Chapter 3 (part 1)
Electrostatic Fields
Outline
1. Coulomb’s law
2. Electric field
3. Electric flux density
4. Gauss’s law
5. Applications of Gauss’s law
6. Electric potential
7. Energy density
Coulomb’s Law (i)

QQ 
F12 = 1 2 2 a R12
4 0 R
Q1, Q2: point charges (C)
R: distance between Q1 and Q2 (m)
0 : permittivity of free space (F/m)
❑ Coulomb’s law states that the force exerted on a point charge Q2 due
to a point charge Q1 is given by the above equation.


 


R12
R12 = r2 − r1 = Ra R12 , a R12 =
R
R

QQ 
QQ
F12 = 1 2 2 a R12 = 1 2
4 0 R
4 0
 
r2 − r1
 3
r2 − r1
Coulomb’s Law (ii)
❑ Originally an empirical law, but later corroborated by further theory.
❑ Two like charges repel on another, whereas two unlike charges
attract. The direction of force has been automatically taken into
account regardless of positive and negative charges.
❑ It is valid only for point charges (i.e., R >> dimensions of Q)
❑ 0, permittivity of free space is 8.8510-12 F/m. It is obtained from
experiments.
❑ One coulomb is approximately equivalent to 6.241018 electrons.
(one electron charge e = -1.601910-19 C)
Principle of Superposition
❑ If we have more than two point charges, we can use the principle of
superposition to determine the force on a particular charge.
 
 
 
 QQ1 r − r1
QQ3 r − r3
QQN r − rN
QQ2 r − r2
F=
 3+
  3+
  3 ++
 
4 0 r − r1
4 0 r − r2
4 0 r − r3
4 0 r − rN 3
 
Q N Qk (r − rk )
=
 
4 0 k =1 r − rk 3

Q2
Q1
Q3
Q4
Q6
Q
Q5
Linear System
If



System

then


 +  
System
 
 
 +  
Electric
Field

❑ The electric field intensity E is the force that a unit positive charge
experiences when placed in an electric field.


F
E = lim
Q →0 Q

❑ The unit of E is newton per coulomb or volt per meter.


Q(r − r )
Q 
E=
=
aR

3
2
4 0 R
4 0 r − r 
❑ For a linear system, we can apply the principle of superposition to
obtain the electric field intensity from N point charges.
 
 
 

Q r −r
Q
Q r −r
Q r −r
E = 1  1 3 + 2   2 3 + 3  3 3 +  + N
4 0 r − r1
4 0 r − r2
4 0 r − r3
4 0
 
N
Qk (r − rk )
1
=
 
4 0 k =1 r − rk 3

 
r − rN
  3
r − rN
Exercise
❑ Florida phosphate ore, consisting of small particles of quartz
and phosphate rock, can be separated into its component by
applying a uniform E-field. Assuming zero initial velocity and
displacement, determine the separation between the particles
after falling 80 cm. Take E=500 kV/m and Q/m=9 μC/kg.
(x direction)

d 2x 
QE = m 2 a x
dt
d2y
(y direction) m 2 = − mg
dt
Initial conditions:
x(t = 0) = 0
dx
=0
dt t = 0
x=
Q
Et 2
2m
x=
Q
Et 2 + C1t + C2
2m
1
y = − gt 2 + C3t + C4
2
y (t = 0) = 0
dy
dt
=0
t =0
1
y = − gt 2
2
When y=-0.8 m, t=0.4 and x=0.367
Thus, 2x = 0.7347 m
Continuous Charge Distribution
❑ We can think of continuous charge distribution along a line, on a
surface, or in a volume.
❑ It is customary to denote the line charge density, surface charge
density, and volume charge density by L (C/m), S (C/m2), and v
(C/m3), respectively.
Electric Fields Due to Continuous
Charge Distribution
❑ The electric field intensities due to line charge, surface charge, and
volume charge can be regarded as the summation of the field contributed
by numerous point charges making up the charge distribution.
❑ The charge element dQ and the total charge Q due to continuous charge
distribution are obtained as
dQ =  L dl  Q =   L dl
(line charge)
L
dQ =  S dS  Q =   S dS
(surface charge)
dQ = v dv  Q =  v dv
(volume charge)
S
v
❑ The electric field intensity by line charge can be expressed by replacing
Q with charge element dQ (= Ldl) as

 L dl 
 L dl 
E=
aR =
R
2
3
4 0 R
4 0 R
L
L


Electric Fields Due to Line Charge
❑ The electric field intensity by line charge in z axis can be expressed as
 L dl
 L dl
a
=
R
2 R
3

4 0 R
4 0 R
L
L
E=
Prime indicates the
source coordinates.
dl = dz 




R = (x y z ) − (0 0 z ) = a x x + a y y + a z (z − z )


= a   + a z ( z − z )

E=


a   + a z ( z − z )
L
L 4 0  2 + (z − z)2


3/ 2
dz 
❑ Solving this integral, we get for a
source
finite line charge
field point

E=


L

− (sin  2 − sin 1 ) a  + (cos  2 − cos 1 ) a z 
4 0 
❑ For infinite line charge,

E=
L 
a
2 0 
Electric Fields Due to Surface Charge
❑ The electric field intensity by surface charge in xy plane can be expressed as
 S dS
 S dS
E=
a =
R
2 R
3
4 0 R
4 0 R
S
S

E=
(0, 0, z)
dS = dxdy  =  d d

R = (0 0 z ) − (x


= −a  + a z z
S 
S 4 0



y  0 ) = − a x x + − a y y  + a z z
(− a  + za )d d =
 
 + z 

2
z
2 3/ 2

S 
za z
d d
2
2 3/ 2
 = 0  = 0 4
0  + z 

 S z 2 
a z
=
d d
4 0  = 0  = 0  2 + z 2 3 / 2

S z 
a z
=
d
2 0  = 0  2 + z 2 3 / 2
2

❑ Solving this integral, we obtain
 S 
E=
az
2 0
Electric Fields Due to Volume Charge
❑ Consider a sphere of radius a centered at the origin. Let the volume of the sphere
be filled uniformly with a volume-charge density v in (C/m3).

E=
dv = r  2 sin   dr  d  d 
 v dv 
v 4 0 R 2 aR
2
R = z 2 + r  2 − 2 zr  cos 
❑ Owing to the symmetry of the charge distribution,
Ex or Ey add up to zero. Thus,
 v dv
Ez =
 cos 
2
4

R
0
v

sin   d 

Ez = v
4 0

2
 =0
d 
RdR
z 2 + R 2 − r 2 1
r
dr 
R= z −r
zr 
2 zR
R2
 
a
r =0
z +r
v
=
4 0 z 2
a
cos 
2
 z 2 − r 2 
r 1 +
 dRdr 
2
R= z −r
R


 
a
r =0
z +r
❑ Solving this integral, we obtain

E=
Q
4 0 z
2

az
4 3 

a v 
Q =
3


Electric Flux Density
❑ The electric field intensity is dependent on the medium in which the
charge is placed.
❑ To study the electric field flux , we need to define electric field flux
density (also known as electric displacement).


D = 0E

❑ The unit of D is coulomb per square meter.

❑ We define the electric flux in terms of D .
 
 = D  dS

S
❑ It is the measure of flow of electric field
through a given area.
Exercise
❑ Determine the electric flux density at (0, 0, z) induced by continuous charge Q
uniformly distributed on a ring (a <  < b).

 S dS 
 S dS 
D=
aR =
R
2
3
4

R
4

R
S
S

=
S
4

= S
4
=
=

 (
dd
+ z2 )
3/ 2
2
S
zdd
 (
+ z2 )
3/ 2
2
S
z S
4
  (
z S
2
 (
2
0
Symmetry
dd
a
2

a
3/ 2 z
+ z2 )
d
2


+ za z )

az
b
b
a
(− a
Q
 (b 2 − a 2 )



R = − a  + za z
S =

a
3/ 2 z
+ z2 )
z S 
1
1
=
−

2  a2 + z2
b2 + z 2

az


R
(0, 0, z)
dS = dd
Chapter 3 (part 2)
Electrostatic Fields
Gauss’s Law

  D = v
❑ Gauss’s law states that the total electric flux through any closed
surface is equal to the total charge enclosed by that surface.
❑ The integral form of Gauss’s law is
 
Q = D  dS =  v dv

S

v
❑ Gauss’s law is useful when the charge distribution is symmetric.
❑ It is important to choose a suitable Gaussian surface over which the
normal component of electric field resulting from a given charge
surface is a constant.
❑ In the case that the charge distribution is not symmetric, we should


resort to Coulomb’s law to determine E or D .
Application of Gauss’s Law
(Point charge)
❑ We choose a spherical surface centered at the origin as a Gaussian

surface. Since D is normal to this surface, we apply Gauss’s law to obtain
 
Q = D  dS = Dr dS = 4 r 2 Dr

S
❑ Thus,

D=
Q 
ar
2
4 r

S
Application of Gauss’s Law
(infinite line charge)
❑ We choose a cylindrical surface having a length l along the z-axis as a

Gaussian surface. Since D is normal to this surface, we apply Gauss’s
law to obtain
 
Q =  L l = D  dS = D dS = 2lD

S
❑ Thus,
 L 
D=
a
2

S
Application of Gauss’s Law
(infinite sheet of charge)

❑ We choose a rectangular box having as a Gaussian surface. Since D is
normal to this surface, we apply Gauss’s law to obtain
 
Q =  S A = D  dS = Dz dS = 2 ADz

S
❑ Thus,
 Q 
S 
D=
az =
az
2A
2
 S 
E=
az
2 0

S
Application of Gauss’s Law
(uniformly charged sphere)
Gaussian
surface
❑ We choose a spherical surface as a Gaussian surface.
❑ For ra, we apply Gauss’s law to obtain
Qenc
❑ Thus,
 
4 r 3
=  v dv =  v
= D  dS =4 r 2 Dr
3
v
S


 r v 
D=
ar
3
Gaussian
surface
❑ For r>a, we also apply Gauss’s law to obtain
4 a 3
Qenc =   v dv =  v
v
3
 
= D  dS =4 r 2 Dr

❑ Thus,
S
 a3 v 
Q 
D=
a
=
ar
r
2
2
3r
4r
v
Potential Difference (i)
❑ Suppose we move a point charge Q from point A to point B in an electric



field E. From Coulomb’s law, the force on Q is F = QE so that the work

done in displacing the charge by dl is
 
 
dW = − F  dl = −QE  dl
❑ The negative sign indicates that the work is being done by an external
agent. The work done in moving Q from A to B per unit charge is known
as the potential difference and expressed as

B 

B 
W − Q A E  dl
V AB =
=
= − E  dl
A
A
Q
Q

Initial
point
E

Final
point
dl
❑ It implies the voltage along the path
(from  to ).
B
Potential Difference (ii)
❑ If VAB is negative, there is a loss in potential energy in moving Q from A to
B. It implies the work is done by the field. If VAB is positive, the work is
done by an external agent.
❑ VAB is independent of the path taken (conservative).
❑ VAB is measured in joule per coulomb, commonly referred to as volt (V).
❑ In the electric field due to a point charge Q, the potential difference
becomes
V AB = −
=−

B
B
 
E  dl
A
A
rB
Q
rA
4 0 r 2



ar  drar
Q 1 1
 − 
4 0  rB rA 
= VB − V A
=
rA
rB
Electric Potential
❑ The potential at any point is the potential difference between that point
and a chosen point (or reference point) at which the potential is zero.
❑ In some cases, the zero potential (reference) point is chosen arbitrarily
to be at infinity.
❑ If we have more than two point charges, we can use the principle of
superposition to determine the electric potential.

Q
Q
Q
Q
1
1
1
1
V (r ) = 1   + 2   + 3   +  + N  
4 0 r − r1 4 0 r − r2 4 0 r − r3
4 0 r − rN
=
N
1
4 0

V (r ) =

k =1
1

Qk
 
r − rk

 L (r )dl 
 
r
− r
L


 S (r )dS 
1
V (r ) =
 
4 0 S r − r 
4 0

(line charges)
(surface charges)
(point charges)
Conservative Static Electric Field
❑ The static electric field is conservative since

 
E  dl = 0
L
❑ This implies that VAB = -VBA. This is Kirchhoff’s voltage law in circuit theory.
❑ Also, applying Stokes’s theorem, we get

 
E  dl =
L
(
)
 
  E  dS =0
S
❑ Comparing the following two expressions,
 
dV = − E  dl = − E x dx − E y dy − E z dz
dV =
V
V
V
dx +
dy +
dz
x
y
z
we have

E = −V

 E = 0
(static)
E and V

E = −V
This relationship is useful in finding
the electric field when the potential
field V is known.
Exercise
❑ A charge Q is uniformly distributed over the wall of a circular tube of radius a and
height h. Determine the electric potential and electric field intensity on its axis.

V (r ) =
=
=
1
4 0
1
4 0


 S (r )dS 
S
8  h  
h
2
0
0
=
=
4 0 h
Q
4 0 h
S =
Q
2ah
Q
a
d  dz 
2
2
2ah a + (z − z )
Q
Q



R = − a  a + a z ( z − z )
 
r − r
S

dS  = a d  dz 

h
0
ln
2
1
a 2 + ( z − z )
1
dz 
2
2
a + ( z − z )
2
0
z
(0, 0, z)
d  dz 

R
h
(a, , z’ )
z + z2 + a2
z−h+
a
(z − h )2 + a 2

 dV
a z Q 
E = −V = −a z
=
dz 4 0 h 

x


−
z 2 + a 2 
(z − h )2 + a 2
1
1
y
Electric Dipole
❑ An electric dipole is formed when two point charges of equal magnitude
but opposite sign are separated by a small distance.
❑ The potential at point P(r, θ, ) is given by
V=
Q 1 1
 − 
4 0  r1 r2 
❑ If r >> d, r2 − r1  dcos θ, r1r2  r2. Thus,
Q d cos
V=
4 0 r 2
Electric dipole moment

 

p
=
Q
d
❑ Since d cos = d  ar , we define
 

p  ar
V (r ) =
4 0 r 2

❑ If the dipole is located at r 
  

p  (r − r )
V (r ) =
 3
4 0 r − r 
❑ The electric field due to the dipole at
the origin is

Qd 
 V  1 V  
ar 2 cos + a sin  
E = −V = − 
ar +
a  =
3
r   4 0 r
 r
Dipole
Electric Flux Line
❑ An electric flux line is an imaginary path or line drawn in such a way that its
direction at any point is the direction of the electric field at that point.
❑ The convention of the electric field: it is pointing out from positive charge and
pointing into negative charge.
Energy Density in Electrostatic Field
❑ The energy present in an assembly of charges can be determined by the amount
of work necessary to assemble them. It can be calculated by moving point
charges one by one from infinite to an initially empty space shown in the figure.
❑ For three charges, the total work done in positioning them is
WE = W1 + W2 + W3 = 0 + Q2V21 + Q3 (V31 + V32 )
(Q1→Q2→Q3)
WE = W3 + W2 + W1 = 0 + Q2V23 + Q1 (V12 + V13 )
(Q3→Q2→Q1)
2WE = Q1 (V12 + V13 ) + Q2 (V21 + V23 ) + Q3 (V31 + V32 )
= Q1V1 + Q2V2 + Q3V3
Position Source
charge
Position
❑ For N charges, we can express
1
WE =
2
N
Q V
k
k
k =1
❑ For a continuous charge distribution
WE =
1
 SV dS
2S

(surface charge)
Electrostatic Energy Density
❑ For a volume charge,
1
(volume charge)
WE =
 vV dv
2v

1
=
  D V dv
2v

1
1 
=
  VD dv −
D  V dv
2v
2v
  1 
1
=
VD dS −
D  V dv
2S
2v

(
)
(
)
(
( )
)
(
)
Gauss’s law
( )

 
  VA = V  A + A  V
Divergence theorem
❑ We choose S to be a very large so that the first term tends to zero. Thus,
1 
1  
WE = −
D  V dv =
D  E dv


2v
2v
D = 0E
1
=
 0 E 2 dv
2v
(
)
(
)

❑ The electrostatic energy density WE (in J/m3) is defined as
1
1
1 2
wE = D  E =  0 E 2 =
D
2
2
2 0
Summary (i)
1. Coulomb’s law

QQ 
F12 = 1 2 2 a R12
4 0 R
2. The electric field is defined as the Coulomb’s force per unit change.
3. The electric field by a unit charge


Q(r − r )
Q 
E=
=
aR

3
2
4

R

4 0 r − r
0
4. We can apply the principle of superposition to Coulomb’s force.


5. The electric flux density is related to the electric field intensity as D =  0 E .

6. Gauss’s law   D =  v
7. Gauss’s law is useful when the charge distribution is symmetric.
8. The electric potential VAB is defined as the work done to move a point
charge from point A to B.
9. The electrostatic field is conservative.

 
 E  dl = 0 or   E = 0
L
Summary (ii)
10. If the potential field is given, the corresponding electric field is found by

using E = −V.
11. If the charge distribution is not known, but the field intensity is given, we
 
find the potential by using V = −  E  dl + C .
L
12. The potential induced by an electric dipole
  

p  (r − r )
V (r ) =
 3
4 0 r − r 
13. The flux density is tangential to the electric flux lines at every point.
14. An equipotential line (or surface) is orthogonal to the electric flux line.
15. The electrostatic energy by N discrete charges
and continuous volume charge distribution
1
2
(
1
WE =
2
)
N
Q V
k
k =1
 
D  E dv
v
k