Post on 17-Aug-2020
Lesson 9Capacitance, Electrostatic Energy
楊尚達 Shang-Da YangInstitute of Photonics TechnologiesDepartment of Electrical EngineeringNational Tsing Hua University, Taiwan
Outline
CapacitanceElectrostatic energy
Sec. 9-1Capacitance
1. Single-conductor capacitors2. Two-conductor capacitors3. Methods to evaluate capacitance
Single-conductor capacitor-1
0=ρ0=E
v
sec10 19−≈t
0=t
Single-conductor capacitor-2
,1
∑=
=n
kkqQ ∑
=
=n
k k
k
RqV
1041πε
On the surface Inside the conductor
Spatial distribution of maintains
Single-conductor capacitor-3
,2QQ =′ VRqV
n
k k
k 224
110
==′⇒ ∑=πε
)(rsvρ
Single-conductor capacitor-4
The constant ratio of the deposited charge to the resulting potential is defined as the capacitance of a single conductor.
VQC ≡
Example 9-1: Conducting sphere (1)
Find the capacitance of a conducting sphere of radius b
b
Q Assume charge Q is deposited
Spherical symmetry, ⇒
is uniformly distributed, ⇒
)(rsvρ
⎪⎩
⎪⎨
⎧
<<
≥=
bR
bRR
QaE R
0 if ,0
if ,4 2
0πεv
v
Example 9-1: Conducting sphere (2)
The surface potential is:
b
∞
( )
Qb
Q
dRaR
Qa
bRVb
RR
∝=
⋅⎟⎟⎠
⎞⎜⎜⎝
⎛−=
=
∫∞
0
20
4
4
)(
πε
πεvv
bVQC 04πε=≡⇒
Two-conductor capacitor-1
∫ =⋅−→→±→2
1 1212 VldEEQVvvv
Ev2 1
any path
Two-conductor capacitor-2
ErrQrVVv
→±→=′ 1212
The constant ratio of the deposited charge to the voltage difference is defined as the capacitance of the conducting pair:
12VQC ≡
Comment
∞→R
Evaluation of capacitance (Method 1)
1. Assume charges are depositedQ±
2. Find by Gauss’s law orEv
∫+
′′′
=21
20 ),(
)(4
1)(SS
sR sd
rrRrarE vv
vvvv ρ
πε
3. Find by( )QV ∝12 ∫ ⋅−=2
1 12 ldEVvv
4. Find by , independent of QC12V
QC =
Evaluation of capacitance-reference figure
Ev2 1
)(2 rsvρ )(1 rs
vρ
3. Find by
Evaluation of capacitance (Method 2)
1. Assume between the conductors
2. Find by solving with BC
Ev
6. Find C by , independent of12V
QC =
12V
)(rV v02 =∇ V
)(rVE vv−∇=
4. Find of either conductor by)(rsvρ
0ερ s
nE =
5. Find deposited Q by ( )12 )( VdsrQ
S s ∝= ∫vρ
12V
Evaluation of capacitance (Method 3)
1. Assume or V for the two conductorsQ±
2. Find and by M1, M2Ev
3. Find the stored energy
22
or iswhich ,
21 VQdvEDW
Ve ∝⎟⎠⎞
⎜⎝⎛ ⋅= ∫ ′
vv
4. Find by orC
Dv
CQWe 2
2
=2
2CVWe =
Example 9-2: Parallel-plate capacitor (1)
1. Assume charges are depositedQ±
2. By Gauss’s law (planar sym.):S
QaE y εvv
−=
Example 9-2: Parallel-plate capacitor (2)
3.
4. dS
VQC ε==12
( )S
QddyaEVd
y ε=⋅−= ∫
0 12vv
Example 9-3: Coaxial cable capacitor (1)
1. Assume charges are depositedQ±
2. By Gauss’s law (cylindrical sym.):rL
QaE r πε2vv
=
( )abL
VQC
ln2
12
πε==
Example 9-3: Coaxial cable capacitor (2)
3.
4.
( ) ⎟⎠⎞
⎜⎝⎛=⋅−= ∫ a
bL
QdraEVa
b r ln2
12 πεvv
Now you know how to evaluate the capacitance per unit length of coaxial TX lines!
( )abLC
ln2πε
=
Sec. 9-2Electrostatic Energy
1. Energy of charges2. Energy of fields
Energy of two charges-1
RQRV
0
1
4)(
πε=
The work done to move Q2 from ∞ to P2 against V(R) due to Q1 is:
120
21222 4 R
QQVQWπε
==
Energy of two charges-2
RQRV
0
2
4)(
πε=′
The work done to move Q1 from ∞ to P1 against V'(R) due to Q2 is:
120
21112 4 R
QQVQWπε
==
Energy of two charges-3
The electrostatic energy stored by a system of two charges Q1-Q2 is:
( )221122112 21 VQVQVQVQW +===
Potential of Q1
at P1 due to Q2
Potential of Q2
at P2 due to Q1
Energy of three charges-1
)()( 23133 RVRVV ′+=The extra work done to move Q3 from ∞ to P3
against V(R), V'(R) due to Q1, Q2 is:
⎟⎟⎠
⎞⎜⎜⎝
⎛+=
=Δ
230
2
130
13
33
44 RQ
RQQ
VQW
πεπε
⎟⎟⎠
⎞⎜⎜⎝
⎛++=Δ+=
23
32
13
31
12
21
023 4
1R
QQRQQ
RQQWWW
πε
Energy of three charges-2
The electrostatic energy stored by a system of three charges Q1-Q2-Q3 is:
Energy of three charges-3
⇒ ( )3322113 21 VQVQVQW ++=
,44 130
3
120
21 R
QR
QVπεπε
+=
Vk denotes the potential of charge Qk at position Pk (k =1, 2, 3) due to the remaining charges, i.e.,
,44 230
3
120
12 R
QR
QVπεπε
+=
230
2
130
13 44 R
QR
QVπεπε
+=
Energy of N charges
The electrostatic energy stored by a system of Ndiscrete charges Q1-Q2-…-QN is:
∑=
=N
kkke VQW
121
∑≠
=kj jk
jk R
QV
041πε
Energy of continuous charge distributions
The electrostatic energy stored by a system of continuous charge distribution over V' is:)(rvρ
∫ ′=
Ve dvrVrW
)()(21 vvρ
Potential at source point due to the total charge distribution
rv
Example 9-4: Sphere of uniform charge density (1)
Find the energy stored in a sphere of radius
with uniform volume charge density
b
bρ
Spherical symmetry, ⇒
ρ
)(REaE Rvv
=
By Gauss’s law, ⇒
if ,
3
0 if ,3
20
30
⎪⎪⎩
⎪⎪⎨
⎧
≥
<<
=bR
Rba
bRRaE
R
R
ερ
ερ
v
v
v
Example 9-4: Sphere of uniform charge density (2)
For any point P inside the sphere
P
Rρ
( )bR <<0
( )
( )22
0
23
0
2
3
0
0
20
3
3623
3
33
)()(
RbRRb
RdRRdRb
RdRRdRb
RdaRERV
R
b
b
R
b
b
R
b
b
R
R
−=⎥⎥⎦
⎤
⎢⎢⎣
⎡ ′+
′−−=
⎥⎦
⎤⎢⎣
⎡′′+′
′−=
⎥⎦
⎤⎢⎣
⎡′
′+′
′−=
′⋅′−=
∞
∞
∞
∞
∫∫
∫∫
∫
ερ
ερ
ερ
ερ
ερ
vv∞
Example 9-4: Sphere of uniform charge density (3)
5
0
52
0
2
154
)4)((21
bb
dRRRVWb
e
∝=
= ∫
επρ
πρ
∫ ′=
Ve dvrVrW
)()(21 vvρByρ
bQ Total charge is:3
4 3ρπbQ =
bbQWe
120
3
0
2
∝=πε
(if ρ = constant)
(if Q = constant)
b
Energy of electric fields-1
In real applications (especially electromagnetic waves), sources are usually far away from the region of interest, only the fields are given.
)(rvρ ∞→RDEvv
,
ROI
Energy of electric fields-2
)(rvρ
( )∫∫ ′′⋅∇==
VVe VdvDdvrVrW
21)()(
21 vvvρ
Dr
⋅∇=ρ
V ′
Av
f
( ) ( ) fAAfAf ∇⋅+⋅∇=⋅∇vvv
By vector identity:
( )∫∫ ′′∇⋅−⋅∇=
VVe dvVDdvDVW 2
1)(21 vv
(1)
(2)
contain all the source charges
Energy of electric fields-3
∫∫ ′′⋅=⋅∇⇒
SVsdDVdvDV
)( vvv( )∫∫ ⋅∇=⋅
VSdvAsdA
,
vvvQ
)(rvρ
V ′ S ′
,VE −∇=v
Q ( ) ( )∫∫ ′′⋅−=∇⋅⇒
VVdvEDdvVD
vvv
( )∫∫ ′′⋅+⋅=
VSe dvEDsdDVW 2
1 21 vvvv
1I 2I
(3)
Energy of electric fields-4
∞→RS ′
S ′
21
RD ∝v
RV 1
∝
Obs. pt.
Energy of electric fields-5
0111
4)()(21
21
22
2
1
→∝⋅⋅∝
⋅≈⋅= ∫ ′
RR
RR
RRDRVsdDVIS
πvvv
∫ ′==⇒
V ee dvrwIW 2 )(v
)mJ( 21 3ED
vv⋅ …energy density
Energy of electric fields-6
∞→RV ′
∫ ′⎟⎠⎞
⎜⎝⎛ ⋅=
Ve dvEDW 2
1 vv
dvEDdWe ⎟⎠⎞
⎜⎝⎛ ⋅=
vv
21
Example 9-5: Parallel-plate capacitor (1)
Planar symmetry, ⇒ uniform E-field:
Find the stored electrostatic energy of:
EDvv
ε=,dVaE y
vv−=
Example 9-5: Parallel-plate capacitor (2)
Total stored energy:
Energy density:2
2
21
21
21
⎟⎠⎞
⎜⎝⎛==⋅=
dVEEDwe εε
vvv
22
21
21 V
dSSd
dVWe ⎟
⎠⎞
⎜⎝⎛=⋅⎟
⎠⎞
⎜⎝⎛= εε
VQC =
QVC
QCVWe 21
221 2
2 ===⇒
Example 9-6: Coaxial cable capacitor (1)
1. Assume charges are depositedQ±
2. By Gauss’s law (cylindrical sym.):rL
QaE r πε2vv
=
Find the capacitance of:
Example 9-6: Coaxial cable capacitor (2)
3.
4.
Instead of evaluating V by line integral of ,Ev
2
21
21 EEDwe
vvvε=⋅=
( ) ( )
⎟⎠⎞
⎜⎝⎛=⎟
⎠⎞
⎜⎝⎛=
⎟⎠⎞
⎜⎝⎛==
∫
∫∫
ab
LQ
rdr
LQ
rdrLLr
QdvEW
b
a
b
ae
ln44
222
121
2
2
22
πεπε
ππε
εεv
( )abL
WQC
e ln2
2
2 πε==5.
Example 9-7: Sphere of uniform charge density (1)
Find the energy stored in a sphere of radius
with uniform volume charge density
b
bρ
Spherical symmetry, ⇒
ρ
)(REaE Rvv
=
By Gauss’s law, ⇒
if ,
3
0 if ,3
20
30
⎪⎪⎩
⎪⎪⎨
⎧
≥
<<
=bR
Rba
bRRaE
R
R
ερ
ερ
v
v
v
Example 9-7: Sphere of uniform charge density (2)
Instead of evaluating V by line integral of ,
and arriving at
Ev
∫ ′=
Ve dvrVrW
)()(21 vvρ
bρ
:0For bR <<
0
52
0
4
0
2
0
22
001
452
92
)4(32
1
επρ
επρ
πε
ρε
bdRR
dRRRW
b
b
e
=⎟⎠⎞⎜
⎝⎛=
⎟⎟⎠
⎞⎜⎜⎝
⎛=
∫
∫dV
2Ev
Example 9-7: Sphere of uniform charge density (3)
bρ
:For bR >
dV
0
52
20
62
22
2
3
02
921
92
)4(32
1
επρ
επρ
περε
bdRR
b
dRRRbW
b
be
=⎟⎠⎞
⎜⎝⎛=
⎟⎟⎠
⎞⎜⎜⎝
⎛=
∫
∫∞
∞
2Ev
bQbWWW eee
0
2
0
52
21 203
154
πεεπρ
==+=(1:5)
Example 9-7: Sphere of uniform charge density (4)
The corresponding “capacitance” is:
bρ
310
20
2 bWQC
e
πε==
Example 9-8: Conducting sphere (1)
Find the stored energy and the capacitance of a conducting sphere of total charge Q
b
Q Spherical symmetry, ⇒
is uniformly distributed, ⇒
)(rsvρ
⎪⎩
⎪⎨
⎧ ≥=
otherwise ,0
if ,4 2
0
bRR
QaE R πε
vv
bQ
dRRR
QWbe
0
2
22
20
0
8
)4(42
1
πε
ππε
ε
=
⎟⎟⎠
⎞⎜⎜⎝
⎛= ∫
∞
Example 9-8: Conducting sphere (2)
Instead of evaluating V by line integral of ,
and arriving at
Ev
∫ ′=
Ve dvrVrW
)()(21 vvρ
:For bR >dV
2Ev
bQ
bWQC
e0
2
42
πε==⇒