Electric Field Theory and
Applications
ENGG1310
2nd semester, 2018-19
Prof. Kenneth Kin-Yip Wong
Department of Electrical and Electronic Engineering
Faculty of Engineering
The University of Hong Kong
Today

Winner applications

Basic properties of electric charges

Coulomb’s law and examples

Gauss’s law and examples

Electric scalar potential

Capacitance and examples
ENGG1310 – Electricity and Electronics
Electric Field Theory and Applications
2
“Nano-ultracapacitors”
Density: 1500m2/g
i.e. 250g = 375,000m2 or
roughly 50 soccer fields.
Q
A
C  
V
d
[IEEE Spectrum, pp. 38 – 42, Nov., 2007]
3
“Nano-ultracapacitors”
Typical capacitor
[IEEE Spectrum, pp. 38 – 42, Nov., 2007]
4
“Nano-ultracapacitors”
Ultracapacitor
Nano-ultracapacitor
[IEEE Spectrum, pp. 38 – 42, Nov., 2007]
5
“Nano-ultracapacitors”
Charge & discharge
[IEEE Spectrum, pp. 38 – 42, Nov., 2007]
6
“Bioelectrics”
[IEEE Spectrum, pp. 18 – 24, Aug., 2006]
7
“Bioelectrics”
[IEEE Spectrum, pp. 18 – 24, Aug., 2006]
Sem 1 2018-19
ENGG1310 – Prof. K. Wong
8
“Bioelectrics”
[Before]
[After]
[IEEE Spectrum, pp. 18 – 24, Aug., 2006]
9
Basic Properties of Electric Charges

Two kinds of charges : positive and negative.

Like charges repel, unlike charges attract.

Positive and negative charges occur in exactly the same
amounts.

In any isolated system, the algebraic sum of the charges is
constant (conservation of charge).

Charge is quantized, i.e. charges come only in discrete
packets which are integer multiples of the basic unit of
charge : the charge on the electron is one unit of negative
charge (-e).
10
Basic Properties of Electric Charges

In the SI units (International System of Units):
•
•
•
•
•
Length
Mass
Time
Temperature
Current
meter (m)
kilogram (kg)
second (s)
Kelvin (K)
ampere (A)

(Defined as the current that flows in opposite directions in two
straight parallel conductors of infinite length and negligible cross
section, separated 1 meter in vacuum, and would produce a repulsive
force of 210-7 newtons per meter length between the two
conductors).

As current = rate of charge flow, this in turn defines
• Charge coulomb (C)
11
Coulomb’s Law

The force F on a point charge Q due to a single point
charge q which is at rest a distance r away is given by
1
qQ
F
r
2
4 o r
q
+
r
Q
+
F
where r is the unit vector along the corresponding
direction and o is permittivity of free space and is given
by
 o  8.854 1012
C2
N.m 2
12
Coulomb’s Law (cont’d)

If there are many point charges q1, q2, …….., qn located at
distances r1, r2……., rn from Q, then according to the
principle of superposition,
qnQ 
q2Q
1  q1Q
F  F1  F2  ... Fn 
 2 rˆ1  2 rˆ2  ....... 2 rˆn 
4 o  r1
r2
rn



or F  QE where E  E( P ) 
1
4 o
n

i 1
qi
rˆi
2
ri
N/C
The vector E is called the electric field (intensity) of the
source charges and is a function of position P = (x,y,z). E(P)
may be defined as the force per unit charge that would be
exerted on the test charge Q placed at P.
13
Example 1

Find the electric field a distance a above the midpoint of
two equal positive charges q at distance d apart:

The horizontal components of the fields from the two
charges cancel, so the net field is vertical as shown.
14
Example 1 (answer)
E2
q
4 o r
2
cos  a z where az is the unit vector along z
r  a   d / 2
2
E

2
cos   ar
and
2aq
3
2
az
2
2

4 o a   d / 2  


2q
a , as the two charges
Note that for a >> d, E 
2 z
4 o a
look more and more like a single charge of 2q.
15
Example 2

Do the above example again, but with the charges of
opposite sign.

Now the vertical components cancel and the net electric
field is horizontal
16
Example 2 (answer)
E2
q
4 0 r
where
sin

a
x
2
r  a   d / 2
2
E
2
and
qd
2
2

4 o a   d / 2  



d /2
sin   r
Note that for a >> d, E 
3
2
ax
qd
4 o a
3
a x , which is the field of
a dipole, and as d  0 the electric field approaches 0, since the
two charges cancel each other.
17
Continuous Charge Distribution

Very often, we have to consider charges that are
distributed continuously over some region.
• line charge  (charge per unit length)
• surface charge  (charge per unit area)
• volume charge  (charge per unit volume)

Thus, qi is replaced by dq   dl ,  da, or  d
respectively and summation is replaced by integration:
n
q
i 1
i
~
 dl ~ 
line
surface
 da ~

 d
volume
18
Continuous Charge Distribution

The electric field of a line charge is
rˆ
E( P ) 
 dl
2

4 o line r
1
where r̂ is the appropriate vector

For a surface charge,
rˆ
E( P ) 
 da
2

4 o surface r
1

For a volume charge,
rˆ
E( P ) 
 d
2

4 o volume r
1
19
Example 3

What is the electric field at a distance a above a line
segment of length 2L, carrying an electric charge of 
coulombs/m.
20
Example 3 (answer)

To solve this, divide the line up into 2 symmetrical lengths on
either side of the point P (from suppl. notes):
E

1
2 L
4 o a a  L
2
2
az
2 L
a
Note that for a >> L, we have E 
2 z , which
4 o a
corresponds to the field from a point charge of q = 2L at a
distance a.

Also as, L  ,
2

E

4 o a 2 o a
which is the field at a distance a from a very long straight line.
21
Example 4

What is the electric field at a distance a above a circular loop
of radius b, carrying an electric charge of  coulombs/m?

At P, the horizontal components of the field from an
elemental length dl cancel - so we only have to consider the
vertical components only (from suppl. notes):
Ez 
1
 2 ab
az
4 o (a 2  b 2 )
1
 ab

az
2
2 32
2 o (a  b )

3
2
What if it is a disc instead of a loop?
22
Example 5


What is the electric field at a distance a above a flat circular disc
of radius b, carrying a surface electric charge of  coulombs/m2 ?
dEz
Consider a charged ring:  .2 r.dr   .2 r
dE z 
1  dr  2 r  a
4 o

a2  r 2

3
2
az
1
1
Ez 
2 a  
2
2
4 o
a
a

b

1
P

 az

a
dr
r
b
2

lim E z 
az 
az
a charged plane (later). b 
4 o
2 o

Note that for b >> a:

What happens to Ez for a >> b? (suppl. notes)
23
24
Field Lines (Stream Lines)

An attempt to give a visual indication of field intensity E.
Direction of field:
tangential to the field lines, with an arrow head
indicating the direction.
Magnitude of field: longer line, thicker line, colour scheme etc. No
simple solution. However, if lines are drawn
uniformly taking into account symmetry, then the
magnitude of the electric field is indicated by the
density of the field lines.
25
Electric Flux

Experiments show that a charge +Q in an inner sphere will
always result in an equal and opposite charge –Q (by
induction) on the outer sphere as though there are imaginary
lines of flux flowing from the inner sphere to the outer
sphere.
Hence introduce concept of flux lines or electric flux 
associated with electric charges such that the direction of
flux lines is the same as the direction of field intensity E at
that point and the amount of flux is proportional to the
charge. In SI units, the proportional constant is 1, i.e.
=Q
The flux  is measured in the same unit as Q, i.e. in
coulombs.
26
Electric Flux (cont’d)

The electric flux may be represented by flux lines with the
following properties:
• They are directional – originating and diverging from positive charges
and converging towards and terminating at negative charges.
• They are elastic and tend towards minimum length (imagine that they
are in tension along their lengths)
• The lines repel each other and they can never cross each other.
• The density of the electric flux is proportional to the density of lines.
From the definition of electric flux, it is obvious that over any
closed surface,
Total flux coming out = Total (+ve) charge enclosed
27
Flux Density D

Corresponding to the electric flux, define electric flux density D
where the magnitude
flux flow perpendicular to ds as ds  0
D
area ds
And direction of D is the same as that of E at the same point. Since
the flux flow perpendicular to the area ds may be represented by,
d = D.ds
where ds = ds n, n being the unit outward normal vector, hence
flux over close surface
=

D.ds = Charge Q enclosed. And this is Gauss’s Law.
surface
28
Gauss’s Law

Consider a point source Q. By symmetry, the flux flow is uniform
and normal to a spherical surface. Hence to determine the flux
density D at a point distance r from the source, construct a
spherical surface of radius r. Then
Flux   Q 

D.ds  D  4 r
D(r)
2
surface
Or
Q
D
 ar
2
4 r
r
Compare with the E at the same point
E
Q
4 0 r
2
 ar
Hence, D = 0E
29
Gauss’s Law (cont’d)

Since for line, surface or volume charges, D and E can be
calculated by superposition in a similar way, the above
relationship always hold.
Gauss’s Law can therefore also be written as
Flux  

 E.ds  Q
surface
This is generally referred to as Gauss’s Law in integral form.
30
Application of Gauss’s Law

The Gauss’s Law provides us an alternative way of finding
the electric field intensity E through the flux density D.
To determine D if charge distribution is known, choose a
closed surface S such that
• D is everywhere either normal or tangential to the closed surface, so
that D.ds becomes either Dds or zero respectively.
• On that portion of the closed surface for which D.ds is not zero,
D = constant.
To achieve this, it is necessary to take advantage of the
symmetry that exists in the system.
31
Application of Gauss’s Law (cont’d)

Examples of situations where symmetry exists:
• A charged sphere or point charge
• An infinite line with uniform line charge density (e.g. coaxial cable)
• An infinite plane with uniform surface charge density (e.g. parallel
plate capacitor)
32
Example 1 – Spherical Symmetry

Find the field outside a uniformly charged sphere (with
total charge q) of radius a.

For r > a:
The Gaussian surface is a sphere of radius r (r > a).
 E  ds 
S
1 q
2
E
ds

E
ds

E
4

r



0
S
S
q
1
 E
rˆ  2
2
4 0 r
r
Note that the result shows that the electric field outside a
charged sphere is the same as if the total charge q had
been concentrated at the center.
33
Example 1 (cont’d)

For r < a:
3
3

r
r
1 q
1 q
2
E

d
s

E
ds

E
ds

E
4

r


S
S
S
0
 a3  0 a3
o
q
 E
rrˆ 
rrˆ  r
3
4 o a
3 o
4
3
4
3
where 0 is the volume charge density.
34
Example 2 – Cylindrical Symmetry

Find the field outside a long wire of uniform line charge
density  and radius ro.
First construct a Gaussian surface: a cylinder of radius r.
Gauss’s Law says:

S
E  ds  1 Qenclosed
0
where Qenclosed = l
35
Example 2 (cont’d)

Because of the symmetry, it is clear that E  E (r )rˆ only, Ex  0.
Also E and ds are pointing in the same direction, and it is also
clear that E is uniform over the Gaussian surface.

For r > ro:
l
1
 E  ds   E ds  E  ds  E 2 rl   0 Qenclosed   0
S
S
S

 E
rˆ ,
2 o r
i.e. E 
1
r
36
Example 2 (cont’d)

For r < ro:
 E  ds 
1Q
E
ds

E
ds

E
2

rl



 enclosed
S
S
0
S
r
 E
rˆ,
2
2 o ro
1 r2

l
2
 o ro
i.e. E  r
37
Example 3 – Planar Symmetry

Find the electric field of an infinite plane carrying a uniform
surface charge density .
Draw a Gaussian surface, in this case a “pillbox” extending equal
distances above and below the surface of the plane.
1Q
1
E

d
s

E
ds

E
2
A






 enclosed 
S
S
o
o
1 A

ds



s
o
where Qenclosed = A and A is the area of top of the pillbox.
38
Example 3 (cont’d)

Why the 2 in E(2A)?
Because the charge Q = A produces an electric field pointing
up and an electric field pointing down. So the effect is spread
over 2A.

 E
n
2 0
where n is a unit vector normal to the plane.
Compare this with the example that deals with the field from
a charged disk.
39
Example 4 – Another Planar Symmetry

Two infinite parallel planes carry equal but opposite uniform
surface charge densities, + and . What is the electric field
between the plates and outside the plates ?
40
Example 4 (cont’d)
On the right hand side of the negative charge plane,


E E E 
( i ) 
(i )  0
2 o
2 o


On the left hand side of the positive charge plane,


E E E 
(i ) 
( i )  0
2 o
2 o





Inside the plates:
E E E 
(i ) 
(i )  i
2 o
2 o
o


41
42
Electric Scalar Potential

Consider moving an elemental test charge qt a distance dL in
an electric field represented by the field intensity E.
Force on qt
F = qt E
E
The work done by the field is:
+
Fappl qt
F
dWField = F.dL = qt E.dL
Alternately external work done on qt is
dWext = - F.dL = - qt E.dL
Total external work done to move qt from B to A is
A
W  qt  E.dL
A (final)
B (initital)
E
B
Fappl
+
dL
dL
+
43
Electric Scalar Potential (cont’d)

Consider a point source Q
E
Q
4 0 r
2
ar
In spherical coordinates
dL = dr ar + rd a + r sin  d a
Hence
A
W  qt  E.dL  qt 
B
A
B
Q
4 0 r 2
qt Q  1 1 
dr 
  
4 0  rA rB 
and is independent of path chosen. Since E  Q and the principle of
superposition applies, the same is true for any field.
44
Example

Given the non-uniform field
E = y ax + x ay
Determine the work done in carrying a charge of 2C from
B(1,0) to A(0.8, 0.6) along
1.
2.
3.

the arc of circle x2 + y2 = 1,
a straight line path from B to A,
from B(1,0) to C(1, 0.6) along x = 1 and then from C(1, 0.6) to
A(0.8, 0.6) along y = 0.6.
In Cartesian coordinates, dL = dx ax + dy ay
A
W  Q  E.dL  2 
B
 2 
0.8
1
A
B
ydx  2 
0.6
0
 y a
x
 x  a y    dx  a x  dy  a y 
xdy
45
Example (cont’d)
1.
Along arc of circle x 2  y 2  1,
W  2 
0.8
1
1 x
2
dx  2 
0.6
0
x  1  y2 ; y  1  x2 ;
1  y 2 dy
0.8
0.6
   x 1  x  sin x    y 1  y  sin y   0.96 J

1 
0
2
2.
1
The line joining AB may be represented by y = -3x + 3
Hence
W  2 
0.8
1
ydx  2 
0.6
0
xdy
y

 2  3  x  1 dx  2  1   dy  0.96 J
1
0
0.6 3 
From B to C along x = 1,
WBC  2  1 dy  1.2 J
0.8

1
2
0.6
From C to A along y = 0.6,
Hence from B to A,
0 0.8
WCA  2  0.6 dx  0.24 J
1
WBA  WBC  WCA  0.96 J
46
Electric Scalar Potential (cont’d)

Since the work done is independent on the path but only on the
end-points, we can therefore define uniquely a potential difference
between two points A and B as the work done in moving a unit
positive charge from point B to point A, i.e.
A
VAB    E.dL
joules/coulomb
B
Note that since VAB = -VBA, therefore
 E.dL  
A
B
B
E.dL   E.dL  0
A
and E is described to be a conservative field.

If point B is an agreed ‘reference point of zero potential’, then VA =
VAB can be regarded as the (absolute) potential at point A. If VP, VQ
are the potentials at points P and Q with respect to the same
reference point, then VPQ = VP - VQ
47
Electric Scalar Potential (cont’d)

Commonly used reference point of zero potential:
• Ground plane
• Point at infinity
• Metal shielding
Potentials due to different charge sources can be added together by
superposition.
Thus if the reference point of zero potential is at infinity, then
For isolated point charges Qi,
V 
i
For volume charge density ,
Qi
4 0 r i
 d
V 
4 0 r
48
Example 1

Find the potential (reference point at infinity) inside and
outside a spherical sphere of radius a carrying a uniform
surface charge  coulombs/m2.
From Gauss’s Law, the field outside the charged sphere is:
2
q
where
q

4

a

E
a
4 o r
r
Vr    E.dr

4 o
r
where the path of integration is along r

q
2
r


1
q 1
dr 
2
r
4 o r
r


q
4 o r
49
Example 1 (cont’d)

In particular, the potential at the surface is given by:
Va 
q
4 o a
The field E inside the sphere is everywhere zero since
there is no charge inside the sphere. The potential inside
the sphere is therefore equal to the potential at the
surface, i.e.
Vr 
q
for r < a
4 o a
50
Example 2

Find the potential on the axis of a charged ring.
Take point at infinity as zero potential.
For the charge element dL, V   dL
4 0 r
Hence for the whole ring,
 dL

V 

4 0 R 4 0 R
ring

 2 a
ring dL  4 0 R
Q
4 0 z 2  a 2
since Q = 2a
51
Example 3

Find the potential difference between the points A(A, A, zA)
and B(B, B, zB) due to an infinite line charge with density .
In cylindrical coordinates,

E
 a
2 0 
dL = d a + d a + dz az
A
VAB    E.dL   
B
A
B

d
2 0 
B


ln
2 0  A
Note that in this case it is not possible to set a point at infinity
to zero potential since as B  , ln B  .
52
Equipotential Surfaces

A surface in an electrostatic field having the same potential at
all points is called an equipotential surface.
Since all points of an equipotential surface are at the same
potential, for two close points on the surface,
dV = E.dL = 0,
i.e.
E  dL,
i.e. the electric field intensity E at any equipotential surface is
normal to that surface.
53
Equipotential Surfaces (cont’d)

Examples of equipotential surfaces:
Q
For a point charge Q, V  r  
4 0 r
Hence for V = V0, the equipotential surface is a sphere with
r
Q
4 0V0
In practice, equipotential surfaces are usually plotted with the
same potential difference V from one surface to the next.
54
Capacitance

A conductor is an equipotential under
static field conditions, and the potential
difference between two charged
conductors is determined by the electric
a
field:
V   E  dl
ab

b
If the two conductors formed an isolated system, (i) the
charges on the conductors will be equal and opposite and (ii)
the electric field E is proportional the charges Q on the
conductors. Thus if Q is doubled, E and hence V will also be
doubled. So V is proportional to charge Q, and the constant of
proportionality is called the capacitance C and the unit is Farad.
The system of the conductors and dielectrics forms a capacitor.
55
Capacitance (cont’d)

 E.ds
Q

Thus, by definition, C 
Farads


V
  E.dl

Consider a charge q on a capacitor with capacitance C,
establishing a potential difference V(q). On further moving a
charge dq from the negatively charged conductor to the
positively charged conductor, the work done is
q
dW  V q  dq  dq
C
Hence the total work done to establish the charge Q on the
Q
capacitor
q
1 Q2 1
1
W 
0
C
dq 
2 C

2
QV 
2
CV 2
and this is the energy stored in the electric field of C.
56
Example 1 – Parallel Plate Capacitor

Suppose that the plates are separated by the distance d and
each has a surface area A and the charge on the plates is +Q
and Q. If d is small compared with the linear dimension of
A, then neglecting edge effects, surface charge density is
uniform and  = Q/A.
57
Example 1 (cont’d)
The field between the plates is given by:
Q

E  
  A
D
and the voltage is:
and:
Q A
C 
V
d
Q
V  Ed 
d
A
For example, if the plates are 1 cm square with 1 mm
separation distance with air dielectric, the capacitance is
(1102 ) 2  8.854 1012
12
C
F

0.885

10
F  0.885 pF
3
110
58
Example 1 (cont’d)
Since breakdown strength of air
Emax = 30 kV/cm
Maximum voltage that the capacitor can withstand
Vmax = 30  0.1 kV = 3 kV
Energy stored in C corresponding to this voltage of 3 kV is
1
W   0.885  1012  (3  103 ) 2 J  4 10 6 J
2
59
Example 2

Find the capacitance of two concentric spherical conducting
shells, of radii a and b and filled with a dielectric with
dielectric constant .
Let there be a charge +Q on the inner shell and Q on the
outer shell. The field between the two shells is:
E
Q
4 r
2
ar
and the potential difference between the shells is:
V  
a
b
1
Q 1 1
b r 2 dr  4  a  b 
Q
ab
C   4
V
ba
Q
E  dl  
4
and the capacitance is:
a
60
Example 3

A parallel plate capacitor of area S and spacing d which is small
compared with the linear dimensions of the plates.
There are two dielectric layers with different permittivities 1
and 2, with the boundary between the dielectrics parallel to
the plates.
At the dielectric interface, E is normal and
DN1 = DN2,
or
1E1 = 2E2
61
Example 3 (cont’d)

Hence

1 
V
V  E1d1  E2 d 2  E1  d1  d 2   E1 
2 
d1   1 /  2  d 2

The charge density on both plates is therefore
V
  D1  1 E1 
d1 / 1  d 2 /  2
The capacitance is then
Q S
1
1
C 


d1
d2
1
1
V
V


1 A  2 A C1 C2
This is the formula for capacitance in series.
62
Example 4

In a parallel plate capacitor with dimensions shown there are
two dielectric layers with relative permittivities R1 and R2.
The boundary between the dielectrics is perpendicular to the
plates and the width of the region containing R1 is 1.2 m. Find
R2 if R1 = 2.5 and the total capacitance is 60 nF.
At each dielectric E is normal and
E1 = E2 = V/d
63
Example 4 (cont’d)

Hence
 1  D1  1 E1 
1V
 2  D2   2 E2 
d
; Q1   1 A1 
 2V
Total charge on plate
Therefore capacitance
d
1 A1
d
; Q2   2 A2 
V
 2 A2
V
d
 1 A1  2 A2 
Q  Q1  Q2  

V
d 
 d
Q 1 A1  2 A2
C 

 C1  C2
V
d
d
and this is same as the formula for capacitance in parallel.
64
Example 4 (cont’d)

Now
C1 
 R1 0 A1
d
2.5  8.854 1012  2 1.2

 26.56 nF
3
2 10
C2 = (60 – 26.56) nF = 33.44 nF
But
C1  R1 A1

C2  R 2 A2
 R2 
 R1 A1C2
A2C1
12 33.44
 2.5  
 4.72
8 26.56
65
Capacitors in the Smartphone
Touchsensor
Accelerometer
LCD Display
CMOS Photodiode Imager
CMOS Transistor
Flash Memory
[6.007, Spring 2011, MIT OpenCourseWare]
66
Capacitive Touch Sensor
Touchsensor
Q  CV
A
C 
d
[6.007, Spring 2011, MIT OpenCourseWare]
67
Supplementary Notes
68
Example 3 (details)

To solve this, divide the line up into 2 symmetrical lengths on
either side of the point P.
 dx
dE  2
cos  a z
2
4 o r
a
cos  
and r  a 2  x 2
r
1 L 2 a
2 a
E
3 dx a z =

4 0 0  a 2  x 2  2
4 o

1
2 L
4 o a a 2  L2

L
x
az
 2 2

2
a
a

x

0
az
69
Example 4 (details)
Ez 
1

 dl
cos  a z
4 o l r
a
2
2
2
where r  a  b and cos 
r
1
a
Ez 
dl a z

4 o (a 2  b 2 ) a 2  b 2 l
=
1
2
a
2 b a z
4 o (a 2  b 2 ) a 2  b 2
1
 2 ab
1
 ab

az 
az
2
2 32
2
2 32
4 o (a  b )
2 o (a  b )
70
Example 5 (details)
and for the whole disc:
Ez 
1
4 o
2 a 
r
b
0
a
2
r
2

3
2
dr a z
1

1

2 a  
az

2
2
4 o
a
a

b


1

Note that for b >> a:
2

lim E z 
az 
az
b 
4 o
2 o
which is the equation for the electric field from a charged plane.
71
Example 5 (details)

What happens to Ez for a >> b:
The disc should start to look like a point charge at a distance.
1
1
1 1  b2 

  1  2 
a
a 2  b2 a a  a 
 12
1 1  1 b2  b2
  1 
 3
2 
a a  2 a  2a
b2
1 2  b 2
Q
 Ez 
2 a 3 a z 
az 
az
2
2
4 o
2a
4 o 2a
4 o a
1
i.e. it looks like the field at distance a from a point charge.
72
Electrostatic Induction

When a conducting body is placed in an electric field, charges
are redistributed inside the conducting body to make the net
internal field zero, a phenomenon known as electrostatic
induction. Consequently the original field is affected or
distorted. In particular, the external field would be normal to
the conductor surface.
73
Faraday’s Cage

Consider an inner area being enclosed by a conducting body.
Because the field inside the conducting body is zero, the
inner area and outer area are cut off. Hence external field
would not affect the inner area and vice versa.
74
Equivalent Sources

For the region of interest, there are many different possible
charge sources that would produce the same field. For example,
for r > a, the field produced by a spherical uniform volume charge
of radius a and total charge Q, a hollow sphere of radius a and
total charge Q, or a point charge Q, would all be indistinguishable.
Hence one technique in solving a given field problem is to replace
a given system of charges by a simpler equivalent source that
would produce the same ‘external ’ field.
75
Boundary condition between two surfaces

Consider the line integral of E over the closed path shown.
V
 E.dL  E
t1
w  Et 2 w  0
1
Dt 2
Hence Et1 = Et2 and Dt1 
2
for h  0
Next consider a Gaussian surface in the form of a cylindrical
pillbox of area s and height h  0. Applying Gauss’s Law,

D.ds   Dn1  Dn 2  s  Q  s
surface
76
Boundary condition between two surfaces
 Dn1  Dn 2  

Notice that  here refers to the surface free charge density
since the bounded surface charge is taken care of by the
relative permittivity. Since normally there will not be any free
charge if it is a surface between two dielectrics,
Hence
Dn1 – Dn2 = 0
and
1En1 = 2En2
There is a change in the direction of D.
with
1
tan 1  tan  2
2
Notice that if 1 > 2, then, D1  D2 , E1  E 2
77