ends to increase capacitance.