Transcript 20. Electric Charge, Force, & Field

23. Electrostatic Energy & Capacitors
1.
2.
3.
4.
Electrostatic Energy
Capacitors
Using Capacitors
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.
W1  q1 V0  0
q1
a
Bringing in q2 takes
W2  q2 V1  r2   q2 k
Bringing in q3 takes
W3  q3 V1  r3   V2  r3  
q 
 q
 q3  k 1  k 2 
a
 a
Total electrostatic energy
U  W1  W2  W2 
k
 q1q2  q2 q3  q3q1 
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.
Far from the edges
Einside  

Q
zˆ  
zˆ
0
0 A
V  Einside   d zˆ  
Q
d
0 A
Capacitance
Parallel-plate capacitor:
V
Q
d
0 A

Q
0 A
d
V
CV
C
C = Q / V = capacitance
C
0 A
d
dQ
dV
Parallel-plate capacitor
See Probs 41 & 42
C  
C
 farad  F
V
Practical capacitor ~ F ( 106 F) or pF ( 1012 F )
C d 


 0    F / m
 A 
Charging / Discharging
dV 
1
dQ
C
Energy Stored in Capacitors
When potential difference between capacitor plates is V,
work required to move charge dQ from  to + plate is
 plate
dW   dQ

E  dr  dQ V plate  V plate 
 V C dV
E  dr < 0
 plate
Work required to charge the capacitor from 0 to V is
V
W  C  V dV 
0
Note:
1
CV2
2
= U = energy stored in capacitor
Q2
1

 QV
2C
2
In a “charged” capacitor, Q is the charge on the + plate.
The total charge of the capacitor is always zero.
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.
 12 10 m 
A
1

d
4  9 109 Vm / C 5.0 103 m
2
(a)
C  0
 0.8  1010 F  80 pF
F
C
V
(b)
Q  C V  80 pF   12 V   960 pC
(c)
1
1
2
U  C V 2   80 pF   12 V   5760 pJ
2
2
 5.76 nJ
2
23.3. Using Capacitors
Computer memories: billions of 25 fF capacitors.
Rectifiers: mF
Fuel-cells: 102 F
220-mF
electrolytic
capacitor
1F
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.
Molecular dipoles
aligned by E0 .
Dielectric layer lowers V between
capacitor plates by factor 1/ ( > 1).
C
Q
A
  0
V
d
 = dielectric constant
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?
Q100F  C V
3
 100 106 F    20 V   2 10 C  2 mC
4
Q1 F  1106 F    300 V   3 10 C  0.3 mC
U100  F
1
1
1
2

Q
V

 2 mC    20 V   20 mJ
 CV
2
2
2
U 1 F 
1
 0.3 mC    300 V   45 mJ
2
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  C V  Q1  Q2  C1 V  C2 V

C  C1  C2
C  C1 or C2
Series: Same I (Q) for both components
V
Q
C

 V1  V2

Q Q

C1 C 2
1 1
1


C C1 C2
C  C1 or C2
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 :
C
0 A
d
in series
in parallel
A = A1 + A 2
 C increases
Vworking
= min(Vw1 ,Vw2 )
d = d1 + d 2
 C decreases
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 :
C  2  10  F  20  F
V1  V2  V
Series :
C

Vworking  15 V
1
 10  F  5  F
2
1
V1  V2  V
2

Vworking  2 15 V  30 V
GOT IT? 23.2.
You have 2 identical capacitors with capacitance C.
How would you connect them to get equivalent capacitances
parallel
series
(a) 2 C, and
(b) ½ C ?
Which combination would have the higher working voltage?
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 ?
C23  C2  C3  3.0  F  1.0  F
 4.0  F
1
1
1
1
1
1
 



C123 C1 C23
12.0  F 4.0  F
3.0  F
C123  3.0  F
V1 
V1 
Q1
C1
Q1  Q123  C123 VAB
 3.0  F  100 V   300 C
300 C
 25 V
12.0  F
( min. working voltage )
Bursts of Power
Capacitors deliver higher energy
much more quickly than batteries.
Flash light:
Battery charges capacitor,
which then discharges to give flash.
San Francisco’s BART train:
KE of deceleration stored as EE in ultracapacitor.
Stored EE is used to accelerate train.
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:
Energy density :
U
uE 
Ad
uE 
1
 0 E2
2
Q2
U
2C

Q2
2 0
A
d
2
1

 0 E2
2 0
2
Q2

2 0 A2
is universal
1
U  0
2

2
E dV
uE   J / m3
E

0
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 1105 V/m.
Find the electric energy contained in this storm.
uE 
1
0 E2
2
U  uE V 
2
2
 5
1
1
3
3
10
V
/
m


10

10
m

10

10
m






9
2
2 
2  4  9  10 Nm / C 
 1.39 1011 J
 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 ?
1
U  0
2

Ek
2
E dV
U  U2  U1
1
 0
2


R2
2

 Q
2

k
4

r
dr


2

R
2
8 k
 r 
1
1
 k Q2
2
R1
 1
1 
  
 R2 R1 
Work need be done to shrink sphere
U  0 for R2  R1
Extra energy stored here

R1


Q
rˆ
2
r
E 4 r 2dr
2
1
k Q2
2

R1
R2
1
dr
2
r
GOT IT? 23.3.
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
doubles
quadruples
decrease
(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?
Negative work done to bring in –q.