Chapter 2 Electrostatics
2.0 New Notations
2.1 The Electrostatic Field
2.2 Divergence and Curl of Electrostatic Field
2.3 Electric Potential
2.4 Work and Energy in Electrostatics
2.5 Conductors
2.0 New Notations
R

2.1 The Electrostatic Field
2.1.1 Introduction
2.1.2 Coulomb’s Law
2.1.3 The Electric Field
2.1.4 Continuous Charge Distribution
2.1.1 Introduction

R
The fundamental problem for electromagnetic to solve is to
calculate the interaction of charges in a given configuration.
That is, what force do they exert on another charge Q ?
The simplest case is that the source charges are stationary.
Principle of Superposition:
The interaction between any two charges is completely unaffected
by the presence of other charges.
F  F1  F2  F3 
Fi is the force on Q due to qi
2.1.2 Coulomb’s Law
The force on a charge Q due to a single point charge q is given
by Coulomb`s law
1
qQ ˆ
F
R
2
4 0 R
R  rQ  rq  R Rˆ
 0  8.85  1012
c2
N  m2
the permittivity of free space
2.1.3 The Electric Field
F  F1  F2  
 q1Q

q
Q
1
2

 2 Rˆ1  2 Rˆ 2   
4 0 R
R2
 1


q3 ˆ
Q  q1 ˆ q2 ˆ

 2 R1  2 R2  2 R3   
4 0 R
R2
R3
 1

F  QE
N
qi ˆ
E ( P) 
Ri

2
4 0
i 1 R1
1
the electric field of the source charge
2.1.4 Continuous Charge Distributions
n
 ( ) qi
i 1
~
 ( )  dl ~  ( )  da ~  ( )  d
volume
surface
line
qi ˆ
E  P 
Ri

2
4 0
i 1 Ri
ˆ
R
E  P 
d

2
4 0
line R
ˆ
R
E  P 
 da

2
4 0
surface R
ˆ
R
E  P 
 d

2
4 0
volume R
1
1
N
1
1
2.1.4 (2)
Example:
the test point is at p   x, y, z  the source point is at  x, y, z 
R   x  x  iˆ   y  y  ˆj   z  z   kˆ
2
2
2
R   x  x    y  y    z  z   


E  x, y , z   1
4 0


volume
1
R
Rˆ 
R
2
 x  x  iˆ   y  y  ˆj   z  z  kˆ
[
2
2
2 32
x  x  y  y   z  z  ]
 
 

 ( x, y, z) dx dy dz
ˆ
R
, y, z )dxdydz 
E ( x, y , z ) 

(
x

4 0 volume R 2
ˆ R
1
R
, y, z)dxdydz



(
x

4 0 volume R 2 R
R  ( x, y, z )dxdydz 
 1 
4 0 volume R3
1
2.1.4 (3)
Example 2.1 Find the electric field a distance z above the midpoint
of a straight of length 2L, which carries a uniform line charge 
Solution:
dE  2
E
1
(1) z>>L
 dx
4 0 R
L
1
4 0 0
2 z

4 0

(
2
) cos  zˆ
2 z
(z  x )
2
2 3 2
dx
L


 z 2


z 2  x 2  0
R
x
2 L
1
4 0 z z 2  L2
E
1
2 L
4 0 z
2
(2)
L
2
E
4 0 z
1
2.2 Divergence and Curl of Electrostatic Fields
2.2.1 Field Lines and Gauss’s Law
2.2.2 The Divergence of E
2.2.3 Application of Gauss’s Law
2.2.4 The Curl of E
2.2.1 Fields lines and Gauss’s law
A single point charge q, situated at the origin
E (r ) 
1
q
4 0 r
2
rˆ
2
Because the field falls off like1 r ,the vectors get shorter as I go
father away from the origin,and they always point radially outward.
This vectors can be connect up the arrows to form the field lines.
The magnitude of the field is indicated by the density of the lines.
strength length of arrow
strength density of field line
2.2.1 (2)
1.Field lines emanate from a point charge symmetrically in all
directions.
2.Field lines originate on positive charges and terminate on
negative ones.
3.They cannot simply stop in midair, though they may extend
out to infinity.
4.Field lines can never cross.
2.2.1 (3)
Since in this model the fields strength is proportional to the
number of lines per unit area, the flux of E (  E  da ) is
proportional to the the number of field lines passing through
any surface .
The flux of E through a sphere of radius r is:
1
q
 E  da   4 0 ( r 2 rˆ)  (r
2
sin  d d rˆ) 
1
0
q
The flux through any surface enclosing the charge is q  0
According to the principle of superposition, the total field is
the sum of all the individual fields:
n
E   Ei
i 1
,
n
n
i 1
i 1
1
 E  da   (  Ei  da )   (  0 qi )
A charge outside the surface would contribute nothing to the
total flux,since its field lines go in one side and out other.
2.2.1 (4)
Gauss’s Law in integral form
1
 E  da   0 Qenc
Turn integral form into a differential one , by applying the
divergence theorem

surface

E  da 
(  E )d
volume

1
0
Qenc 

volume
(
1
0
 )d
Gauss’s law in differential form   E 
1
0

2.2.2 The Divergence of E
Calculate the divergence of E directly
E 
Rˆ
1
  ( 2 )  (r )d

4 0
R
,
R  r  r
The r-dependence is contained in R
E (r ) 
1
4 0
From
(
 E 
1
all space R
)d 

(
r
2
Rˆ
R
Thus

Rˆ
3
)

4

( R)
2
4

4 0
3
(r  r)  (r)d  
1
0
 (r )
2.2.3 Application of Gauss’s Law
Example 2.2
Sol:

Find the field outside a uniformly charged sphere
of radius a
E  da 
surface
1
0
Qenc

(a) E point radially outward ,as does d a

E  da   E da
surface
(b)E is constant over the Gaussian surface

surface
Thus

E da  E
da  E 4 r 2
surface
E 4 r 
2
1
0
q
E
1
q
4 0 r
2
rˆ
2.2.3 (2)
1. Spherical symmetry.
Make your Gaussian surface a
concentric sphere (Fig 2.18)
2. Cylindrical symmetry. Make your Gaussian surface a
coaxial cylinder (Fig 2.19)
3. Plane symmetry.
Use a Gaussian surface a coaxial
the surface (Fig 2.20)
2.2.3 (3)
Example 2.3 Find the electric field inside the cylinder which
contains charge density as   kr
Solution:

surface
E  da 
1
0
Qenc
The enclosed charge is
2
Qenc    d   (kr )(r dr d dz )  2 k l  r dr    klr 3
0
3
(by symmetry)
 E  da   E da  E  da  E 2 rl
r
thus
1 2
E 2 rl 
 klr 3
0 3
2
1
E
kr 2rˆ
3 0
2.2.3 (4)
Example 2.4 An infinite plane carries a uniform surface charge. 
Find its electric field.
Solution: Draw a ”Gaussian pillbox
Apply Gauss’s law to this surface

E  da 
surface
1
0
Qenc
By symmetry, E points away from the plane
thus, the top and bottom surfaces yields
 E  da  2 A E
(the sides contribute nothing)
2A E 
1
0
A

E
nˆ
2 0
2.2.3 (5)
Example 2.5 Two infinite parallel planes carry equal but opposite
uniform charge densities  .Find the field in
each of the three regions.
Solution:
The field is (σ/ε0 ), and points to the right, between the plane
elsewhere it is zero.
(
(
2 0
2 0
) (
2 0
) ( 2 )
0
) ( 2 ) ( 2 )
0
0
2.2.4 The Curl of E
A point charge at the origin. If we calculate
the line integral of this field from some point
a to some other point b :
b
a E  dl
In spherical coordinate
dl  dr rˆ  rd ˆ  r sin  d ˆ
b
1
b
a E  dl  4 0 a
E  dl 
1 q
dr 
2
4 0 r
r
q
rb
ra
1
q
dr
4 0 r
1 q q

  
4 0  ra rb 
2
This integral is independent of path. It depends on the
two end points
 E  dl  0
( if ra = rb )
Apply by Stokes’ theorem,  E  0
2.2.4
The principle of superposition
E  E1  E2 
so
 E   (E1  E2  )  ( E1)  ( E2 ) 
0
 E  0
must hold for any static charge distribution whatever.
2.3 Electric potential
2.3.1 Introduction to Potential
2.3.2 Comments on Potential
2.3.3 Poisson’s Equation and Laplace’s Equation
2.3.4 The Potential of a Localized Charge Distribution
2.3.5 Electrostatic Boundary Conditions
2.3.1 introduction to potential
Any vector whose curl is zero is equal to the gradient of some
scalar. We define a function:
p
V ( p)    E  dl

Where is some standard reference point ; V depends only on the
point P. V is called the electric potential.
a

V (b)  V (a )    E  dl     E  dl

 
b
   b E  dl

a

The fundamental theorem for gradients
b
V (b)  V (a)   (V )  dl
a
so
b
b
 a (V )  dl   a E  dl
E  V
2.3.2 Comments on potential
(1)The name
Potential is not potential energy
F  qE  qV
V : Joule/coulomb
U  F  X
U : Joule
2.3.2 (2)
(2)Advantage of the potential formulation
V is a scalar function, but E is a vector quantity
V (r )
E  E x xˆ  E y yˆ  E z zˆ
If you know V, you can easily get E: E  V .
Ex , E y , Ez are not independent functions
 E  0
so
1
E x 1 E y

,
y
x
E z 2 E y

,
y
z
2
E x 3 E z

z
x
 y  z Ex   z  y Ex   z  x E y   x z E y   x y Ez   y x Ez
3
 z Ex   x Ez
2.3.2 (3)
(3)The reference point 
Changing the reference point amounts to adds a constant to
the potential
p

p
V ( p)   E  dl   E  dl  E  dl  K  V ( p)






(Where K is a constant)
Adding a constant to V will not affect the potential difference
between two point:
V (b)  V (a )  V (b)  V (a )
Since the derivative of a constant is zero: V  V 
For the different V, the field E remains the same.
Ordinarily we set V ()  0
2.3.2 (4)
(4)Potential obeys the superposition principle
F  F1  F2 
,
F  QE
Dividing through by Q
E  E1  E2 
Integrating from the common reference point to p ,
V  V1  V2 
(5)Unit of potential
Volt=Joule/Coulomb
F : newton
F  x : Joule
qV
F  qE  qV 
X
FX
V:
 Joule/Coulomb
q
2.3.2 (5)
Example 2.6 Find the potential inside and outside a spherical
shell of radius R, which carries a uniform surface
charge (the total charge is q).
solution:
Eout 
Ein  0
q
ˆ
r
2
4 0 r
1
for r>R:
q
1
V  r    E  d 

4 0 r 
r
r
q
1

4 0 r

_
for r<R:
V  R 
q
4 0 R
1
q
V r   V  R 
4 0 R
1
Ein  0
2.3.3 Poisson’s Eq. & Laplace’s Eq.
  E       V   2V
0
E   V
Poisson’s Eq.  2V  
 0
2V  0

0
Laplace’s eq.
2.3.4 The Potential of a Localized Charge Distribution
r
V  V    Edr 
E  V

V (r )  1
4 0
q
1 q

dr

  r 2
4 0 r 
r
r

V  0
q
 1
4 0 r
• Potential for a point charge
q
V ( P) 
4 0 R
1
R
R  r  rp
• Potential for a collection of charge
n
qi
V ( P) 
4 0  Ri
i 1
1
Ri  ri  rp
Ri
2.3.4 (2)
• Potential of a continuous distribution
for volume charge
 q   d
V ( P) 

4 0  R
1
d
for a line charge
for a surface charge
 q  d
 q   da
V ( P) 
• Corresponding electric field
E ( P) 
d
4 0  R
1
ˆ
1
R
[  
R R2
rˆ  d
rˆ  d E ( P)  1
4 0  R 2
4 0  R 2
1
V ( P) 
 da
4 0  R
1
]
E ( P)  1  rˆ2 da
4 0 R
2.3.4 (3)
Example 2.7 Find the potential of a uniformly charged spherical
shell of radius R.
Solution:
1

2
2
2
r

V (r ) 
da
,

r

R

z

2
Rz
cos

4 0  r
R 2 sin  d d 
4 V ( z )  
R 2  z 2  2 Rz cos  

 2 R 
2
 2 R 2
2 R
z
 2 R
z



0
( Rz1
sin  
R 2  z 2  2 Rz cos  
R 2  z 2  2 Rz cos  
d 
)

0
(
R 2  z 2  2 Rz  R 2  z 2  2 Rz )
[
( R  z )2  ( R  z )2
]
2.3.4 (4)
R
R 2
V ( z) 
[( R  z )  ( z  R)] 
, outside
2 0 z
0z
R
R
V ( z) 
[( R  z )  ( R  z )] 
, inside
2 0 z
0
2.3.5 Electrostatic Boundary Condition
Electrostatic problem
•Superposition
•Coulomb law
The above equations are differential or integral.
For a unique solution, we need boundary conditions. (e.q. , V()=0 )
(boundary value problem. Dynamics: initial value problem.)
2.3.5 (2)
B.C. at surface with charge
:
surface
=

A
Q
E  da  enc  
0
0
E above  A  E below  A
Eabove  Ebelow  
0
∥:
 E  dl  0

  E  0
E‖ above= E‖ below
 +∥
Eabove  Ebelow   nˆ
0
2.3.5 (3)
Potential B.C.
Vabove  Vbelow   
b
a
ab  0
E  d  0
 Vabove  Vbelow
Eabove  Ebelow   nˆ
E  V
0
 n Vabove  n Vbelow   
0
 V   V  nˆ
n
2.4 Work and Energy in Electrostatics
2.4.1 The Work Done in Moving a Charge
2.4.2 The Energy of a Point Charge Distribution
2.4.3 The Energy of a Continuous Charge Distribution
2.4.4 Comments on Electrostatic Energy
2.4.1 The Work Done in Moving a charge
A test charge Q feels a force Q E (from the stationary source charges).
To move this test charge, we have to apply a force F  QE
conservative
The total work we do is
W   F  d  Q  E  d  Q V  b   V  a  
b
b
a
a
V b   V  a   W
Q
So, bring a charge from  to P, the work we do is
W  Q V ( P)  V ()  QV ( P)
V ()  0
2.4.2 The Energy of a Point Charge Distribution
It takes no work to bring in first charges
W1  0
for q1
Work needed to bring in q2 is :
q1
q1
1
1
W2  q2 [
]
q2 (
)
4 0 R12
4 0
R12
Work needed to bring in q3 is :
q1
q2
q1
q2
1
1
1
W3  q3[

]
q3 (

)
4 0 R13 4 0 R23
4 0
R13 R23
Work needed to bring in q4 is :
q3
q1
q2
1
W4 
q4 [


]
4 0
R14 R24 R34
2.4.2
Total work
W=W1+ W2+ W3 +W4
 q1q2 q1q3 q2q3 q1q4 q2q4 q3q4 
1







4 0  R12
R13
R23
R14
R24
R34 
n n qq
i j
 1 
4 0
Rij
Rij  R ji
i 1 j 1
n
n
qi q j
1
8 0   Rij
i 1 j 1
j i
n
W  1  qiV  Pi 
2
i 1
 n

q
j

 1  qi   1
2
4 0 Rij 

i 1  j 1

j i
n
n
qj
V  Pi   
4 0 Rij
1
j 1
j i
Dose not include the first charge
2.4.3 The Energy of a Continuous Charge Distribution
 q   d
1
W    Vdτ
2
Volume charge density
   EV      E V  E   V 
ρ   0  E
0
  E Vdτ 


2

0 



EV
d


E


V
d







2 
F.T.for  
E
 VE  da
0
2
surface   
surface
(


VE  da 
surface
W
E 2d )
volume
0
2

all space
E 2 dτ
2.4.3
Example 2.8 Find the energy of a uniformly charged spherical
shell of total charge q and radius R
1
q
Sol.1： q  4 R 
V
4 0 R
1
1
1 q 1 q
1 q2
W   Vd   Vda  
da 
2
2
2 A 4 0 R
8 0 R
2
Sol.2：E 
W
1
q
4 0 r 2
0
2
 E d 
q2
2
E2 
rˆ
0
outer
q2
(4 0 )2 r 4
q2
2 space (4 
o)
1
2
r
4

 1
1 q2



2  2  2 dr  
2 
R r
32  0 
 8 0 R
r 2 sin  d d dr

2.4.4 Comments on Electrostatic Energy
1 A perplexing inconsistent
W
0
2 all space
E 2d
 Energy  0
or
1 n
W   qiV ( Pi )
2 i 1
W
1
4 0
n
n
 
 Energy  0 or  0
qi q j
i 1 j 1
j i
rij
Only for point charge
q1  q , q2   q  W 
1
4 0
 0
q 2
r12
( q1 and q2 ) are attractive
2.4.4 (2)
W
0
2
all space
E 2dτ
is more complete
the energy of a point charge itself,


 1
0
q2
q2 2
W
( ) r sin θdθdφdr 
dr  

2 4
2
8 π 0 0 r
2(4 0 )
r
1 n
W   qi V  Pi 
2 i 1
1
W    Vd
2
, V ( Pi )
does not include qi
, V ( P)
is the full potential
There is no distinction for a continuous distribution,
because
d  0
 (d ) p
0
2.4.4 (3)
(2)Where is the energy stored? In charge or in field ?
Both are fine in ES. But,it is useful to regard the energy
as being stored in the field at a density
E2
 Energy per unit volume
2
0
(3)The superposition principle,not for ES energy
W1 
0
Wtot 

2

0
2
0
E12d
W2 
0
2
2
E
 2 dτ
2
(
E

E
)
 1 2 d
2
2
(
E

E
 1 2  2E1  E2 )d
2
 W1  W2   0  ( E1  E2 )d
2.5 Conductors
2.5.1 Basic Properties of Conductors
2.5.2 Induced Charges
2.5.3 The Surface Charge on a Conductor;
the Force on a Surface Charge
2.5.4 Capacitors
2.5.1 Basic Properties of Conductors
e  are free to move in a conductor
(1) E  0 inside a conductor
otherwise, the free charges that produce E
will move to make E  0 inside a conductor
(2) ρ = 0 inside a conductor

E 
0
E 0

 0
(3) Any net charge resides on the surface
(4) V is constant, throughout a conductor.
b
V (b)  V (a )    E  dl  0
a
V (b)  V (a)
2.5.1 (2)
(5) E is perpendicular to the surface, just outside a conductor.
Otherwise, E will move the free charge to make E  0
in terms of energy, free charges staying on the surface have
a minimum energy.
uniform 
1 q2
Energy 
8 0 R
Ein  0 Win  0
uniform 
Ein  0 Win  0
3 q2
Energy 
20 0 R
2.5.1 (3)
Example : A point charge q at the center of a spherical conducting
shell. How much induced charge will accumulate there?
Solution :
Ein  0
q induced
1 q
a  
4 a 2
 4 a   a  q
2
charge conservation
4 b   b
Ein  0
2
 4 a 2
a
Qenc  q  qinduced  0
qinduecd  q
1 q
b 
4 b 2
2.5.2 Induced Charge
Example 2.9 Within the cavity is a charge +q.
What is the field outside the sphere?
-q distributes to shield q and to make Ein  0
i.e., Vsurface  constant
from charge conservation and symmetry.
+q uniformly distributes at the surface
 E ( P) 
1
q
4 0 r
2
rˆ
 E  dl  0
b
 a E  dl  0
Eincavity  0
a,b are arbitrary chosen
 a Faraday cage can shield out stray E
2.5.3 The Surface Charge on a Conductor

Eabove  Ebelow  nˆ
0
Ein  0 or Ebelow  0

V  Eabove  nˆ
0
or
if we know V, we can get .
V


n
0
2.5.3
Force on a surface charge,
f  Eaverage
1
  ( Eabove  Ebelow )
2
why the average?
 ˆ
Eabove  Eother 
k
2 0
Eother
 ˆ
Ebelow  Eother 
k
2 0
 0
1
 ( Eabove  Ebelow )

Eaverage
2
In case of a conductor
electrostatic pressure
1
2
f   Eabove 
nˆ
2
2 0
 2 0 2
P
 E
2 0 2
2.5.4 Capacitors
Consider 2 conductors (Fig 2.53)
The potential difference

V  V  V  
E  dl

E
1
rˆ

 d

2
4 0 r

(V is constant.)
double   double Q  double E  double V
Define the ratio between Q and V to be capacitance
Q
C
V
a geometrical quantity
in mks 1 farad(F)= 1 Coulomb / volt
inconveniently large ;
106 F : microfarad
1012 F : picofarad
2.5.4 (2)
Example 2.10 Find the capacitance of a “parallel-plate capacitor”?
+Q
d  A
Solution:
-Q

Q
E

 0 A 0
Q
V  E d 
d
A 0
A 0
C
d
2.5.4 (3)
Example 2.11 Find capacitance of two concentric spherical shells
with radii a and b .
+Q -Q
Solution:
E
1
Q
4 0 r
a
ˆ
r
2
V    E  dl  
b
Q
a
4 0 b
Q
ab
C   4 0
V
b  a 
Q 1 1
dr 
  
2
4 0  a b 
r
1
2.5.4 (4)
The work to charge up a capacitor
q
dW  Vdq  ( )dq
C
W 
Q
0
1 Q2 1
1
q
( )dq 
 QV  CV 2
C
2 C 2
2