Capacitors & Energy Storage

Lecture 7 Capacitors & Energy Storage

• Capacitors: what they are and how to calculate their capacitance

(Finish electric potential and gradient)

Application: flash device for a

camera (stores charge and energy)

Charge up a flash device in a camera and then short out the two leads in the 18.5 µ F cap

CAPACITOR• A capacitor is device formed with two or

more separated conductors that store charge

and electric energy.

• Consider any two conductors and we put

+Q on a and –Q on b. Conductor a has constant

Va and conductor b has constant Vb , then

• The electric field is proportional to the charges ±Q. If we double the

charges ±Q, the electric field doubles. Then the voltage difference is

Va-Vb proportional to the charge. This proportionality constant

depends on size, shape and separation of the conductors.

ldEVVbr

arba

rrr

r⋅∫=−

( )ba VVconstQ −×=

Y&F fig. 24.1

• If we call this constant, Capacitance, C,

and the voltage difference, V= Va-Vb, then,

• Capacitance, depends on the geometry of the

two conductors (size, shape, separation) and capacitance is always

a positive quantity by its definition (voltage difference and charge

of + conductor) [N.B. C does not depend on voltage or charge]

• UNITs of capacitance, Coulomb/Volts or Farads, after Michael Faraday

V

QC =Or CVQ =

CAPACITOR, continued

Example: Parallel Plate Capacitor

• Calculate the capacitance. We assume +σσσσ, -σσσσ charge densities on each plate with potential difference V:

d

A

- - - - -

+ + + +

V

QC ≡

• Need Q:

• Need V: from def’n:

– Use Gauss’ Law to find E

AQ σ=

∫ •−=−b

a

ab ldEVVrr

CLICKER QUESTION

What is the electric potential difference, Va-Vb ?

dEqd

dEqc

dEb

dEa

r

r

r

r

0

0

)

)

)

)

−

+

−

+

Test charge moves to lower potential

Recall: Two Infinite Sheets(into screen)

• Field outside the sheets is zero

• Gaussian surface encloses zero net charge

E=0 E=0

E

σσσσ

+

+

++

+

+

+

σσσσ

+

+

+

-

-

--

-

--

-

-

-+

+

---

A

AAQ σ=

insideAESdE =•∫rr

• Field inside sheets is not zero:

• Gaussian surface encloses non-zero net charge

0εσ

=E

Example: Parallel Plate Capacitor

d

A

- - - - -

+ + + +

• Calculate the capacitance:

• Assume +Q, -Q on plates with potential difference V.

• As expected, the capacitance of this capacitor depends only on its GEOMETRY (A,d). DOES NOT DEPEND on Q or on V !!!!!

• Note that C ~ length; this will always be the case!

00 εεσ

A

QE ==

dA

QEdldEVV

b

a

ab

0ε==•−=− ∫

rr

⇒⇒⇒⇒ d

A

V

QC 0ε

=≡

Cylindrical Capacitor Example

• Calculate the capacitance:

• Assume +Q, -Q on surface of

cylinders with potential difference V.a

bL

r

• Gaussian surface is cylinder of radius r (a < r < b) and length L

⇒⇒⇒⇒Lr

QE

02πε=• Apply Gauss' Law:

0

2ε

πQ

rLESdE ==•∫rr

==−=•−= ∫∫∫ a

b

L

Qdr

rL

QEdrldEV

b

a

a

b

a

b

ln22 00 πεπε

rr

If we assume that inner cylinder has +Q, then the potential V is positive if we take the zero of potential to be defined at r = b:

⇒⇒⇒⇒

=≡

a

b

L

V

QC

ln

2 0πε

Spherical Capacitor Example• Suppose we have 2 concentric spherical shells

of radii a and b and charges +Q and –Q.

• Question: What is the capacitance?

• E between shells is same as a point charge +Q. (Gauss’s Law):

• What is the capacitance of the earth?

-Q

+Q

ab

)11

(44

4

00

2

0

ba

Q

r

Q

drr

QdrE

ldEldEVVV

b

a

b

a

b

a

r

b

a

a

b

baab

−=−=

==

•=•−=−=

∫∫

∫∫

πεπε

πε

rrrr)

11(

4 ba

Q

Q

V

QC

o

ad −=≡

πε

⇒⇒⇒⇒ab

abC

−= 04πε

24

1

r

QE

o

r πε=

Summary

• A Capacitor is an object with two spatially separated conductingsurfaces used to store charge or energy.

• The definition of the capacitance of such an object is:

V

QC ≡

• The capacitance depends on the geometry :

d

A

- - - - -

+ + + +

Parallel Plates

a

bL

r

+Q

-Q

Cylindrical

a

b

+Q

-Q

Spherical

ab

abC

−= 04πε

=

a

b

LC

ln

2 0πε

d

AC 0ε

=

CAPACITORs in parallel

• Voltage is same across each capacitor

• Total charge and voltage ratios for parallel capacitor,

ababab

V

QC

V

QC

C

Q

C

QV 2

21

12

2

1

1 ;; ====

2121 CCC

V

QQparallel

ab

+==+

Note; 2 parallel capacitors C, doubles capacitance.

To find the equivalent capacitance Ceq:

∑=i

iparallel CCFor more parallelcapacitors:

CAPACITORs in series

• If a voltage is applied across a and b, thena +Q appears on upper plate and –Q on lowerplate.

• A –Q charge is induced on lower plate of C1and a +Q charge is induced on upper plate ofC2. The total charge in circuit c is neutral.

22

11 ;

C

QV

C

QV ==

2121

C

Q

C

QVVV +=+=

21

111

CCCQ

V

series

+==

Note; 2 series capacitors C, halves capacitance.

Want to find the equivalent capacitance Ceq:

Examples:Combinations of Capacitors

C1 C2

C3

C

a

b

a b≡≡≡≡

• How do we start??

• Recognize C3 is in series with the parallel combination on C1 and C2. i.e.,

213

111

CCCC ++= ⇒⇒⇒⇒

321

213 )(

CCC

CCCC

+++

=

Two identical parallel plate

capacitors are shown in an end-

view in A) of the figure. Each

has a capacitance of C.

4) If the two are joined together as shown in B), forming a

single capacitor, what will be the final capacitance?

a) C/2 b) C c) 2C

UI7PF7: CLICKER PROBLEM

6) Which of these configurations has the lowest overall capacitance?

Three configurations are constructed using identical capacitors

a) Configuration A

b) Configuration B

c) Configuration C

CC

C

C C

Configuration AConfiguration B Configuration C

UI7PF7: CLICKER PROBLEM

UI7ACT 3: CLICKER PROBLEM

• What is the equivalent capacitance, Ceq, of the combination shown?

(a) Ceq = (3/2)C (b) Ceq = (2/3)C (c) Ceq = 3C

o

o

C CC

Ceq

• What is the equivalent capacitance, Ceq, of the combination shown?

(a) Ceq = (3/2)C (b) Ceq = (2/3)C (c) Ceq = 3C

o

o

C CC

Ceq

CCC ≡ C C1

CCC

111

1

+=2

1

CC = C

CCCeq

2

3

2=+=

UI7ACT 3

CLICKER QUESTION

Which Circuit has the LARGEST capacitance between connections

a and b? assume all capacitors have the same value.

(A) (B) (C) (D)

A circuit consists of three unequal capacitors C1, C2, and C3 which are

connected to a battery of emf ε. The capacitors obtain charges Q1 Q2,

Q3, and have voltages across their plates V1, V2, and V3. Ceq is the

equivalent capacitance of the circuit.

8) Check all of the following

that apply:

a) Q1= Q2 b) Q2= Q3 c) V2= V3 d) ε = V1

e) V1 < V2 f) Ceq > C1

Not a clicker problem

• Energy storage in capacitors

• Improving capacitors by adding dielectrics

Hints for Homework (remind me)

UI8ACT1/Clicker Problem

(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


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


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!!

CCeq2

1=

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 ===

εε ( ) 00 3

2

2/3V

C

Q

C

QV

eq

===

Energy storage in CAPACITORs

( ) 2

22

002

1

22CV

C

CV

C

Qdq

C

qdUU

QU

===== ∫∫

V

dq+q

-q

Charge capacitor by transferring bits of charge dq at a time from bottom to top plate.Can use a battery to do this. Battery does work that increases potential energy ofcapacitor.

q is magnitude of charge on plates

V= q/C V across plates

dU = V dq increase in potential energy

two ways to express the result

Question!—constant Q

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:1

1

dU U

d=C

d

dC

1

1 = Vd

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—constant V• Suppose the battery (V) is kept

attached to the capacitor.

• Again pull the plates apart from dto d1.

• Now what changes?

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

Answers: 1

1

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 ( )212

U CV=

must decrease ( , )V

ED

=0

EE

σ=

1

1

dU U

d=

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

Clicker questions:

Two parts

Clicker problem

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

UI8ACT2

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: 0

2

0220

1

2

01

2

1

2

12

2

1UVCUU

C

QU ====Later:

Where is the Energy Stored?• 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 εεσ

== ⇒⇒⇒⇒ 2

0

1

2U E Adε=

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

2

0

1

2

U Uu E

volume Adε= = =

3m

JUnits:

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

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

2 2

0

1 1

2 2 ( / )

Q QU

C A dε= =++++++++ +++++++

- - - - - - - - - - - - - -- 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 field2

02

1Eu ε=

2

2

1CVW =

2 2

0 0

1 1.

2 2U E dV E r dr dl etcε ε π= = =∫ ∫ ∫

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

U

Clicker problem

• 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

r

r

LC

ln

2πε

1

1~

1.1ln

1

C

2

1~

1.2ln

1

C = = =2

2 2 1

2

1 1 2

/ 2 1.2ln( )

/ 2 1.1

U Q C C

U Q C C

Hint: what is C for a cylindrical capacitor ?

DIELECTRICS

Y&F Figure 24.13

Consider parallel platecapacitor with vacuumseparating plates (left)

Suppose we place a materialcalled a dielectric in betweenthe plates (right)

The charge on the platesremain the same, but adielectric has a property ofhaving induced charges on its surface that REDUCEthe electric field in between and the voltage difference.

Since C = Q/V, the resulting capacitance will INCREASE.

DIELECTRICS

Suppose the charges on the plate and thedielectric are, σ and σi. The electric Fields before and after are

i

i

E

EKEE

σσσ

εσσ

εσ

−=≡

−== 0

000 ;;

We define the ratio of the original field over the new field as the dielectric constant, K.

Hence, the voltage difference changes by1/K and the capacitance, Co=Q/V, changesby C=KQ/V=K Co

C = KCo E = Eo/K V = Vo/KFor same Q:

C = KCoBut General

DIELECTRICS Materials

Glass, mica, plastics are very good dielectrics

DIELECTRICS and permittivity

Define ε, called permittivity, as

ε = Κ ε0

Consider a parallel plate capacitor with no dielectric

d

AC 00 ε=

A capacitor with a dielectric becomes simply,

d

A

d

AKKCC εε === 00

The change in capacitance can be accounted forby changing permittivity.

Y&F Problems 24.72 and 24.71

A parallel plate capacitor has two

dielectrics, side by side, find the

capacitance

221

0KK

d

AC

+= ε

A parallel plate capacitor has two

dielectrics, stacked, find the

capacitance

21

210

2

KK

KK

d

AC

+= ε

A

A

EXAMPLE

Two parallel plate capacitors, C1 = C2 = 2 µF, are connected acrossa 12 V battery in parallel.

a.) What energy is stored?

JUJCVUU T µµ 2881442

1 2

21 ====

b.) A dielectric (K = 2.5) is inserted between the plates of C2. Energy?

JUJVCU

FFKCC

T µµ

µµ

5043602

1

525.2

2'

2

'

2

2

'

2

===

=×==

Note: a dielectric increases amount of energy stored in C2.

EXAMPLE of parallel plate capacitor problem

A parallel plate capacitor is made by placing polyethylene (K = 2.3)between two sheets of aluminum foil. The area of each sheet is 400 cm2, and the thickness of the polyethylene is 0.3 mm. Find thecapacitance.

0.3 x 10-3 m

C =K εo A/d = (2.3) (8.85 x 10-12 C2/Nm2) (400 cm2)(1m2/104 cm2)

= 2.71 nF

Two identical parallel plate capacitors are connected to a battery.

Remaining connected, C2 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 charges on the plates of the capacitors.

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

UI8PF 8: TWO PART CLICKER PROBLEM

Note: Unlike constant Q case, here V and E remain the same

but C = K Co still.

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 both

capacitors.

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

CLICKER PROBLEM:


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


d

(Area A)

V0

+Q

-Q

conductor

(Area A)

V

+Q

-Q

d/3

d/3


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


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!!

CCeq2

1=

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 ===

εε ( ) 00 3

2

2/3V

C

Q

C

QV

eq

===

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Capacitors & Energy Storage