1
Chapter 4. Potential and energy
EMLAB
2
Solving procedure for EM problems
Known charge
distribution
Vector
calculation
Coulomb’s law
 Rˆ d
E
40 R 2
V
Vector
calculation
Known boundary
condition
Gauss’ law differential
form
 D    E  D/
Scalar
calculation
Known charge
distribution
Integration of
Coulomb’s law
V 
V
Scalar
calculation
Known charge
distribution
 d
40 R
E  V
Poisson equation
 2V  

0
EMLAB
3
Gravitational field
G
F
GM
  2 rˆ
m
r
Moon
GMm
F   2 rˆ
r
Earth
EMLAB
4
Gravitational potential
Instead of field lines, potential energy levels can imply the direction and magnitude
of gravitational force.
EMLAB
5
Work in a gravitational field
r2
W   ( F)  dr
GMm
F   2 r̂12
r
r2
r2
mass : m
To move an object in the gravitational
field, an external force must be
applied that compensates the force
due to gravity.
r2
r1
mass : M
r2
GMm
rˆ  rˆ dr
2
r2
r
W F
r
O
2
1 1
 1
 GMm     GMm  
 r  r2
 r2 r2 

GMm
( r2  r2 )  mgr.
r22

GM 
 g  2 
r2 

EMLAB
6
Potential energy in a gravitational field
U (r2 )
r2
•
The scalar field quantity of potential energy is
introduced to represent energy levels inherent to
positions in the space.
•
Differences between the energy levels can be
obtained 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.
U (r2 )
r2
r1
U  U (r2 )  U (r2 )  W
O
EMLAB
7
Potential energy in an electric field
Electric field
Electric potential
Electric potential (3D)
EMLAB
8
Electric potential energy (V)
•
•
•
•
+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)
+q
rB
rB
rA
rA
W   ( F)  dr   (  qt E)  dr  qtV
rB
V   (  E)  dr
rA
EMLAB
9
Electric potential due to a point charge
+q1
r̂
rA
rB
 q 
q1
q1
   E  ds   
rˆ  rˆ dr   1  

 VB  VA
2
4

r
4

r
4

r
4

r
0
0  rA
0 B
0 A

A
rA
rB
B
VAB
rB
r
 q1 
q1
ˆ
ˆ
   E  ds   
r

r
dr


 4 r 
40 r 2
40 r
0 

A

B
VAB
q1
r
q1
If the position A is infinitely distant from the charge q1, VA approaches zero.
EMLAB
Distribution of electric potentials due to a point charge
10
+q1
+q1
r
V   (E)  dr 

q1
40 r
EMLAB
11
Potential distribution due to a dipole
+q
V
q
40 R
-q

q
40 R
R  r  r , R  r  r
EMLAB
12
Movement of a charge in a potential field
A positive charge moves from the position of high potential energy to that of lower potential
energy.
+q1
+q1
EMLAB
Structure of a cathode ray tube
13
EMLAB
14
Scalar field V
•Rectangular coordinate
VB
B
VA
V    E  dr
B
A
A
dr
•To obtain the potential difference V, we should
integrate the electric field from A to B.
dr  d ( x xˆ  y yˆ  z zˆ )
 xˆ dx  yˆ dy  zˆ dz
•Cylindrical coordinate
dr  d (  ˆ  z zˆ )
 ˆ d   dˆ  zˆ dz
 ˆ d  ˆ d  zˆ dz
•Spherical coordinate
B
V    E  dr  VB  VA
A
• For every position, the potential V is defined. So the
potential is a scalar field quantity.
dr  d ( r rˆ)  rˆ dr  r drˆ
 drˆ
drˆ 
 rˆ dr  r 
d 
d 
d 
 d
 rˆ dr  ˆ r d  ˆ r sin  d
• Only the potential difference has physical significance.
So voltage reference point should be specified always.
EMLAB
15
Example 4.1
ŷ
E  y xˆ  x yˆ  2 zˆ
B
B
VAB    E  ds     y xˆ  x yˆ  2 zˆ   xˆ dx  yˆ dy  zˆ dz 
A
1
B (0.8, 0.6,1)
VB
ds V
A
A
B
0.8
    y dx  x dy  2 dz     1  x dx 
2
A
1
0.6

1  y 2 dy  0.48V
1 A
0
x̂
A (1,0,1)
E  y xˆ  x yˆ  2 zˆ
B
B
A
A
VAB    E  ds     y dx  x dy  2 dz 
0.8
0.6

    3( x  1)dx   1 
1
0
y
 dy  0.48V
3
• 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 difference
is the same. A field that has this property is called as a conservative
field.
ŷ
1
B (0.8, 0.6,1)
VB
ds
VA
1 A
x̂
A (1,0,1)
EMLAB
16
Conservative field
VB
C3
VA
B
C2
A
VAB    E  ds    E  ds
C1
C1
C2
C1
VA
VB
A
B
  E  0    E   E  ds  0
C
 E  ds   E  ds   E  ds
C
C1
C2
C1
  E  ds 
C1
• In conservative field, the voltage difference
depends on only the start and the end point. The
result of the integral is independent of the path.
• The condition for a conservative field is that curl of
the field should be zero.
B
; Faraday' s law
t
  E  0 이면 conservati ve field
E  
E y 
 E
 E
E 
E x 
 E
  yˆ  x  z   zˆ  y 

  E  xˆ  z 
z 
x 
y 
 z
 y
 x
C2
  E  ds 
 C2
• All electric field in electrostatic problems are
conservative field.
 E  ds  0
C2
• The electric field in the example 4.1 is
conservative.
 E  ds
C2
EMLAB
17
Circulating
electric field is
non-conservative.
Conservative field
Non-conservative field
EMLAB
18
Relation between E and V
• We have learned how to obtain the voltage difference from an electric field. The reverse
process is also possible. That is, we can obtain the electric field from the potential
distribution.
VAB   E  ds
• 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.
C
VA
C1
A
VB
B
• 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.
ds
VB
VA dV
VB  VA  dV  E  ds  E x dx  E y dy  E z dz 
V
V
V
dx 
dy 
dz
x
y
z
V
V
V
 Ex  
, Ey  
, Ez  
x
y
z
h1  h3  h3  1
dV ( x, y , z ) 
E  V
•This operation is called “gradient V”.
EMLAB
Gradient operator in other curvilinear coordinates
19
•Gradient operators in other coordinate systems can be derived from the following relation.
dV  E  dr
Cylindrical coordinate
Spherical coordinate
V
V
V
d 
d 
dz


z
V
1 V
V
 E  
, E  
, Ez  

 
z
V
1 V
V
 V 
ρˆ 
φˆ 
zˆ

 
z
h1  1, h2   , h3  1
V
V
V
dr 
d 
d
r


V
1 V
1 V
 E  
, E  
, E  
r
r 
r sin  
V
1 V ˆ
1 V
 V 
rˆ 
θ
φˆ
r
r 
r sin  
h1  1, h2  r, h3  r sin 
VB  VA  dV  E  ds  E d  E  d  E z dz 
VB  VA  dV  E  ds  Er dr  E r d  E r sin  d 
dV 
dV 
VB  VA  dV  E  ds   Ei dsi
dV 
V
dui , dsi  hi dui
 ui
 Ei dsi   Ei hi dui 
 Ei  
1 V
,
hi  ui
V
dui
 ui
V
hi  ui
V i  1
EMLAB
20
Properties of gradient operator
Equi-potential surface
0V
ds
+q1
V
-q1
E
 5V
 5V
V (r )  C (constant)
V
V
V
dx 
dy 
dz  V  dr  0
x
y
z
 Because the inner product is zero ,
V is perpendicu lar to dr.
 3V
• Because the derivatives of voltage on the equi-potential surfaces are zero, Gradient V is perpendicular
to those surfaces.
• Electric field line is directed from the higher potential region to lower region.
• Gradient V is directed to higher potential region.
Using chain rule, gradient operation becomes simpler.
For the function f of argument R
y  f ( R ), R  r  r '
y
df ˆ
R
dR
EMLAB
21
Potential of multiple charge distribution
Point charges
+q1
40 r  r1
a1 
2
R
R


V    E  ds    
+q2
+q3
q1
E(r ) 
V 
+qn
V (r )
q1
40 r  r1
i

q2
40 r  r2
2
a2   
qn
40 r  rn
2
an
qi
qi
rˆ  rˆi dr  
2 i
40 ri
i 40 ri
q2
40 r  r2

qn
40 r  rn
• If potential contributions of separate charges are added, the potential
of the multiple charges are obtained.
E(r ) 
Continuous charge distribution
ˆ
(r' ) R
 4
V'
0
r  r'
2
d'




ˆ
ˆ
(r' ) R
(r' ) R
(r' )
V    E  ds     
d

'

d
s



d
s
d'


 d'  
2
2


4

r
4

r

r
'
4

r

r
'



0
C
C
V
'
V
'
C
V
'
0
0




(r' )
V  
d'
40 r
V'
(r )
V (r )
• If the charge distribution is continuous, the sum (Σ) symbol is replaced with integral (∫)
symbol.
• Line charge:
V
L (r' )
 4 r dl '
C'
0
•Surface charge:
S (r' )
da '
40 r
S'
V
EMLAB
22
Example : potential due to a charged annular disk
Find voltage on the z-axis, and electric field using the voltage.
R  r  r , r  z zˆ , r    ρˆ 
 S (r' )
da '
4

r

r
'
0
S'
V (   0, z )  
ẑ
S
b
a

2
0

b
a

S
40 z   
b
 d '
2
S
2 0 a z 2   2 1 / 2
z b


 S 
dt  S
2 0 z a
2 0

2
2
2
2

2 1/ 2
 d d '
z
2
 b2  z 2  a 2

 t  z 2   2
E(   0, z )  V  zˆ

S z 
1
1


2 0  z 2  a 2
z 2  b2 
EMLAB
23
Electric dipole
(Equi-potential surface)
ẑ
P (0, r sin , r cos )
θ
Q (0, 0, d / 2)
+q
d
-q
V 
Q (0, 0,  d / 2)

d (0, 0, d)
p  qd ; dipole moment .
q
q
q  1
1 
q  PQ   PQ  

 





40 PQ  40 PQ  40  PQ  PQ   40  PQ   PQ  
q  PQ   PQ   qd cos 
p  rˆ



2
2
40 
r
40 r 2
 40 r
 E   V 

qd cos 
rˆ 2 cos   θˆ sin 
40 r 3

 PQ   (0, r sin , r cos   d / 2)  (r sin ) 2  (r cos   d / 2) 2  r 2  rd cos   (d / 2) 2
PQ   (0, r sin , r cos   d / 2)  (r sin ) 2  (r cos   d / 2) 2  r 2  rd cos   (d / 2) 2
1 x 1
1
x
2
PQ   PQ   r 2  rd cos   (d / 2) 2  r 2  rd cos   (d / 2) 2

f ( x )  f ( x0 )  f ( x0 )( x  x0 ) 
r 2  rd cos   r 2  rd cos   r 1 
d
d
cos   r 1  cos   d cos 
r
r
1
1
f ( x0 )( x  x0 ) 2  f ( x0 )( x  x0 )3  
2!
3!
EMLAB
Poisson’s & Laplace’s equations
24
D   , D   E
E  V
  ( E)    ( V )  
for homogeneous medium
  V  2 V   
2 V  


( Poisson' s equation )
Laplace operator has different forms
for different coordinate systems.
 V
V
V
  V     xˆ
 yˆ
 zˆ
y
z
 x



 

   V
V
V
  xˆ
 yˆ
 zˆ    xˆ
 yˆ
 zˆ
y
z   x
y
z
 x
 2V  2V  2V
 2  2  2
x
y
z



EMLAB
Laplace’s equations
25
The differential equation for source-free
region becomes a Laplace equation.
2 V  0 (Lapace' s equation )
 2V  2V  2V
V  2  2  2
x
y
z
2
1   V
V 

   
2
 1  2V  2V
  2 2  2
z
  
(rectangular coordinate)
(cylindrical coordinate)
1   2 V 
1
 
V 
1
 2V
 V  2 r

 sin 

r r  r  r 2 sin   
  r 2 sin 2   2
2
(spherical coordinate)
EMLAB
26
Example 1 : Laplace eqs.
 2  2  2  2
  2  2  2  2 0
x
y
z
z
2
d
S
V0
If the plates are wide enough to ignore the variation of
electric field along x and y directions
ẑ


0,
0
x
y
x̂
0V
(1)  ( z ) 

V0
z
d
Using the boundary conditions on the two plates,
 (d )  Ad  B  V0 ,  (0)  B  0
V0
d
V
(3) D   E   zˆ 0
d
(2) E     zˆ
A
(4)  s ( top)  nˆ  D   zˆ  D 
V0 S
S
( 6) C 
Q V0 S  S


V
dV0
d
d
V0
V
  ( z)  0 z
d
d
V0
d
 s (bottom)  nˆ  D  zˆ  D  
(5) Q    s da   s S 

 A    Az  B ( A, B constant)
z
V0
d
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.
EMLAB
Example 2
27
 1  2V  2V 1   V 
  2

0
 2 
2
z
    
  
 The cylinder is assumed to be infinite in the z  direction and axially symmetric
 2V 
1   V

   
 2V
 2V
0, 2 0
 2
z

V
  0  
 A , V  A ln   B ( A, B constants)


Using the boundary conditions ,

r
  V

  
V (b)  A ln b  B  0 , V (a )  A ln a  B  V0
A
V0
V ln b
V ln( b /  )
,B 0
 V ( )  0
ln( a / b)
ln( a / b)
ln( b / a )

a
b
ŷ
V0 ln( b /  )
ln( b / a )
V
1
(2) E   V  0
ρˆ
 ln( b / a)
V
1
(3) D   E  0
ρˆ
 ln( b / a)
(1) V (  ) 
x̂
V0
0V
V0
1
a ln( b / a )
V
1
 s (   b)  nˆ  D  ρˆ  D   0
b ln( b / a )
V
1
(5) Q    s da  0
 2aL
a
ln(
b
/
a
)
S
(4)  s (   a )  nˆ  D  ρˆ  D 
(6) C 
Q
2L

V ln( b / a )
EMLAB
28
Charge storage
+
V
-
Q
V
If charges are accumulated, potential difference increases.
EMLAB
29
Capacitor
Q
C
V
S
Volume
h
EMLAB
30
Charging capacitor
Q0
V 0
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.
EMLAB
31
EMLAB
Potential distribution near parallel plates
32
EMLAB
33
Electrostatic energy
•Vn is a potential due to N-1
charges other than n-th charge
+q1
+q2
1 N
W   qnVn
2 n 1
+q3
•The work to assemble
charges q1, q2~ qn.
Vi , j 
qj
40 ri  ri
W2  q2V2,1
W1  q1V1, 2  q1V1,3  q1V1, 4    q1V1, N
W3  q3V3,1  q3V3, 2
W2  q2V2,3  q2V2, 4    q2V2, N

n 1
n 1
Wn   qnVn ,i  qn Vn ,i
i 1

Wn 
N
n 1
Wtotal  Wn   qnVn ,i
n 2
The magnitudes of works
to assemble or
disassemble are the same.
WN 1  qN 1VN 1, N
i 1
N
Potential energy of qi due to qj.

W4  q4V4,1  q4V4, 2  q4V4,3
i  n 1
n n ,i
 qn
N
V
i  n 1
n 1
N
2Wtotal   qnVn ,i  
n 1 i 1
n ,i
N 1
N 1
n 1
n 1 i  n 1
Wtotal  Wn  
n  2 i 1
N
N
q V
•The work for charges q1,
q2,~,qn to be separated to
infinitely distant points.
N
q V
n n ,i
N
q V
n 1 i  n 1
n n ,i
N
N
 n 1

    qnVn ,i   qnVn ,i 
n 1  i 1
i  n 1

N
N
N
n 1
i 1
i n
n 1
  qn Vn ,i   qnVn
1 N
Wtotal   qnVn
2 n 1


N


Vn  Vn ,i 
i 1


i n


EMLAB
34
Electrostatic energy
W 

1
 S (r ' )V (r ' ) d '
2 V'
1
1
(  D)V d '     (VD)  V  Dd '

2V'
2V'
1
1
1
  (VD)  da'  V  D d '   ( E  D) d '
2S
2V'
2V'
  (uA)  u  A  u  A
1
W   (E  D) d'
2 V'
•If the product V*D becomes zero on the
surface S, the surface integral becomes zero.
0
V
q
(1) The integral is performed on the
surfaces marked by red lines.
(1) V ( r )   
r
b
qrˆ  rˆdr
q

2
40 r
40
1 1
  
r b
1
q2  1 1 
qV ( a ) 
  
2
80  a b 
q
( 2) D 
rˆ ,
4r 2
b
2  b
1
1
W   E  D d '     D  D r 2 sin drdd
2a
2 00a
W 

4
32 2
b
q2 2
q2
r
dr

a r 4
8
 (uAx )  (uAy )  (uAz )


x
y
z
A
u
A
u
u
A

Ax  u x 
Ay  u y 
Az  u z
x
x y
y
z
z
 u  A  u  A
  (uA ) 
0
V
q
b
r2
q2  1 1 
dr

  
a r 4
8  a b 
(2) The integral is performed over the
volume marked by blue lines.
EMLAB
35
Capacitance
•The magnitude of an electric field is proportional to charges, and voltages are proportional
to electric field. Hence, charges are proportional to voltages. This proportionality constant
is called capacitance.
V  E  Q  Q  V  Q  CV
Q
C 
V
 E  da  E  da
S

  E  ds

S
V  V

Example: Capacitance of a parallel plate capacitor
E   zˆ
ẑ
x̂
S
 S
 S
d
S


S 
 S  da
  (  zˆ da )
S

S
S
S





d
d
d
1
 ˆ S  ˆ
  E  ds
    z   z dz
 S  dz
 
 0

0
S
C  
d
Q
C 
V
 E  da     zˆ 
EMLAB
36
Capacitance from electrostatic energy
(1) We 
1
1
D  E d   E  E d

2V
2V
( 2) We 
1
1
1
1
1 Q2
2

V
d


V

d


QV

CV

2 V
2 V
2
2
2 C
 Q  CV 
 E  E d
2We V
C  2 
2
V


  E  ds 


Example : parallel plate capacitor
E  zˆ
S

1
1  S 
S2

E

E
d



d


Sd


2 V
2 V   
2
2
ẑ
x̂
S
 S
 S
d
We 

  
V      zˆ S   zˆ dz  S d
 

0
d
S2
Sd
2W
S
C  2e   2  
V
d
 S 
 d
  
S
C  
d
EMLAB