ELECTRIC FLUX DENSITY ,
GAUSS LAW, & MAXWELL’S
EQUATION
Electric Flux Density
• Electric flux density is measured in
coulombs per square meter (sometimes
described as "lines per square meter," for
each line is due to one coulomb), is given
the letter D, which was originally chosen
because of the alternate names of
displacement flux density or displacement
density.
Electric Flux Density
• The electric flux density D is a vector field
and is a member of the "flux density" class
of vector fields, as opposed to the "force
fields" class, which includes the electric
field intensity E.
• The direction of D at a point is the
direction of the flux lines at that point, and
the magnitude is given by the number of
flux lines crossing a surface normal to the
lines divided by the surface area.
Electric flux density (displacement
vector) in free space:
• The electric field intensity is dependent on
the medium in which the charge is placed
(free space in this case). A new vector can
be defined as: D   0 E
• The flux of D is:
   Dds
s
• This vector is known as the electric flux
density vector or displacement vector.
Electric flux density (displacement
vector) in free space:
• In SI units the electric flux is measured in
coulombs and the electric flux density in
coulombs per meter square.
• Gauss Law for electric flux density can be
written as:
  D   v (Divergence of D)
• This is the first of the four Maxwell’s
equations to be derived.
Electric flux density (displacement
vector) in free space:
• This equation states that the divergence of
the electric flux density is equal to the
volume charge density.
Note that:
• 1) Gauss Law is the alternative statement
of the Coulomb’s Law.
• 2) Gauss Law provides an easy means of
finding E and D for the for symmetrical
charge distributions.
Electric Flux Density
• Electric flux density in free space:
Q
Q
D  0E  0
a 
a
2 r
2 r
4 0 r
4r
• Volume electric flux density in free space:
 v dv
 v dv
D  0E  0 
ar  
ar
2
2
vol 4 R
vol 4R
0
Electric Flux Density
• Uniform Line Charge Density:
L
D
aR
2R
• Uniform Surface Charge Density:
S
D
aR
2
Example
A 25 μC point charge located at the origin.
Calculate the electric flux passing through
(a) that portion of the sphere r = 20 cm
bounded by 0 < θ < π and 0 < Φ < π/2; (b)
the closed surface defined by ρ =0.8 m, z
= ±0.5 m; (c) the plane z = 4 m
Solution
Area of the strip

 2
0
0
A

 
A 
0
2
0
r 2 sin dd
(0.2) sin dd
2

 2
0
0
A  0.04  sin d 

d
A  0.04  cos 0  0

 2

A  0.04 cos  cos 0 2  0
A  0.04( )  0.125664 m
2
Solution
The flux through the strip
A
 net 
Q
2
4r
0.125664
6
 net 
(25  10 )
2
4 (0.2)
 net  6.25  10 C  6.25C
6
Solution
 net   Ds  dS
S
(b)   Ds 0.5 2 ddz
net
0.5 0
 net
 net
 net
 net
 net
Q

2
0.5
 
2
 0.5 0
ddz
0.5
25  10 6 2

d  dz

0
 0.5
2
25  10 6
2
0.5
 0 z 0.5

2
25  10 6
2  00.5  0.5

2
 25  10 6 C  25 C


Example
Find D (in cartesian coordinates) at (6,8,10) caused by: (a) a point charge of 30
mC at the origin; (b) a uniform line charge
ρL = 40 μC on the z-axis; (c) a uniform
surface charge density ρS = 57.2 μC /m2
on the plane x = 9.
Solution
R  (6  0)ax  (8  0)ay  (10  0)az
R  6ax  8ay  10az
| R | 6 2  8 2  10 2  14.142
6ax  8ay  10az
aR 
 0.424 ax  0.566 ay  0.707 az
14.142
Q
30  10 3
D
aR 
(0.424 ax  0.566 ay  0.707 az)
2
2
4R
4 (14.142 )
D  5.06ax  6.75ay  8.44azC / m 2
Solution
R  (6  0)ax  (8  0)ay
R  6ax  8ay
| R | 6 2  8 2  10
6ax  8ay
aR 
 0.6ax  0.8ay
10
L
40  10 6
D
aR 
(0.6ax  0.8ay)
2R
2 (10)
D  0.382 ax  0.509 ayC / m 2
Solution
R  (6  9)ax  3ax
| R | 3  3
 3ax
aR 
 ax
3
6
S
57.2  10
D
aR 
(ax)
2
2
D  28.6C / m 2
2
Example
Find the volume charge density associated
with D = xy2ax + yx2ay + zaz C/m2
Solution
xy
x y z
divD 


x
y
z
2
2
divD  y  x  1 C / m
2
2
3
Seatwork
1. Given a 60 μC point charge located at the
origin, find the total electric flux passing
through (a) that portion of the sphere r =
26 cm bounded by 0 < θ < π/2 and 0 < Φ
< π/2; (b) the closed surface defined by ρ
=26 cm and z = ± 26 cm; (c) the plane z =
26 cm
Answer: 7.5 μC; 60 μC; 30 μC
Seatwork
2. Calculate D in rectangular coordinates at
point P(2,-3,6) produced by: (a) a point
charge QA = 55 mC at Q(-2,3,-6); (b) a
uniform line charge ρLB = 20 mC/m on the
x-axis; (c) a uniform surface charge
density ρSC = 120 μC /m2 on the plane z =
-5m.
Answer: 6.38ax – 9.57ay +19.14az μC /m2;
– 212ay +424az μC /m2; 60az μC /m2
Seatwork
3. Determine an expression for the volume
charge density associated with D =
zsinΦaρ + zcosΦaΦ + ρsinΦaz
Answer: 0
GAUSS LAW
• Gauss Law constitutes one of the
fundamental laws of electromagnetism.
• The total flux ψ of the electric field intensity
over any closed surface in free space is
equal to the total charge enclosed in the
surface divided by ε0.
Qenc ;
  d  E  d s

0

s

s
GAUSS LAW
• Integral Form of the Gauss Law:
 E ds 
Qenc
0
s

s
• Qenc is the total charge contained in
volume V bounded by surface S.
• Applying the Divergence Theorem:
 v dv

Qenc v
;
E  d s    E dv    E dv 


v


v

0
0
GAUSS LAW
• So, the differential form (point form) of the
Gauss’ Law is:
v
E 
0
Applications of Gauss’s Law:
• Calculate the electric field intensity
involves whether symmetry exists.
• Once it has been found that the symmetry
charge distribution exists, construct a
mathematical closed surface (known as
the Gaussian Surface) such that the
electric field intensity is the same
everywhere on that surface and is
perpendicular to that surface.
Example
Let D = r ar / 3 nC/m2 in free space (a) find
E at r = 0.2 m; (b) find the total charge
within the sphere r = 0.2 m; (c) find the
total electric flux leaving the sphere r = 0.3
m.
Solution
9
0.2 / 3  10
E

 7.53V / m
12
 0 8.854  10
D
Solution

2
0
0
A



A    (0.2) sin dd


A  0.04  sin d  d
r 2 sin dd
2
0
2
0
2
0

0
A  0.04  cos 0  

A  0.08(2 )  0.503m
2

2
0
A  0.04 cos  cos 02  0
Q  DA  (0.2 / 3  10 9 )(0.503)
Q  33.5  10 12 C  33.5 pC
Solution
 net   Ds  dS
S
 net  Ds 

0

2
0
r 2 sin dd
 net  0.1  10
9

2
0
0
  sin dd


(0.09)  d  sin d
 net  0.3 / 3  10 (0.3)
9
2
2
0

0
 cos 0 
 net  0.1  10 9 (0.09)2  0 cos  cos 0
 net  0.1  10 (0.09)  
9
2
0
 net  1.131  10 10 C  113 .1 pC

Example
Find the total electric flux leaving the
spherical r = 2.5 m given the charge
configuration: surface formed by the six
planes x,y,z = ±5 if the charge distribution
is: (a) Q = 2-x2 nC on the x-axis at x = 0,
±1, ±2 m (b) a line charge ρL = 1/(z2+1)
nC/m on the z-axis
Solution
  Q   Qn
At
x  0 Q  2 0  10 9  1  10 9
1
9
At
x  1 Q  2  10
At
x  1 Q  2 1  10 9  0.5  10 9
At
x  2 Q  2  4  10 9  0.0625  10 9
At
4
x  2 Q  2  10
 0.5  10
9
9
 0.0625  10
Total,  2.125  10 9 C  2.125 nC
9
Solution
At the side ( no z component at the side)
  Q  Ds  dS
L
1  Q 
2
4R
1  10 9
1 
(0  1)4
9
2

0
0
 


0

R sin dd
2
2
sin d  d
0

1  10
 cos 0  02  1  10 9
1 
(0  1)4
Solution
At the top
 3  Q    L dL
 2  Q    L dL
9
1  10
2  
dz
2
 2.5 ( 2.5)  1
2.5
1  10 9
3  
dz
2
 2.5 ( 2.5)  1
2.5
2.5
 2  0.138  10 z 2.5  3  0.138  10 z  2.5
9
 2  0.689  10 C
9
2.5
9
 3  0.689  10 9 C
Solution
 T   1  2  3
 T  1  10  0.689  10  0.689  10
9
9
 T  2.38  10 C  2.38nC
9
9
Seatwork
Given the electric flux density, D = 0.3r2ar
nC/m2 in free space (a) find E at point P( r
= 2, θ = 25O, Φ = 90O); (b) find the total
charge within the sphere r = 3; (c) find the
total electric flux leaving the sphere r = 4.
Answer: 135.5 ar V/m, 305 nC; 965 nC
Seatwork
Calculate the total electric flux leaving the
cubical surface formed by the six planes
x,y,z = ±5 if the charge distribution is: (a)
two point charges, 0.1 μC at (1, -2,3) and
1/7 μC at (-1, 2,-2); (b) a uniform line
charge of π μC/m at x = -2, y = 3; (c) a
uniform surface charge of 0.1 μC/m2 on
the plane y = 3x.
Answer: 0.243 μC; 31.4 μC; 10.54 μC