ELECTROMAGNETICS THEORY(SEE 2523)
ELECTROSTATIC FIELD
INTRODUCTION
An electrostatic field is produced by a
static charge distribution.
A magnetostatic field is produced by the moving charges or a constant current
flow (DC current)
AC current produced the electromagnetic field.
Charges are measured in Coulombs (C). The smallest unit of electric charge is found on the
negatively charged electron and positively charged proton. The electron mass, qe= -1.602x10-19 (C).
So, 1 C would represent about 6x1018 electrons.
In real world, electric charges
can be found
at a point
along a line
on a surface
in a volume or
combination the above distributions
2.1: ELECTRIC CHARGES
2.1.1 A POINT CHARGE
Fig. 2.1: A point charge
The concept of a point charge is used when the dimensions of an electric charge distribution are very small compared to the distance to neighboring electric charges.
2.1.2 A LINE CHARGE
A line charge denotes as the electric charge distribution along a thin line.
Fig 2.2: A line charge
)/(lim0
mCQ
A line charge density :
Total charge along the line :
dQ
l
Q
2.1.3 A SURFACE CHARGE
A surface charge is defined as charges distribution on a thin sheet.
Fig. 2.3: A surface charge
Q
S)2(C/msQ
ss
lim
0
A surface charge density :
Total charge on the surface :
dss s
Q
2.1.4 A VOLUME CHARGE
A volume charge means electric charges in a volume. This volume charge can be viewed as a cloud of charged particles.
Fig. 2.4: A volume charge
Q
V
A volume charge density :
Total charge in the volume :
)3(C/mvQ
vv
lim
0
dvv v
Q
Expressed mathematically,
21QQF 2
12
1R
F 212
21R
QkQF and or
2.2 COULOMB’S LAW
Coulomb’s law states that the force F between two point
charges Q1 and Q2 is : Directly proportional to the
magnitude of the both products of the charges and
inversely proportional to the square of the distance R
between them.
Fig. 2.5: Coulomb’s law
12ˆ
212
212 R
aR
QkQF
ok 4
1
(N)12
ˆ212
421
2 Ra
Ro
QQF
where
o is known as the permittivity of free space
A unit vector, then the force :
The force, on Q1 due to Q2 :
12
12
12ˆ
R
R
Ra
1F
(N)123
124
212
RRo
QQF
(N)2213
214
211
FRRo
QQF
Things need to considered in Coulomb’s law :
>> The force expressed by Coulomb’s law is a mutual force, for
each of the two charges experiences a force of the same magnitude,
although of opposite direction.
>> Coulomb’s law is linear, if the charge multiply by n, the force is
also multiplied by the same factor, n.
>> If there are more than two point charges, the principle of
superposition is used. The force on a charge in the presence of several
other charges is the sum of the forces on that charges due to each of the
other charges acting alone.
2.3 ELECTRIC FIELD INTENSITY DUE TO POINT CHARGE
Consider one charge fixed in position, Q1 and test charge,
Qt that exists everywhere around Q1 , shown in Fig. 2.6.
From the definition of Coulomb’s law, there will exist force
on Qt cause by Q1.
Fig. 2.6
The electric field intensity E is the force per unit charge when placed in the electric field.
V/m)(N/C,1
ˆ21
41
tRa
tRo
Q
tQtFE
tF
For n point charges, the electric field intensity at a point in
space is equal to the vector sum of the electric field intensities
due to each charge acting alone. (Following superposition
theorem). Shown in Fig. 2.7.
Fig. 2.7
The total field at point P :
kRa
n
k kRo
kQ
nRanRo
nQRa
Ro
QRa
Ro
QRa
Ro
Q
n
k kEnEEEEE
ˆ1 24
ˆ24
....3
ˆ23
43
2ˆ
22
42
1ˆ
21
41
1...
321
E
We have :
Thus, the electric field for a line charge :
dQ
Ra
R
dE ˆ
20
4
2.4 ELECTRIC FIELD INTENSITY DUE TO THE LINE CHARGE
Fig. 2.8
Ex:4. Find the electric field intensity about the finite line
charge of uniform distribution along the z axis as shown
in Fig. 2.8.
Ra
Ro
dEd ˆ
24
2-'2
ˆ-'-ˆ
zzrR
zzzrrR
Solution:
∫ 2/32)-'(24
'ˆ)-'(-∫ 2/3
2)-'(24
'ˆ
∫ ˆ)-'(-ˆ2/3
2)-'(24
'
b
azzro
dzzzzb
azzro
dzrr
b
azzzrr
zzro
dzE
Using 2/1222
∫ 2/322
,2/1
22
1-∫ 2/3
22
xcc
x
xc
dx
xcxc
xdx
Hence :
2)-(22cos,
2)-(21cos
2)-(2)-(
2sin,
2)-(2)-(
1sin
zbr
r
azr
rzbr
zb
azr
az
mVrz
rr
oE /
1cos-
2cosˆ
1sin
2sinˆ
4
Hence :
2/1)2)-(2(1-
2/1)2)-(2(1ˆ
2/1)2)-(2()-(-
2/1)2)-(2()-(ˆ
4
2/1)2)-'(2(ˆ
2/1))2-'(2(2)-'(ˆ
4
zarzbrz
zarza
zbrzb
rr
o
b
azzrz
zzrrzzrr
oE
If the line is infinite :
(V/m)rr
E ˆ0
2
1
2 2/= =
mVrz
rr
oE /
1cos-
2cosˆ
1sin
2sinˆ
4
We obtain :
We have :
Thus, the electric field for a surface charge
dsS S
Q ∫
Ra
s R
dsSE ˆ∫ 20
4
2.5 ELECTRIC FIELD INTENSITY DUE TO THE SURFACE CHARGE
)(C/m2
s
Fig. 2.9
Ex:6. Find the electric field intensity along z axis on a disc with
a radius a(m) which is located on xy plane with uniform charge
distribution as shown in Fig. 2.9.
s R
aR
dssE ˆ2
04
22ˆˆ
rzR
rrzzR
Using cylindrical coordinate :
Solution:
szr
rrdrrdss
zr
zzdrrds
srrzz
zr
drrdsE
2/322
04
ˆ2/3
220
4
ˆ
ˆˆ2/3
220
4
)(
The component is cancel because the charge distribution is
symmetry.
Only component exists.
r
z
r
2/1
221
2ˆ1
2/122
12
ˆ
za
zoszz
zao
zsz
a
zro
zszza
zr
rdrdo
zsE
0
2/122
1)2(4
ˆˆ0 2/3
22
2
04
02
ˆ szE For an infinite sheet of charge :
02
ˆ snE In general, it can be
summarized to be :
The electric field is normal to the sheet and independent of
the distance between the sheet and the point.
In the parallel plate capacitor, the electric field existing
between the two plates having equal and opposite charge is
given by
The electric field is zero for both above and below plates.
nsnsnsE ˆ0
)ˆ-(0
2-
ˆ0
2
We have
Thus, the electric field intensity for a volume charge is
dvv vQ ∫
Ra
v R
dvvE ˆ∫ 20
4
2.6 ELECTRIC FIELD INTENSITY DUE TO THE
VOLUME CHARGE
)3(C/mv
Fig. 2.10
Ex:8. Find the electric field intensity for a sphere with a radius
a(m) and with uniform charge density, shown in
Fig. 2.10.
The electric field at P due to elementary volume charge is
where
Ed
Ra
Ro
ddrdrv
Ra
Ro
dvvEd
ˆ24
)sin2(
ˆ24
rzaR
ˆsinˆcosˆ
Solution:
Due to the symmetry of the charge distribution, the Ex and Ey
becomes zero.
Only Ez exists. Thus,
Using Cosine law,
cosˆ dEzEzE
cos2-222cos2-222
zRRzr
zrrzR
It is convenient to evaluate the integral in terms of R and r, so
Differentiating (keeping z and r fixed) then
zRrRz
zrRrz
22-22
cos
22-22
cos
dcos
zrRdRd sin
0 )( )-( rzRrz where and
Ra
Ro
ddrdrvEd ˆ24
)sin2(
Substituting all the equations into electric field equation yields
Ra
Ro
ddrdrvEd ˆ24
)sin2(
vazo
drra
zo
v
draa
RrzRr
zo
v
a
r
rz
rzRdRdr
Rrzr
zo
v
da
r
rz
rzR RzRrRzdrzr
RdRov
zE
334
2412
04
24
00
)2-2(-24
0 22-21
28
)2(
2
0 0 21
22-22
4
cosˆ dEzEzE
Because and then electric field at
P(0,0,z) is given by
Due to the symmetry of the charge distribution, the electric field at any point can be written as
vQv
3
34 av
zzo
QE ˆ24
),,( rP
rro
QE ˆ24
va
zozE
3
34
241
Which is identical to the electric field at the same point due
to a point charge Q located at the origin.