23. Electrostatic Energy & Capacitors

1. Electrostatic Energy

2. Capacitors

3. Using Capacitors

4. Energy in the Electric Field

The lifesaving jolt of a defibrillator requires a large amount of energy delivered in a short time. Where does that energy come from?

Capacitor

23.1. Electrostatic Energy

Electrostatic Energy = work done to assemble the charge configuration of a system.

Reference ( 0 energy): when all component charges are widely separated.

Bringing q1 in place takes no work.

2 2 1 2W q V rBringing in q2 takes

1 1 0 0W q V

12

qq k

a

3 3 1 3 2 3W q V V r rBringing in q3 takes

1 23

q qq k k

a a

Total electrostatic energy1 2 2U W W W 1 2 2 3 3 1

kq q q q q q

a

23.2. Capacitors

Capacitor: pair of conductors carrying equal but opposite charges.

Usage: store electrical energy

Parallel-Plate Capacitor:2 conducting plates of area A separated by a small distance d .

Plates are initially neutral.

They’re charged by connecting to a battery.

Charge transfer plates are equal but oppositely charged.

Large A, small d E 0 outside.

0

ˆinside

E zFar from the edges0

ˆQ

A z ˆinsideV d E z

0

Qd

A

Capacitance

0

QV d

AParallel-plate capacitor: 0 AQ V

d

CV

0 ACd

C = Q / V = capacitance

Parallel-plate capacitor

See Probs 41 & 42

CC

V farad F

Practical capacitor ~ F ( 106 F) or pF ( 1012 F )

0

C d

A

/F m 1dV dQ

C

Charging / Discharging

dQC

dV

Energy Stored in Capacitors

When potential difference between capacitor plates is V,

work required to move charge dQ from to + plate is

Work required to charge the capacitor from 0 to V is

0

VW C V dV 21

2CV = U = energy stored in capacitor

Note:

In a “charged” capacitor, Q is the charge on the + plate.

The total charge of the capacitor is always zero.

plate

plate

dW dQ d

E r V C dV plate platedQ V V E dr < 0

2

2

Q

C 1

2QV

Example 23.1. Parallel-Plate Capacitor

A capacitor consists of two circular metal plates of radius R = 12 cm,

separated by d = 5.0 mm. Find

(a) Its capacitance,

(b) the charge on the plates, and

(c) the stored energy when the capacitor is connected to a 12-V battery.

0

AC

d

100.8 10 F

22

9 3

12 101

4 9 10 / 5.0 10

m

Vm C m

80 pF

(a)

(b)

(c)

Q CV 80 12pF V 960 pC

21

2U CV 21

80 122

pF V 5760 pJ

5.76 nJ

CF

V

23.3. Using Capacitors

Computer memories: billions of 25 fF capacitors.

Rectifiers: mF

Fuel-cells: 102 F

220-mF electrolytic capacitor

1 F

43 pF to 2.2 mF

Practical Capacitors

Inexpensive capacitors:

Thin plastic sandwiched between aluminum foils

& rolled into cylinder.

Electrolytic capacitors (large capacitance):

Insulating layer developed by electrolysis.

Capacitors in IC circuits (small capacitance):

Alternating conductive & insulating layers.

Dielectrics

Dielectrics: insulators containing molecular dipoles but no free charges.

Dielectric layer lowers V between

capacitor plates by factor 1/ ( > 1).

0

A

d

QC

V

= dielectric constant

Molecular dipoles aligned by E0 .

Dipole fields oppose E0.Net field reduced to E = E0 / .

Hence V = V0 / .Q is unchanged, so C = C0 .

: 2 ~ 10 mostly

Working voltage V = Max safe potential < Ebkd d

Example 23.2. Which Capacitor?

A 100-F capacitor has a working voltage of 20 V,

while a 1.0-F capacitor is rated at 300 V.

Which can store more charge? More energy?

100 FQ CV 6100 10 20F V 32 10 C 2 mC

61 1 10 300FQ F V

43 10 C 0.3mC

2100

1

2FU CV 12 20

2mC V

1

2QV 20 mJ

1

10.3 300

2FU mC V 45mJ

GOT IT? 23.1.

You need to replace a capacitor with one that can store more energy.

Which will give you greater energy increase:

(a) a capacitor with twice the capacitance and same working voltage as the old one,

or

(b) a capacitor with the same capacitance and twice the working voltage?

Connecting Capacitors

Two ways to connect 2 electronic components: parallel & series

Parallel: Same V for both components

Q CV 1 2Q Q 1 2C V C V

1 2C C C

Series: Same I (Q) for both components

QV

C 1 2V V

1 2

Q Q

C C

1 2

1 1 1

C C C

1 2C C or C

1 2C C or C

Conceptual Example 23.1. Parallel & Series Capacitors

Using parallel-plate capacitors, explain why capacitance should increase with capcitors in parallel an decrease with capacitors in series.

What happens to the working voltage in each case?

Parallel-plate capacitor : 0 ACd

in parallel

A = A1 + A 2 C increases

in series

d = d1 + d 2 C decreases

Vworking = min(Vw1 ,Vw2 )

Vworking < Vw1 +Vw2

Making the Connection

You’ve got two 10-F capacitors rated at 15 V.

What are the capacitances & working voltages of their parallel & series combinations?

Parallel : 2 10 20C F F

15workingV V

Series :1

10 52

C F F

2 15 30workingV V V 1 2

1

2V V V

1 2V V V

GOT IT? 23.2.

You have 2 identical capacitors with capacitance C.

How would you connect them to get equivalent capacitances

(a) 2 C, and

(b) ½ C ?

Which combination would have the higher working voltage?

parallel

series

Example 23.2. Connecting Capacitors

Find the equivalent capacitance of the combinations shown in the Figure.

If the maximum voltage to be applied between points A and B is 100 V,

what should be the working voltage of C1 ?

23 2 3C C C 3.0 1.0F F 4.0 F

123 1 23

1 1 1

C C C

1 1

12.0 4.0F F

123 3.0C F

1

3.0 F

11

1

QV

C

1 123Q Q 123 ABC V

3.0 100F V

1

300

12.0

CV

F

25V ( min. working voltage )

300 C

Bursts of Power

San Francisco’s BART train:

KE of deceleration stored as EE in ultracapacitor.

Stored EE is used to accelerate train.

Capacitors deliver higher energy

much more quickly than batteries.

Flash light:

Battery charges capacitor,

which then discharges to give flash.

Other examples:

Defibrillator, controlled nuclear fusion, amusement park rides, hybrid cars, …

23.4. Energy in the Electric Field

Charging a capacitor rearranges charges energy stored in E

Energy density = energy per unit volume

Parallel-plate capacitor:2

2

QU

C

2

02

QAd

E

Uu

A dEnergy density :

2

202

Q

A

2

02

20

1

2E

0

E

is universal2

0

1

2Eu E 3/Eu J m

2

0

1

2 U dVE

Example 23.4. A Thunderstorm

Typical electric fields in thunderstorms average around 105 V/m.

Consider a cylindrical thundercloud with height 10 km and diameter 20 km,

and assume a uniform electric field of 1105 V/m.

Find the electric energy contained in this storm.

111.39 10 J

20

1

2Eu E

EU u V 2 25 3 39 2 2

1 110 / 10 10 10 10

2 4 9 10 /V m m m

Nm C

139 GJ

~ 1400 gallons of gasoline.

Example 23.5. A Shrinking Sphere

A sphere of radius R1 carries charge Q distributed uniformly over its surface.

How much work does it take to compress the sphere to a smaller radius R2 ?

2ˆ

Qkr

E r2

0

1

2 U dVE

1

2

22

1 1

2

R

Rk Q dr

r 1

2

22

2

14

8

R

R

Qk r dr

k r

2

2 1

1 1 1

2k Q

R R

2 10U for R R Work need be done to shrink sphere

2 1U U U 2 1

2 20

14

2 R Rr dr

E

Extra energy stored here

You’re at point P a distance a from a point charge +q.

You then place a point charge q a distance a on the opposite

side of P as shown.

What happens to

(a) the electric field strength and

(b) the electric energy density at P ?

(c) Does the total electric energy U = ∫ uE dV of the entire field

increase, decrease, or remain the same?

GOT IT? 23.3.

doubles

quadruples

decrease

Negative work done to bring in –q.