CapacitanceEnergy & Dielectrics

E0

Q

E

V 0

V

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

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

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

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

qSdE

0

( )q

dW V q dq dqC

2

2

1CVW

202

1Eu

Q

Today…• Review items from last lecture:

• Capacitors (series and parallel)

• Calculate Energy Stored in Capacitor

• Calculate Energy Density in Electric Field

• Define Dielectric Constant

• Modify Gauss’ Law to include Dielectrics

Summary of important capacitor geometries

• The definition of the capacitance relates Q to V via C:

V

QC

• The capacitance depends on the geometry:

d

A

- - - - -

+ + + +

Parallel Plates

d

AC o

a

b L

r

+Q

-Q

Cylindrical

ab

LC o

ln

2

a

b

+Q-Q

Spherical

ab

abC o

4

In SI unit system: C has units of “Farads” or F (1F = 1C/V)o has units of F/m 1

Lecture 8, ACT 1

(a) V = (2/3)V0 (b) V = V0 (c) V = (3/2)V0

• What is the relationship between V0 and V in the systems shown below?

d

(Area A)

V0

+Q

-Q

conductor

(Area A)

V

+Q

-Q

d/3

d/3

Lecture 8, ACT 1

• The electric field in the conductor = 0.• The electric field everywhere else is: E = Q/(A0)• To find the potential difference, integrate the electric field:

EdldEV

0

30

3

dE

dEldEV

EdV3

2

(a) V = (2/3)V0 (b) V = V0 (c) V = (3/2)V0

• What is the relationship between V0 and V in the systems shown below?

d

(Area A)

V0

+Q

-Q

conductor

(Area A)

V

+Q

-Q

d/3

d/3

Lecture 8, ACT 1, Another Way

(a) V = (2/3)V0 (b) V = V0 (c) V = (3/2)V0

• What is the relationship between V0 and V in the systems shown below?

d

(Area A)

V0

+Q

-Q

conductor

(Area A)

V

+Q

-Q

d/3

d/3

• The arrangement on the right is equivalent to capacitors (each with separation = d/3) in SERIES!!

CCeq 21

conductord/3

(Area A)

V

+Q

-Qd/3

d/3+Q

-Qd/3

000

2

3

2

3

3/2

1C

d

A

d

ACeq

00 3

22/3

VC

QCQ

Veq

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

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

Calculate incremental work dW needed to add charge dq to capacitor at voltage V (there is a trick here!):

dqC

qdqqVdW

)(

- +

• In terms of the voltage V:

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

2

2

1CVW

C

Qqdq

CW

Q 2

0 2

11 Look at this!

Two ways to write W

Capacitor Variables

• In terms of the voltage V: 21

2U CV

• The total work to charge to Q equals the energy U stored in the capacitor: 2

0

1 1

2

QQ

U qdqC C

Look at this!Two ways to write U

CVQ

C

QV

• 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

Question!

d

A

- - - - -

+ + + +• Suppose the capacitor shown here is

charged to Q and then the battery is disconnected.

• Now suppose I pull the plates further apart so that the final separation is d1.

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

• How much do these quantities change?.. exercise for student!!

Answers: 11

dU U

dC

d

dC

11 V

d

dV 1

1

• Q:• C:• E:• V:• U:

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

increases.. add energy to system by separating

decreases.. since 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

Related Question• Suppose the battery (V) is kept

attached to the capacitor.

• Again pull the plates apart from d to d1.

• Now what changes?

• How much do these quantities change?.. exercise for student!!

Answers: 11

dC C

d 1

1

dE E

d

• C:• V:• Q:• E:• U:

decreases (capacitance depends only on geometry)

must stay the same - the battery forces it to be V

must decrease, Q=CV charge flows off the plate

d

A

- - - - -

+ + + +V

must decrease ( )212U CV

must decrease ( , )V

ED

0

EE

11

dU U

d

2

Two identical parallel plate capacitors are connected to a battery, as shown in the figure. C1 is then disconnected from the battery, and the separation between the plates of both capacitors is doubled.

2) What is the relation between the charges on the two capacitors ?

a) Q1 > Q2 b) Q1 = Q2 c) Q1 < Q2

3) How does the electric field between the plates of C2 change as separation between the plates is increased ? The electric field:

a) increases b) decreases c) doesn’t change

Preflight 8:

Two identical parallel plate capacitors are connected to a battery, as shown in the figure. C1 is then disconnected from the battery, and the separation between the plates of both caps is doubled.

5) What is the relation between the voltages on the two capacitors?

a) V1 > V2 b) V1 = V2 c) V1 < V2

Preflight 8:

Lecture 8, ACT 2

d dV 2d 2dVC1C2

• Two identical parallel plate capacitors are connected to a battery.

– What is the relation between the U1, the energy stored in C1, and the U2, energy stored in C2?

– C1 is then disconnected from the battery and the separation between the plates of both capacitors is doubled.

(a) U1 < U2 (b) U1 = U2 (c) U1 > U2

Lecture 8, ACT 2

d dV 2d 2dVC1C2

• Two identical parallel plate capacitors are connected to a battery.

– What is the relation between the U1, the energy stored in C1, and the U2, energy stored in C2?

– C1 is then disconnected from the battery and the separation between the plates of both capacitors is doubled.

(a) U1 < U2 (b) U1 = U2 (c) U1 > U2

• What is the difference between the final states of the two capacitors?• The charge on C1 has not changed. • The voltage on C2 has not changed.

• The energy stored in C1 has definitely increased since work must be done to separate the plates with fixed charge, they attract each other.• The energy in C2 will actually decrease since charge must leave in order to reduce the electric field so that the potential remains the same.

21 CC Initially: 02

02201

20

1 2

1

2

12

2

1UVCUU

C

QU Later:

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

Think of the energy needed to charge the capacitor as being the energy needed to create the field.

• The electric field is given by:

A

QE

00

20

1

2U E Ad

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

20

1

2

U Uu E

volume Ad

3m

JUnits:

This is the energy density, u, of the electric field….

• To calculate the energy density in the field, first consider the constant field generated by a parallel plate capacitor, where

2 2

0

1 1

2 2 ( / )

Q QU

C A d ++++++++ +++++++

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

+Q

Energy Density

• Example (and another exercise for the student!)

– Consider E- field between surfaces of cylindrical capacitor:

– Calculate the energy in the field of the capacitor by integrating the above energy density over the volume of the space between cylinders.

is general and is not restricted to the special case of the constant field in a parallel plate capacitor.

Claim: the expression for the energy density of the electrostatic field

202

1Eu

2

2

1CVW

2 20 0

1 1.

2 2U E dV E r dr dl etc

– Compare this value with what you expect from the general expression:

3

Lecture 8, ACT 3

• Consider two cylindrical capacitors, each of length L.

– C1 has inner radius 1 cm and outer radius 1.1cm.– C2 has inner radius 1 cm and outer radius 1.2cm.

(a) U2 < U1 (b) U2 = U1 (c) U2 > U1

If both capacitors are given the same amount of charge, what is the relation between U1, the energy stored in C1, and U2, the energy stored in C2? 1

C2

1.2

1

C1

1.1

Lecture 8, ACT 3

• Consider two cylindrical capacitors, each of length L.

– C1 has inner radius 1 cm and outer radius 1.1cm.– C2 has inner radius 1 cm and outer radius 1.2cm.

(a) U2 < U1 (b) U2 = U1 (c) U2 > U1

If both capacitors are given the same amount of charge, what is the relation between U1, the energy stored in C1, and U2, the energy stored in C2? 1

C2

1.2

1

C1

1.1

The magnitude of the electric field from r = 1 to 1.1 cm is the same for C1 and C2. But C2 also has electric energy density in the volume 1.1 to 1.2 cm. In formulas:

inner

outer

o

rr

LC

ln

2

1

1~

1.1ln

1

C

2

1~

1.2ln

1

C 2

2 2 12

1 1 2

/ 2 1.2ln( )

/ 2 1.1

U Q C C

U Q C C

Dielectrics• Empirical observation:

Inserting a non-conducting material between the plates of a capacitor changes the VALUE of the capacitance.

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

0C

C

– values are always > 1 (e.g., glass = 5.6; water = 78)

– They INCREASE the capacitance of a capacitor (generally good, since it is hard to make “big” capacitors)

– They permit more energy to be stored on a given capacitor than otherwise with vacuum (i.e., air):

220

2 2

C VCVU U

Parallel Plate Example• Deposit a charge Q on parallel plates

filled with vacuum (air)—capacitance C0.

• Disconnect from battery

• The potential difference is V0 = Q / C0.

• Now insert material with dielectric constant .

– Charge Q remains constant

– Capacitance increases C = C0

– Voltage decreases from V0 to:

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

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

E0V0

0

0

VQ QV

C C

0 0V EVE

d d

– Electric field decreases also:

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

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

V E

+

- +

-

+

-+

-

+

-

+

-+

-

Note: The field only decreases when the charge is held constant!

Parallel Plate Example, assuming Q constant• HOW ABOUT GAUSS' LAW?

– How can field decrease if charge remains the same?

– Answer: the dielectric becomes polarized in the presence of the field due to Q. • The molecules partially align with

the field so that their negative charge tends toward the positive plate.

• The field due to this rearrangement inside the dielectric (a la dipole) opposes the original field and is therefore responsible for the reduction in the effective field.

• Some of the original field lines now end on charges in the dielectric

κ

VV 0

0E

E

Q

Q

E

V0

V

E0

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

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

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

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

++

-

+

- +

-

+

- +

-

+

-

+

-+

-+

-

4

modify Gauss Law, see Appendix

Two identical parallel plate capacitors are given the same charge Q, after which they are disconnected from the battery. After C2 has been charged and disconnected it is filled with a dielectric.

7) Compare the voltages of the two capacitors.

a) V1 > V2 b) V1 = V2 c) V1 < V2

8) Compare the electric fields between the plates of bothcapacitors.

a) E1 > E2 b) E1 = E2 c) E1 < E2

Preflight 8:

10) When we insert the dielectric into the capacitor C2 we do:

a) positive work

Preflight 8:

Recall the meaning of negative work the energy of the system is reduced. The dielectric is sucked into the capacitor. When the charge is constant, the total energy of the capacitor decreases because the presence of the dielectric increases the capacitance . It turns out that the dielectric is pulled in even if the voltage is held constant, e.g., via a battery. On a microscopic scale the force on the dielectric arises due to “fringing fields” at the edges of the capacitor.

b) negative work c) no work

Lecture 8, ACT 4• Two parallel plate capacitors are identical

(same A, same d) except that C1 has half of the space between the plates filled with a material of dielectric constant as shown.

C1

C2

E1=?

E2=?

+Q

+Q

-Q

-Q

– If both capacitors are given the same amount of charge Q, what is the relation between E1, the electric field in the air of C1, and E2, the electric field in the air of C2

(a) E1 < E2 (b) E1 = E2 (c) E1 > E2

Lecture 8, ACT 4• Two parallel plate capacitors are identical

(same A, same d) except that C1 has half of the space between the plates filled with a material of dielectric constant as shown.

C1

C2

E1=?

E2=?

+Q

+Q

-Q

-Q

– If both capacitors are given the same amount of charge Q, what is the relation between E1, the electric field in the air of C1, and E2, the electric field in the air of C2

(a) E1 < E2 (b) E1 = E2 (c) E1 > E2

• The key here is to realize that the electric field in the air in C1 must be equal to the electric field in the dielectric in C1!

• WHY?• The top plane is a conductor; therefore it is an equipotential surface.• The bottom plane is a conductor; therefore it is also an equipotential surface.• For this to happen, the charge density on each plane must be non-uniform to create equal electric fields!!

• Since C1 > C2, for the same charge, V1 < V2.• Consequently, E1 < E2. NOTE: If the voltage were held constant, E would NOT have changed!!

Why Salt Dissolves

Normally NaCl is in a rigid crystal structure, maintained by the electrostatic attraction between the Na+ and Cl- ions.

Water has a very high dielectric constant (78). This reduces the field between the atoms, hence their attraction to each other. The crystal lattice comes apart and dissolves.

Note: This should leave lots of ions in the water

Summary• Capacitors in parallel and in series• energy stored in a charged capacitor

– work done to assemble charges on plates

• energy stored in electric field– capacitor example– field energy density J/m3

• electric response of dielectric materials– atomistic view: polarized atoms -> dielectric constant– modified Gauss’ Law– dielectric constant

C

QCVW

22

2

1

2

1

enclosedo d qES

21 1

2 2E E

0 0u E

Replace with !oo

Modified Gauss’ Law(in presence of dielectric)

Reinterpret if in the presence of dielectric:

This form of Gauss' Law can be used in vacuum ordielectric alike where q represents only the "free" charge.

- free charge is the charge you can move around

qE Sd

0

In vacuum, Gauss' Law is: qSE

d00

In a dielectric, the field is reduced:

0EE

Rewrite Gauss' Law: qSE

d

0

0

Replace with !oo