chapter 4 electrostatics part 3
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this field theory lecture in our universityTRANSCRIPT
UNIVERSITY OF HORMUUD
FIELD THEORY
CHAPTER FOUR:
ELECTROSTATICSPart 3
Engr Burhan Omar Sheikh Ahmed
M.Eng (Electrical-Electronics and Telecommunications)
Universiti Teknologi Malaysia (UTM
ELECTROSTATICS
CHAPTER FOUR
3
ELECTROSTATICS
2.1 COULUMB’S LAW
2.2 ELECTRIC FIELD INTENSITY
2.3 LINE, SURFACE & VOLUME CHARGES
2.4 ELECTRIC FLUX DENSITY
2.5 GAUSS’S LAW
2.6 ELECTRIC POTENTIAL
2.7 BOUNDARY CONDITIONS
2.8 CAPACITANCE
2.9 POISSON'S AND LAPLACE'S EQUATIONS
4
2.4 ELECTRIC FLUX DENSITY
Consider an amount of charge
+Q is applied to a metallic
sphere of radius a.
Enclosed this charged sphere
using a pair of connecting
hemispheres with bigger
radius.
5
ELECTRIC FLUX DENSITY (Cont’d)
The outer shell is grounded. Remove the ground
then we could find that –Q of charge has
accumulated on the outer sphere, meaning the
+Q charge of the inner sphere has induced the
–Q charge on the outer sphere.
6
Faraday’s experiment found that the total charge on the outer sphere was equal in magnitude to the original charge placed on the inner sphere.
He concluded that there is some sort of displacement from the inner sphere to the outer which was independent of the medium, and we refer to this flux as displacement, displacement flux or simply electric flux.
Faraday showed, that there is direct proportionality between the electric flux and charge.
ELECTRIC FLUX DENSITY (Cont’d)
7
From previous Electric Field equations we have shown that the electric field intensity is dependent on the medium in which the charge is placed.
Anew vector field D independent of the medium is defined by
The electric flux density, D in is:2mC
ED
ELECTRIC FLUX DENSITY (Cont’d)
8
Electric flux, extends from the positive
charge and casts about for a negative charge.
It begins at the +Q charge and terminates at
the –Q charge.
psi
ELECTRIC FLUX DENSITY (Cont’d)
We can find the total flux over a surface as:
dS D
9
This is the relation between D and E, where
is the material permittivity. The advantage of
using electric flux density rather than using
electric field intensity is that the number of flux
lines emanating from one set of charge and
terminating on the other, independent from the
media.
ELECTRIC FLUX DENSITY (Cont’d)
ED
10
We could also find the electric flux density, D
for:
Infinite line of charge:
Where
aE02L So,
aD
2L
ELECTRIC FLUX DENSITY (Cont’d)
11
Where
So, NS aD
2
N
S aE02
Infinite sheet of charge:
Volume charge distribution:
RV dV
aR
E2
04
So, RV dV
aR
D2
4
ELECTRIC FLUX DENSITY (Cont’d)
12
EXAMPLE 1
Find the amount of electric flux through
the surface at z = 0 with
and
mymx 3050 ,
243 mCxxy zx aaD
13
SOLUTION TO EXAMPLE 1
The differential surface vector is
zdxdyd aS
We could have chosen but
the positive differential surface vector is
pointing in the same direction as the flux, which
give us a positive answer.
zdxdyd aS
14
SOLUTION TO EXAMPLE 1 (Cont’d)
Therefore,
C
xdxdy
dxdyxxy
dS
x y
zzx
aaa
D
150
4
43
5
0
3
0
Why?!
15
EXAMPLE 2
Determine D at (4,0,3) if there is a point
charge at (4,0,0) and a line
charge along the y axis.
mC 5
mCm 3
16
SOLUTION TO EXAMPLE 2
How to visualize ?!
17
Let total flux, LQTOTAL DDD
Where DQ is flux densities due to point charge
and DL is flux densities due to line charge.
Thus, 0 0 20
2
4
4
Q R
R
QD E a
R
Q a
R
SOLUTION TO EXAMPLE 2 (Cont’d)
18
SOLUTION TO EXAMPLE 2 (Cont’d)
Where,
za
R
3
3,0,00,0,43,0,4
Which has magnitude, and a unit vector,
zz
R aa
a 3
3
3R
19
So,
2
3
2
138.0
94
105
4
mCm
Q
z
z
RQ
a
a
aR
D
SOLUTION TO EXAMPLE 2 (Cont’d)
20
And then0 2
LLD E a
Where, 5
34
0,0,03,0,4
0,0,03,0,4 zx aaa
So,
218.024.0
5
34
52
3
mCmzx
zxL
aa
aaD
SOLUTION TO EXAMPLE 2 (Cont’d)
02LE a
::::::::::::::
21
Therefore, total flux:
242240
18.024.0318.0
mC
zx
zxz
LQTOTAL
aa
aaa
DDD
SOLUTION TO EXAMPLE 2 (Cont’d)