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EMLAB Chapter 4. Potential and energy 1. EMLAB 2 Solving procedure for EM problems Known charge...
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Transcript of EMLAB Chapter 4. Potential and energy 1. EMLAB 2 Solving procedure for EM problems Known charge...
EMLABEMLAB
Chapter 4. Potential and energy
1
EMLABEMLAB
2Solving procedure for EM problems
Known charge distribu-tion
Coulomb’s law
Known boundary condi-tion
Gauss’ law differential form
/DED
Vector cal-culation
Vector cal-culation
V R
d2
04
ˆ
R
E
Known charge distribu-tion
Integration of Coulomb’s law
Scalar cal-culation
V R
dV
04
Known charge distribu-tion
Poisson equationScalar cal-culation
0
2
V
VE
EMLABEMLAB
Gravitational field
rF ˆ2r
GMm
rF
G ˆ2r
GM
m
Earth
Moon
3
EMLABEMLAB
4Gravitational potential
Instead of field lines, potential energy levels can imply the direction and magnitude of gravitational force.
EMLABEMLAB
2r
2r
Work in a gravitational field
2
2
)(r
rrF dW
122rF
r
GMm
1r
To move an object in the gravitational field, an external force must be ap-plied that compensates the force due to gravity.
O
22
2222
22
2
.)(
111
ˆˆ
2
2
2
2
r
GMgrmgrr
r
GMm
rrGMm
rGMm
drr
GMmW
r
r
r
r
rrFMmass :
mmass :
5
EMLABEMLAB
Potential energy in a gravitational field
• The scalar field quantity of potential energy is intro-duced to represent energy levels inherent to positions in the space.
• Differences between the energy levels can be ob-tained from the work that must applied for the object to move from the initial position to the final position.
• The position that corresponds to zero energy level is one that is located far away from the earth.
)( 2rU
)( 2rU
2r
2r
1r
O
WUUU )()( 22 rr
6
EMLABEMLAB
7Potential energy in an electric field
Electric fieldElectric potential Electric potential (3D)
EMLABEMLAB
Electric potential energy (V)
+q
VqdqdW tt
B
A
B
A
r
r
r
rrErF )()(
+qt
• As in the gravitational field, a potential energy for the electric field can be introduced.
• The potential energy in an electric field is defined as the energy levels of the test charge with +1C.
• The unit of potential energy is “voltage” named after the physicist Volta.
• The position far away from the source charge has zero potential energy.
Electric potential energy defined as the work to move a test charge with (+1C)
B
A
dVr
rrE)(
8
EMLABEMLAB
Electric potential due to a point charge
ABAB
r
r
r
r
B
A
AB VVr
q
r
q
r
qdr
r
qdV
B
A
B
A
0
1
0
1
0
12
0
1
444ˆˆ
4 rrsE
+q1
ArBr r
r
q
r
qdr
r
qdV
rrB
A
AB0
1
0
12
0
1
44ˆˆ
4
rrsE
9
If the position A is infinitely distant from the charge q1, VA approaches zero.
EMLABEMLAB
+q1
+q1
r
qdV
r
0
1
4)(
rE
Distribution of electric potentials due to a point charge 10
EMLABEMLAB
+q -q
Potential distribution due to a dipole
R
q
R
qV
00 44
11
rrrr RR ,
EMLABEMLAB
A positive charge moves from the position of high potential energy to that of lower potential energy.
+q1
Movement of a charge in a potential field
+q1
12
EMLABEMLAB
13Structure of a cathode ray tube
EMLABEMLAB
B
A
dV rE
rd
Scalar field V
dzzdyydxx
zzyyxxdd
ˆˆˆ
)ˆˆˆ(
r
dzzdd
dzzdd
zzdd
ˆˆˆ
ˆˆˆ
)ˆˆ(
r
drdrdrr
dd
rdd
d
rdrdrr
rdrdrrrrdd
sinˆˆˆ
ˆˆˆ
ˆˆ)ˆ(
r
• Rectangular coordinate
• Cylindrical coordinate
• Spherical coordinate
• To obtain the potential difference V, we should in-tegrate the electric field from A to B.
AB
B
A
VVdV rE
AAV
BBV
• For every position, the potential V is defined. So the potential is a scalar field quantity.
• Only the potential difference has physical signifi-cance. So voltage reference point should be speci-fied always.
14
EMLABEMLAB
Example 4.1
Vdyydxxdzdyxdxy
dzdydxxydV
xy
B
A
B
A
B
A
AB
48.0112
ˆˆˆˆ2ˆˆ
ˆ2ˆˆ
6.0
0
28.0
1
2
zyxzyxsE
zyxE
AAV
BV
1
1
x
y
)1,0,1(A
)1,6.0,8.0(B
AAV
BV
1
1
x
y
)1,0,1(A
)1,6.0,8.0(B
Vdyy
dxx
dzdyxdxydV
xyB
A
B
A
AB
48.03
1)1(3
2
ˆ2ˆˆ
6.0
0
8.0
1
sE
zyxE
sd
sd• In this example, a different integration path is used with the
same start point A and end point B.
• Although the integration paths are different, the voltage differ-ence is the same. A field that has this property is called as a con-servative field.
15
EMLABEMLAB
Conservative field
21 CC
AB ddV sEsE
AAV
BBV
1C
2C
3C
• All electric field in electrostatic problems are conservative field.
• In conservative field, the voltage difference de-pends on only the start and the end point. The re-sult of the integral is independent of the path.
• The condition for a conservative field is that curl of the field should be zero.
fieldveconservati0
laws'Faraday;t
이면
E
BE
AAV
BBV
1C
2C
21
21
21
0
00
CC
CC
CC C
C
dd
dd
ddd
d
sEsE
sEsE
sEsEsE
sEEE
2C
y
E
x
Eˆ
x
E
z
Eˆ
z
E
y
Eˆ xyzxyz zyxE
• The electric field in the example 4.1 is conservative.
16
EMLABEMLAB
17
Conservative fieldNon-conservative field
Circulating elec-tric field is non-conservative.
EMLABEMLAB
Relation between E and V
C
AB dV sE
AAV
BBV
1C
• We have learned how to obtain the voltage difference from an electric field. The re-verse process is also possible. That is, we can obtain the electric field from the poten-tial distribution.
• As in the derivation process for divergence operator, the relation between the electric field and the potential can be derived from the integral equation with the integration path infinitesimally small.
• Compared with electric field calculations which contain vector operations, voltage calculations are easier and simpler as the voltage is scalar quantity.
• For ease of operation, we calculate first potential functions and then electric field can be derived from potentials.
sd
AV
BV
VE
1
,,
),,(
331
hhh
z
VE
y
VE
x
VE
dzz
Vdy
y
Vdx
x
VzyxdV
dzEdyEdxEddVVV
zyx
zyxAB sE
• This operation is called “gradient V”.
dV
18
EMLABEMLAB
ii
iii
i
ii
iiiii
iiiii
iiAB
u
V
hV
u
V
hE
duu
VduhEdsE
duhdsduu
VdV
dsEddVVV
1,
1
,
sE
• Gradient operators in other coordinate systems can be derived from the following relation.
1,,1
ˆˆ1
ˆ
,1
,
321
hhh
z
VVVV
z
VE
VE
VE
dzz
Vd
Vd
VdV
dzEdEdEddVVV
z
zAB
zφρ
sE
sin,,1
ˆsin
1ˆ1ˆ
sin
1,
1,
sin
321 rhrhh
V
r
V
rr
VV
V
rE
V
rE
r
VE
dV
dV
drr
VdV
drEdrEdrEddVVV rAB
φθr
sE
Cylindrical coordinate Spherical coordinate
Gradient operator in other curvilinear coordinates
rE ddV
19
EMLABEMLAB
Properties of gradient operator
+q1 -q1
Equi-potential surface
. lar toperpendicu is
,zero isproduct inner theBecause
0
(constant))(
r
r
r
dV
dVdzz
Vdy
y
Vdx
x
V
CV
sd
EV
• Because the derivatives of voltage on the equi-potential surfaces are zero, Gradient V is perpendicu-lar to those surfaces.
• Electric field line is directed from the higher potential region to lower region.
• Gradient V is directed to higher potential region.
V5V3
RdR
dfy
RRfy
ˆ
'),(
rr
V0
V5
Using chain rule, gradient operation becomes simpler. For the function f of argument R
20
EMLABEMLAB
Potential of multiple charge distribution
+q1
+q2
+q3n0
n
20
2
10
1
i i0
iii
R
i2
i0
iR
n2
n0
n22
20
212
10
1
4
q
4
q
4
qV
r4q
drˆˆr4
qdV
4
q
4
q
4
q)(
rrrrrr
rrsE
arr
arr
arr
rE
+qn
)(V r
Continuous charge distribution
'dr4
)(V
'dr4
)('dd
'4
ˆ)(d'd
'4
ˆ)(dV
'd'4
ˆ)()(
'V 0
'V 0'V C2
0C 'V2
0C
'V2
0
r'
r's
rr
Rr's
rr
Rr'sE
rr
Rr'rE
)(r
)(V r
'dr4
)(V
'C 0
L l
r'
• If potential contributions of separate charges are added, the poten-tial of the multiple charges are obtained.
• If the charge distribution is continuous, the sum (Σ) symbol is replaced with integral (∫) symbol.
'dar4
)(V
'S 0
S
r'• Line charge: • Surface
charge:
Point charges
21
EMLABEMLAB
22
Sb
a
z
ρrzrrrR ˆ,ˆ, z
2222
00
2/1220
2
0 2/1220
' 0
22
'
2
'4
''4
)(),0(
22
22azbzdt
z
d
ddz
dazV
Sbz
az
S
b
a
S
b
a
S
S
S
rr
r'
Example : potential due to a charged annular disk
Find voltage on the z-axis, and electric field using the voltage.
22 zt
22220
11
2ˆ),0(
bzaz
zVz S
zE
EMLABEMLAB
Electric dipole
+q
-q
)cosr,sinr,0( P
)2/d,0,0(Q
)2/d,0,0(Q
sinˆcos2ˆr4
cosqdV
r4
ˆ
r4
cosqd
r
PQPQ
4
q
PQPQ
PQPQ
4
q
PQ
1
PQ
1
4
q
PQ4
q
PQ4
qV
30
20
20
20
0000
θrE
rp
cosdcosr
d1rcos
r
d1rcosrdrcosrdr
)2/d(cosrdr)2/d(cosrdrPQPQ
)2/d(cosrdr)2/dcosr()sinr()2/dcosr,sinr,0(PQ
)2/d(cosrdr)2/dcosr()sinr()2/dcosr,sinr,0(PQ
22
2222
2222
2222
)d,0,0(d
d
.momentdipole;qdp
(Equi-potential surface)
θ
z
23
xx2
111
300
200000 ))((
!3
1))((
!2
1))(()()( xxxfxxxfxxxfxfxf
EMLABEMLAB
24Poisson’s & Laplace’s equations
EDD ,
)equationsPoisson'(2
V
VE
)()( VE
VV 2
for homogeneous medium
2
2
2
2
2
2
ˆˆˆˆˆˆ
ˆˆˆ
z
V
y
V
x
V
z
V
y
V
x
V
zyx
z
V
y
V
x
VV
zyxzyx
zyx
Laplace operator has different forms for different coordinate systems.
EMLABEMLAB
25
The differential equation for source-free region becomes a Laplace equation.
)equations'Lapace(0V2
2
2
2
2
22 11
z
VVVV
2
2
2
2
2
22
z
V
y
V
x
VV
2
2
2222
22
sin
1sin
sin
11
V
r
V
rr
Vr
rrV
(rectangular coordinate)
(cylindrical coordinate)
(spherical coordinate)
Laplace’s equations
EMLABEMLAB
26Example 1 : Laplace eqs.
Sd
zx
0V
V0
Unlike the procedures in the previous chapters, the potential V is first obtained solving Laplace equation. Then, using the potential, E, D, , Q, C are obtained.
02
2
2
2
2
2
2
22
zzyx
d
S
dV
SV
V
QC
d
SVSdaQ
d
Vd
Vd
Vd
V
zd
Vz
s
S
s
s
s
0
0
0
0
0
0
0
0
)6(
)5(
ˆˆbottom)(
ˆˆtop)()4(
ˆ)3(
ˆ)2(
)()1(
DzDn
DzDn
zED
zE
If the plates are wide enough to ignore the variation of electric field along x and y directions
0,0
yx
constant),( BABAzAz
zd
Vz
d
VA
BVBAdd
00
0
)(
0)0(,)(
Using the boundary conditions on the two plates,
EMLABEMLAB
27Example 2
)/ln(
)/ln()(
)/ln(
ln,
)/ln(
ln)(,0ln)(
,conditionsboundary theUsing
constants),(ln,0
0,0
symmetricaxially anddirectionz in the infinite be toassumed iscylinder The
0111
000
0
2
2
2
2
2
2
2
2
22
ab
bVV
ba
bVB
ba
VA
VBaAaVBbAbV
BABAVAVV
z
VV
V
z
VVVV
)/ln(
2)6(
2)/ln(
1)5(
)/ln(
1ˆˆ)(
)/ln(
1ˆˆ)()4(
ˆ)/ln(
1)3(
ˆ)/ln(
1)2(
)/ln(
)/ln()()1(
0
0
0
0
0
0
ab
L
V
QC
aLaba
VdaQ
abb
Vb
aba
Va
ab
V
ab
VV
ab
bVV
S
s
s
s
DρDn
DρDn
ρED
ρE
x
y
r
a
b
V00V
EMLABEMLAB
28Charge storage
Q V
+V-
If charges are accumulated, potential difference increases.
EMLABEMLAB
29
V
QC
h
VolumeS
Capacitor
EMLABEMLAB
30
0V
0Q
Due to potential difference, positive charges rush to the capacitor. As the amount of charges increases, the voltage increases.
If the voltage difference between the terminals of the capacitor is equal to the supply voltage, net flow of charges becomes zero.
Charging capacitor
EMLABEMLAB
31
EMLABEMLAB
32Potential distribution near parallel plates
EMLABEMLAB
Electrostatic energy
+q1
+q2
+q3
N
n
n
iinn
N
nntotal
n
iinn
n
iinnn
VqWW
VqVqW
VqVqVqW
VqVqW
VqW
2
1
1,
2
1
1,
1
1,
3,442,441,444
2,331,333
1,222
1
1 1,
1
1
1,
1,
,111
,224,223,222
,114,113,112,111
N
n
N
niinn
N
nntotal
N
niinn
N
niinnn
NNNN
N
N
VqWW
VqVqW
VqW
VqVqVqW
VqVqVqVqW
N
nnntotal
N
n
N
nii
innnn
N
n
N
nii
inn
N
n
N
niinn
n
iinn
N
n
N
niinn
N
n
n
iinntotal
VqW
VVVqVq
VqVq
VqVqW
1
1 1,
1 1,
1 1,
1
1,
1 1,
1
1
1,
2
1
2
N
nnnVqW
12
1
• The work to assem-ble charges q1, q2~
qn.
• The work for charges q1, q2,~,qn to be sepa-
rated to infinitely dis-tant points.
• Vn is a potential due to N-1
charges other than n-th charge
The magnitudes of works to assemble or disassem-ble are the same.
33
ii
jji
qV
rr
0, 4
Potential energy of qi due to qj.
EMLABEMLAB
Electrostatic energy
')(2
1'
2
1')(
2
1
')(2
1')(
2
1
')'()'(2
1
''
''
'
ddVdV
dVVdV
dVW
VVS
VV
V
S
DEDaD
DDD
rr
AAA uu)u(
'd)(2
1W
'V
DE
• If the product V*D becomes zero on the surface S, the surface integral becomes zero.
0
Vq
ba
qdr
r
rqdrr
r
q
ddrdrdW
r
q
ba
qaqVW
br
q
r
drrrqrV
b
a
b
a
b
a
b
a
r
b
11
8832
4
sin2
1'
2
1
,ˆ4
)2(
11
8)(
2
1
11
44
ˆˆ)()1(
2
4
222
4
2
2
2
0 0
2
2
0
2
02
0
DDDE
rD
0
Vq
(1) The integral is performed on the surfaces marked by red lines.
(2) The integral is performed over the vol-ume marked by blue lines.
34
AA
A
uu
z
AuA
z
u
y
AuA
y
u
x
AuA
x
u
z
uA
y
uA
x
uAu
zz
yy
xx
zyx )()()()(
EMLABEMLAB
35Capacitance• The magnitude of an electric field is proportional to charges, and voltages are propor-
tional to electric field. Hence, charges are proportional to voltages. This proportionality constant is called capacitance.
CVQVQQEV
VV
d
d
d
V
QC SS
aE
sE
aE
SS
S
d
d
SC
d
S
dz
da
dz
da
d
d
V
QC d
S
S
S
dS
S
S
S
S
00
1ˆˆ
)ˆ(ˆ
ˆ
zz
zz
sE
aE
zEz
x
Example: Capacitance of a parallel plate capacitor
EMLABEMLAB
36Capacitance from electrostatic energy
22
22
2
2
1
2
1
2
1
2
1
2
1)2(
2
1
2
1)1(
sE
EE
EEED
d
d
V
WC
CVQC
QCVQVdVdVW
ddW
Ve
VV
e
VV
e
SS
S
dzx
d
SC
d
S
d
Sd
V
W2C
ddzˆˆV
Sd2
d2
1d
2
1W
ˆ
2
S
2S
2e
Sd
0
S
2S
V
2
S
V
e
S
zz
EE
zE
Example : parallel plate capacitor