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Transcript
EXAM 1, September 22, Thursday
Ch.21-23, Lectures 1-6.
1. Finding E(x,y,z) or E(r) using

N
 kq 

kQ
r
i
a) E of a point charge E  2 r0 and Etotal   2 0i
r i
i 1
superposition principle, r
  Qencl
qencl
E  dA 
E
b) or Gauss’s law

0
0 A
 surface
 

c) or E  i Ex  j E y  k Ez
Ex  
V
V
V
, Ey  
, Ez  
x
y 
z
2. Finding a force a)FQq  qEQ

N

b) or F  kQi r0i q
total

i 1
ri
3. Finding potential using


a) V  V   E  dl
kq
b) or potential of a point charge V (r )  r
N
and superposition principle Vtotal (r )   kQi
i 1 r i
4) Finding the work and potential energy
b
a
b
a


 q  E  dl  q(Va  Vb )  U a  U b  U
b
Wab
a
Preparation for Exam
1. Look in advance at the formula sheet on the
website of the course
2. Problems worked out in Lecture notes
3. Previous midterm exams on my website
4. WebCT and assigned homework
At the Exam
• You’ll get 3 problems+ formula sheet
• Ask me the questions if you do not
understand the formulation of the problem
• Start with the figure
• Get a general (algebraic) solution
• Indicate the units
• Get numerical results if required
(bring your calculator)
Lectures 7 (Ch. 24)
Capacitance and capacitors
1. Definitions and functions
2. Types of capacitors
3. Effective capacitance
(capacitors in series and parallel)
4. Dielectrics
5. Energy stored in capacitor
Capacitor is a device to store the charge and energy.
It consists of two conductors (+Q,-Q), separated by a dielectric.
Store Q and V in order
1. To use it on demand
(camera flash, energy back
up,…)
2. To block surges of Q and V
protecting sensitive
devices
3. Part of tuner of a radio,
selecting specific
frequency of radio waves
4. Binary code of a memory
Q
does not depend neither on Q no on V.
Capacitance: C 
V
It is defined by the geometrical sizes and by a dielctric inside
Types of capacitors
1. II plate capacitor
Q
C
; V  Ed ;
V

Q
E

0
A 0
A 0
Qd
V 
;C
A 0
d
Spherical capacitor
Q
C
; Vab
Vab
 
  E dr ;
b
a
b
a
b kQ
 kQ 
1 1
ba
E  2 r0 ; Vab   2 dr  kQ(  )  kQ(
)
r
a b
ab
a r
40 ab  0 A
C

ba
d
A
4a 2 4b 2  4ab; d  b  a
Cylindrical capacitor
b 

Q
C
; Vab   E dr ;
Vab
a
a
b
b 2 k

2k 
b
E
r0 ; Vab  
dr  2k ln
r
r
a
a
20 L
Q
 ;C
b
L
ln
a
In the case d  a, b ( where d  b  a )
b
d
d
ln
 ln( 1  ) 

a
a
a
0 A
C
; A  2aL
d
SI unit of C
[C]=[Q]/[V]=1C/1V=1F(farad)
Symbol:
Typically C~1pF-1

F
d  1mm
A=?
Michael Faradey
(1791-1867)
!
Supercapacitors: C~1F-1kF
1. Using porous conductors (carbon) to
increase an effective A
EL
2. Electrolyte (EL) as 1 of 2 conductors
oxide layer with an effective d~1nm
d
EL
EL
Commercial capacitors
Capacitors in parallel: C=C1+C2
Q Q1  Q2
C 
 C1  C2
V
V
Capacitors in series:
1
1
1


C C1 C2
1 V V1  V2 1
1
 


C Q
Q
C1 C2
Capacitors networks
Example 1.
Example 2. You have 3 capacitors with capacitances 5 nF, 2nF and 2nF.
Your circuit needs 6 nF. What do you do?
Capacitors networks
Example 1.
Example 2. You have 3 capacitors with capacitances 5 nF, 2nF and 2nF.
Your circuit needs 6 nF. What do you do?
2nF
5 nF
2nF
Dielectrics
different from vacuum or air are
typically used in capacitors
•
to increase the capacitance: C=KC0, K>1
• to improve a mechanical stability (avoid touching between
conducting plates)
• to increase the dielectric strength compared to air (for air:
Ecr=3MV/m)
Polarization of dielectric in E

Suppose capacitor was charged, σ is the surface charge density, E0 
0
Then a battery was removed and a dielectric was
inserted. Polarization of dielectric led to induced charge
on the plates,σi, reducing E between plates:
 i E  i
i 1
E
;

 1   1
0
E0

 K
K≥1 is a dielectric constant, ε=Kε0 is a dielectric permittivity
E0
V0


E

 ; V  Ed  V  ;
K K 0 
K
i
1
1
 1    i   (1  )

K
K
A 0 K A
Q Q
C   K  C0 K 

V V0
d
d
Suppose capacitor is connected
to a battery with voltage, V.
Then a dielectric is inserted.
Q0 Q
Q
V
 

C0 C KC0
v
Q  KQ0 ; Q  Q0  Q0 ( K  1)
Qi
1
1
 1   Qi  Q(1  )  Q  Q0
Q
K
K
Battery sup plies an additional ch arg e
to compensate for Qi and to keepV  const 
V
  0K  0
E   const  

d
 0K 0
!
For charges interacting in a
dielectric medium
Coulomb’s Law

1 Qq 
Fe 
r
2 o
4 r
  Qencl
 E  dA  
Gauss’s Law
Q
q
Potentical energy stored in the capacitor
q
dW  dqV , V 
C
V
dq
Q
E
-q
q
q
Q2
U  W   dq 
C
2C
0
2
2
Q
CV
QV
Q  CV ; U 


2C
2
2
Density of Energy
2
2 2
U
CV
AE d
U
E
u




Volume Ad 2 Ad
d 2 Ad
2
2
U can be viewed as energy stored in E inside capacitor
Insert dielectric, C=KC0
2
2
Q
CV
QV
U


2C
2
2
Does U ↑or↓?
1. Charged Capacitor. Battery is removed. Then a dielectric is inserted.
Q
-Q
+
-
F
-
dl
F
+
2
2
U0
Q
Q
Q  const ; U 


 U0
2C 2 KC0 K
2
2
V0
KC0V0
U0
CV
or V  ; U 


 U0
2
K
2
2K
K
final

U 0  U  W   Fel dl  0
0
2. Capacitor is connected to battery with V=const. Then
a dielectric is inserted
2
2
KC0V
CV
V  const ; U 

 KU 0  U 0
2
2
or Q  KQ0 (Q  VC  VC0 K  Q0 K )
2
V
2
2
K Q0
Q
U

 KU 0  U 0
2C 2 KC0
Energy is supplied by the battery, increasing the charge on the
plates and keeping V=const
Find
1)Charge and energy of
the 1st capacitor before the
switch was closed.
2) Charge of each
capacitor, voltage and total
energy after closing the
switch .
Q0V0
 0.058 J
2
Q  Q1
Q
Q Q
1
1
V 1 2  0
 Q1 (  )  0
C1 C2
C2
C1 C1
C2
Q0  V0C1  960C ; U initial 
Q1 
Q0C1
2
 Q0  640C
C1  C2 3
Q2 
Q0C2
1
 Q0  320C
C1  C2 3
Q1V Q2V Q0V
Q0
V
 80V ;U final 


 0.038 J
C1  C2
2
2
2
Example 24.15
Data: C1= C2 = C3 = C4 =C=4ϻF; Vab =28V
,
, ,
Find: Q1 Q2 , Q3,Q4 V1 V2 , V3,V4
C12  C / 2, C123  3C / 2,
1
C1234
1
1
3C 12F

 , C1234 

 2.4F
C123 C4
5
5
Q4  28VX 2.4F  67.2C ;
Q4 67.2C
V4 

 16.8V
C4
4F
V3  28V  16.8V  11.2V , Q3  11.2VX 4F  44.8C
V3
V1  V2   5.6V
2
Q1  Q2  5.6VX 4F  22.4C