al capacitor p.34. a capacitor is an electrical device for storing electric charge and energy. -...
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AL Capacitor
P.34
P.34
P.34
P.34
Capacitor
Resistor
P.34
A capacitor is an electrical device for storing electric charge and energy.
- consists of two parallel metal plates with an insulator (dielectric) between the plates
Dielectric – air, paper, wax, ceramic, mica
Capacitor
P.34
Air capacitor Disc ceramic capacitor Film capacitor
Dipped mica capacitor Metallized paper c
apacitor
Electrolytic capacitor
Types of capacitors
P.34
fixed capacitor
variable capacitor
Types of capacitors
P.34Charging
+++
---
Vx
Q σ E=σ/ε V=Ed
P.35Discharging
+++
---
Vx
Q σ E=σ/ε V=Ed
P.35
ε/R
-ε/R
P.35Capacitance C
Q σ E=σ/ε V=Ed
VQ
V
QC Unit : Farad (F)
a
QV
4
1For a point charge,a
V
QC 4
The capacitance of a capacitor is one farad (1 F), if the capacitor stores one coulomb (1 C) of charge when there is a potential difference of one volt (1 V) applied across it.
P.36 Experiment
+++
- - -
Rmax I= V/R
Set R to max I is min.
Reduce R Keep I constant Vc
VR
ε = VC + VR
P.37 If Q is constant,
++++++++++
----------
V
QC
A +Q -Q B
A
Q
d
VE E is constant,
If d is reduced, V is reduced
C is increased
V
QC
If A is increased, E is reduced
C is increased
V is reduced
d
A
V
QC
P.37
Increase C by:(1) increasing overlapping area (A)(2) placing two plates closer (d)(3) replacing dielectric with higher permittivity ()
P.38 Dielectric material
External E field
E field by dipole
Resultant E field
For all insulators, r > 1.0dielectric increases capacitance (charge-storing ability)
- molecules of dielectric are polarized- one end of each molecule has excess +ve charge- other end has excess –ve charge- charges appear on surface- potential difference is induced- By C = Q/V, C increases
P.38 Dielectric material
V+V-
VC = V+ - V-More and more charges are needed to increase VC to ε
Qd
A
V
QC
P.38
Material Relative permittivity (r)
Vacuum 1.0000
Air 1.0005
Oil 2 – 5
Paper 2 – 6
Glass 8
Ceramic 80 – 1 200
Capacitance of capacitors with different materials between plates
Conducting material
External E field
E field by induced charges
Zero resultant E field
P.38
P.39
CVQ
)(1025.0
10100 46
F
dtdVdtdQ
C
P.39
d
VE
)/(108101
800 53
mVE
P.39
d
AC o
Cd
A
d
A
d
AC ooo 22
2/''
VV '
a.
b.
c. QVCVCQ 2))(2('''
d. EEoo
22'
'
P.39
V
QC
+++
+++
A +? +? B
++++
++
+900μ +100μ
Q=400μ)(100
4
400F
P.39
''
d
AC
C
d
A
d
A
6
5
6
5
56
d
AC
C
QV
P.40
Q is constant Overlapping area is decreased
C is decreased
V is increased
P.40
Q is constantd
AC
d is decreased C is increased
C
QV V is decreased
A
QE E remains unchanged
P.40
VC is constant d
AC
d is increased
C is decreased CVQ Q is decreased
A
QE
E is decreased
VP is decreased
W is decreased
P.41 Use of reed switch for measuring capacitance
Charging
+-
Discharging
Vibrating switch- vibrates between A and B at frequency f- at A, capacitor is charged Q = CV- at B, capacitor discharges
A B
fV
ICfCV
t
QI ,
P.41 Use of reed switch for measuring capacitance
Charging
Discharging
VC
Protective resistor ->reduce I
Mean VC
Mean IC
P.42
Gradient)(1
f
fV
I
V
QC
If C = 2 F , V = 6 V and the average current I = 0.5 mA, what should be the frequency f ?
CV
If
6102
105.06
3
)(7.41 Hz
Capacitance exists in (1) an insulated conductor and metal framework(2) two conductors in an electric cable(3) the turns of a coil of wire
electric field lines leak to conductors- gives rise to stray capacitance
C=Co+Cs
Determination of capacitance by reed switch (Stray capacitance)
P.42
P.43
d
AC
3
12
101
)1085.8)(5)(4.0(
CVfI )(10062.1)50)(120)(1077.1( 48 A
)(1077.1 8 F
P.43
d
AC
)1085.8)(5.2)(1040(
)10105.1)(102(123
326
d
wl
w
Cdl )(33 m
P.44
CRO measures VC + VR = E during charging
Reading of CRO is constant during charging
CRO measures VC + VR = 0 during discharging
Reading of CRO is zero during discharging
P.44Parallel Connection
21 QQQ
V is the same
V +Q C
- Q
VCVCCV 21
21 CCC
If C1 < C2
then Q1 < Q2
P.44Parallel Connection
321 QQQQ
V is the same
V
+Q - Q
C
VCVCVCCV 321
321 CCCC
QCCC
CQ
321
11
P.44Series Connection
21 VVV
Q is the same
21 C
Q
C
Q
C
Q
V
-Q + Q
C
21
111
CCC
21
21
CC
CCC
V
CCCC
CV
21
21
11
1
V
CC
C
21
2
P.44Series Connection
321 VVVV
Q is the same
321 C
Q
C
Q
C
Q
C
Q
V
-Q + Q
C321
1111
CCCC
P.45
)102())105.0()105.0((
)102))(105.0()105.0((666
666
C
6103
2 C
)(1046103
2 66 CCVQ
)(104 62 CQQ
)(1022
1 65.0 CQQ
2
2
C
C
V
V )(2)6(
102
1032
6
6
22 VV
C
CV
5.0
5.0
C
C
V
V
)(4)6(101
1032
6
6
5.05.0 VV
C
CV
P.45
)(103301010 46)(10 CCVQ i
What is the connection ?
V is the same Parallel connection
)(50)(10)(10 ffi QQQ
VCVCQ i 5010)(10
)(510501010
10366
4
5010
)(10 VCC
QV i
P.45
)(108.1200109 36)(9 CCVQ i
V is the same
)(3)(9)(9 ffi QQQ
VCVCQ i 39)(9
)(1035.1103109
108.1)109( 366
36
)(939
9)(9 CQ
CC
CQ if
)(939
9)(9 i
n
nf QCC
CQ
n
12
95.0 41.2n 3
P.45
CVQQ oldp )(
What is the type of connection ?Apoldp QQQQ )(
Parallel connection
A B
+ - + -
+ -
QA QB
QP
Bpoldp QQQQ )(
BA QQ 0
BA QQ
BA VV
BAp VVV
BAp QQQ
Ap QQ 2
QQp 3
2
P.45
A. Yes. Charges cannot be destroyed or created
B. Yes. Voltage will be the same. By the equation. Q = C V, comparing the capacitance, C2 > C1, then Q2 > Q1
C. Yes. C1 shares charges with C2, so V1 decreases as Q1 decreases.
D. Yes. Both of them are fully charged. No p.d. between them, so V1 = V2
E. No. There is charge flowing through R. P=I2R , there is energy loss.
P.46
)(42222 FQ
Q22 and Q8 are the same
822 QQ
outVCVC 82222
outVV 622
6 108104
outVV 222
outVV 2245
outout VV 245
)(15 VVout
P.46
C
QQVCVE
22
2
1
2
1
2
1
Q=I t, x-axis relates to Q
2QE
P.46
Connected to X
212
C
CC
CCC
1 and 2 will be charging up
1
2 3
4
Connected in series
Charge in C12 CC
Q 3)6(212
Connected to Y 3 and 4 will be charging up 1 will be discharging
234
C
CC
CCC
341 VV
34
341
C
Q
C
Q 341 2QQ
By conservation of charges 3413 QQC CQ 21
1Q1Q
)(21 VV
21 QQ
P.46
212
C
CC
CCC
1 2
3
CCC
C2
3
2123
P.47
Connected in series
3
10
105
10510,5
C 20
3
106105
QQ )(2
10
2010 VVVp
)(1651
11 V
kk
kVV kQ
Switch K is on )(1 VVp
1010 nQ 25555 nQ
5μF 10μF
25205 Q
102010 Q
+5μ +10μ
P.47
VC/2(1) , V2C(2) and E are the same when steady state is obtained.
2
C
2
R
R2
C2
22
)1(2 4
1
22
1CEE
CEc
22
)2(2 4
142
2
1CEECE C
4
1
)2(2
)1(2 C
C
E
E
Whole circuit is independent of R when steady state is obtained.
P.47
(1) Yes. By E = (1/2) Q V
(2) Yes. By W = (1/2) F e
(3) No. By W = P V
P.48
P.48
P.49
P.48
P.48
P.50Charging mechanism of capacitors
(1) electrons from –ve terminal of battery accumulate on one plate of capacitor(2) equal amount of +ve charges induced on opposite plate(3) until p.d. across the capacitor = e.m.f. of battery
Charge stored in capacitor
P.50Charging at constant rate
+++
---
R set to maximum value
R I=V/R Q=I t = C V t = R C
Keep I constant
P.50Charging at varying current
+++
---
R is not fixed, I will be changed
RCQ
V
R
VV
R
V
dt
dQI CR
00
QCVdt
dQRC 0
Qt
QCV
dQdt
RC 00
0
1
QQCVtRC 00ln1
0
0ln1
CV
QCVt
RC
QCVeCVt
RC
0
1
0
tRC
tRC eQeCVQ
1
0
1
0 11
P.51 VR
tRCeVV1
0 1
P.51Discharging at varying current
+++
---
RC
Q
R
V
R
V
dt
dQI CR
dt
dQRCQ
Q
Q
t
Q
dQdt
RC 00
1
QQQt
RC 0ln
1
tRCeQQ1
0
tRCeVV1
0
P.51
tRCeQ
RCdt
dQI
1
0
1
tRC
tRC eIe
R
VI
1
0
10
P.52Time constants τ
Charging
Discharging
tRCeQQ1
0 1
t
eQ 10
If t=τ, then
10 1 eQQ 0632.0 Q
tRCeQQ1
0
t
eQ
0
If t=τ, then1
0 eQQ 0368.0 Q
If Q=0.5Q0, then
21
005.0
t
eQQ
21
2lnt
2ln21 RCt
P.52
e.m.f. of battery (E)= p.d. across R + p.d. across C= VR + V = IR + Q/C
1. Variation of I
At t = 0:00 RI
C
QIRE
R
EI 0
At t:CR
t
eII
0
Charging and Discharging
P.52
2. Variation of Q
CR
t
eCECVQ 1
CR
t
eQQ 10
3. Variation of V
VReIVIRE CR
t
0
CR
t
eEV 1
Charging and Discharging
P.52Energy of a charged capacitor
C
Qdq
C
qdWW
Q2
0 2
1
QVCVC
QW
2
1
2
1
2
1 22
- Charge on each plate (q) = CV- Suppose a charge of +dq is moved from –ve plate to +ve plate:
dqC
qVdqdW
P.52Energy of a charged capacitor
Difference in energy
If capacitor is charged by a battery of e.m.f. (E)- energy stored in capacitor = ½ QE
QEQEQE2
1
2
1
energyin Difference
The energy escapes in the form of HEAT in the connecting wires and EM WAVEs are emitted
P.53Energy changes (Constant V)
V is the same
V +Q C
- Q
V = constant = E d
A
QE
QVCVC
QW
2
1
2
1
2
1 22
Energy returns to battery
P.53Energy changes (isolated capacitor)
Q is the same
V +Q C
- Q
Q = constant
A
QE
QVCVC
QW
2
1
2
1
2
1 22
Energy supplied by battery to overcome the attractive force between plates
= constant
V = E d
P.53Energy changes (inserting dielectric material with constant V)
V is the same
V +Q C
- Q
V = constant = E d
A
QE
QVCVC
QW
2
1
2
1
2
1 22
Energy supplied by battery to accumulate more charges inside the capacitor
P.53Energy changes (inserting dielectric material into isolated capacitor)
Q is the same
V +Q C
- Q
Q = constant
A
QE
QVCVC
QW
2
1
2
1
2
1 22
Energy returns to battery because +ve work has been done by electric force to attract the dielectric material into the capacitor.
V = E d
P.54
C
QE
2
2
1
C should be constant for an isolating conducting sphere
P.54
QA and QB are the same
r
QV
04
1
rA < rB
VA > VB
QVE2
1 EA > EB
V
QC CA < CB
When they are connecting by wire, there is p.d.. The charges flow from high potential (A) to low potential (B)
QA < QB
P.54Energy loss on joining capacitors
2
22
1
21
2
1
2
1
C
Q
C
QEi
21
221
2
2
1
2
1
CC
C
QE f
21
221
2
22
1
21
2
1
2
1
2
1
CC
C
Q
C
QE f
2121
22112
2
1
CCCC
QCQC
0
P.55Energy loss on joining capacitors
0 fE
2121
22112
2
1
CCCC
QCQC
02112 QCQC
2
2
1
1
C
Q
C
Q
21 VV
No movement of charges, so no energy loss
P.55Types of capacitors
Series or parallel connection ?
Parallel connection
Greater total capacitance
P.56Electrolytic capacitors
Greater capacitance
Polarity of capacitor is fixed, for d.c. only
P.56Variable capacitors
Varying capacitance by
changing the overlapping area
P.57Spooning charge
Extra high internal resistance
P.57As voltmeter
No other connection
If the meter is full scale deflection (f.s.d.), then V=1 (V)
P.57As ammeter
R= 1010 F V= 1V
If the meter is full scale deflection (f.s.d.), then I = V/R = 10-10 A
If V=0.5 V, then I = V/R = 0.5 x 10-10 A
P.57As meter to measure charge
C= 10-8 F V= 1V
If the meter is full scale deflection (f.s.d.), then Q = CV = 10-8 C
If V=0.5 V, then Q = CV = 0.5 x 10-8 C
P.59Spooning charge
For each spoon, a certain amount of +ve charge will be transfer to electrometer, the corresponding reading will be shown on the voltmeter
Different size of spoon will have different amount of +ve charge carried.
Different size of p.d. of EHT will have different amount of +ve charge carried.
P.59
P.d. across voltmeter equals to p.d. across capacitor
VCd
AQ V
Charges are isolated
22
11 VC
d
AVC
d
AQ VV
25102
)1.0(7.4100
102
)1.0(3
20
3
20
VV CC
VV CC 320
320 102)1.0(7.4108)1.0(4
20
3 )1.0(7.0106 VC
1110244.3 VC
P.60
Before K is closed
Capacitors are fully charged
Capacitors are connected in series
Q1 : Q2 = 80μ : 80μ
C1 : C2 = 10μ : 20 μ
V1 : V2 = 2 : 1 = 8 : 4
VR1 : VR2 = 4 : 8
P T
After K is closed
VP = 8 , VT = 4
VP = 8 = VT
Q2f = (20 μ)(8) = 160 μ
Q1f = (10 μ)(4) = 40 μ
80 μ ->40 μ
80 μ ->160 μ
80μ
40μ
120μ
Charge flowing through K is 120 μC
P.60
Initially, C is fully charged
When P breaks, C starts to discharge
When Q breaks, discharging stops
VC = 12 (V)
VC drops
VC = 3.2(V)
P.60
When P breaks, C starts to discharge
)7.4)(21(122.3 kk
t
e
)(01864.0 st
t
dv
01864.0
1v
)/(64.53 smv
P.60
QVCVC
QW
2
1
2
1
2
1 22
Half of W is dissipated
2
02
1
Q
Q
02
1QQ
)1)(2(0
M
t
eQQ
2
2
1 t
e
)(693.0 st
P.60
2
1
22
11 CR
CR
1
2
2
1 R
R
2
1
1
2
2
1 C
C
4
1
2
1 C
C
P.60
RC
t
e1
1106
))6.0ln()4.0(ln(221 RCtt
Charging to 6V by 10V battery
RC
t
e2
1104
Charging to 4V by 10V battery
RC
t
e1
4.0
RC
t14.0ln
4.0ln1 RCt
RC
t
e2
6.0
RC
t26.0ln
6.0ln2 RCt
)(9.4 sRC
P.61
RC
t
eVV 10
)2)(1(
5.0
15 MeV
)(106.1 VV
Charging for 0.5s by 5V battery
Charging for t s to 5V by 10V battery
)2)(1(
2
1105 M
t
e
)(386.12 st Charging for t s to 1.106V by 10V battery
)2)(1(
1
110106.1 M
t
e
)(2344.01 st )(1516.12344.0386.112 sttt
P.61
Slope = VAC / t = Q / (Ct) = I / C
Slope = const => I = const => VR = const
Charging with const. I
Fully charged, I=0
Discharging with const. I in opposite direction
P.61
(1) C = Q / V C = (1 C) / (1 V) = 1(F)
Yes
(2) E = (1/2) Q2 /C = (1/2) (J)
No
(3) Q = I t = (1 A) (1 s) = 1 (C)
No
P.62
RC
t
eVV
0
)47(
8.7
82 Re
)47(
8.7
25.0 Re
)47(
8.7)25.0ln(
R
)25.0ln()47(
8.7
R
kR 7.119
P.62
(1) Q = C V = (100000μ) (20) = 2(C)
Yes
(2) Time const RC = (10) (100000 μ) = 1(s)
No
(3) E = (1/2) CV2 = (1/2) (0.1)(20)2 = 20 (J)
No
mean I= Q / t = 2 / 2000 = 1 (mA)
After 1s, only 63% of initial charge has discharged
P.62
Time constant t = RC
2 x 10-3 = (1 x 103) C
C = (2 x 10-6) F
C = 2 μ F
P.62
I1 = V /R
No
NoNo
I2 = V /(2R) = (1/2) I1
V1 = V2
E dissipated in R independent of R
E dissipated in R = E in Cap = (1/2) CV2
Total charge Q stored depends on V and C only
RC
t
eQQ
0
RC
t
eQQ1
01
CR
t
eQQ )2(02
2
2ln1 RCt 2ln22 RCt
Yes
P.62
P.63
P.63
P.63
P.63
P.63
P.64
P.64
P.64
P.65
P.65
P.65
P.65
P.65
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P.65
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