about electromagnetic engineering

1. Presentation

2. Table of contents
Applications of gauss’s law
Symmetric charge distribution
Differential volume element
Example & problems

3. Gauss’s law applications
symmetric charge distribution
We know that Gauss’s law is given by
Q = ∮DS · dS
Choose a closed surface
Satisfying two conditions:
1.DS is everywhere either normal or tangential to the
closed surface, so that
DS · dS becomes either DSdS or zero.
2.On that portion of the closed surface for which DS ·
dS is not zero, DS =constant.

4. charge Q at the origin of a
spherical coordinate system
DS is everywhere normal to the surface so,

5. Solution (a)
Drill (3.5): A point charge of 0.25 µC is located at r = 0, and uniform surface charge
at r 1cm and s  .6mC / m at r  1..8cm
2
Calculate D at: (a) r = 0.5 cm; (b) r = 1.5 cm;
(c) r = 2.5 cm.(d) What uniform surface charge density should be established at r = 3cm to
cause D = 0at r = 3.5 cm?
densities are located as follows:s  2mC/ m 2
2 5  1 0  6
4  0 .5  1 0  2
2
r
Q
4  r 2
D  a  Simplify yourself.
Given is surface charge density so, find charge,
(b)
2
2 s s
Q = S   4r Put values & getAns.
 Q2
a
4 r2
D 
Q1
r Put values & getAns.
(c)
r
4 r2
Find Q3 & D same as part (b).
 Q2  Q3  Q4
a ......(1)
D 
Q1

6. (d) Here
But it is given that D = 0 so equation (1) can be written as
4  r 2
 Q 2  Q 3  Q 4
0 
Q 1
r
a . . . . . . . . ( 2 )
2
4 s s
But Q =  S   4  r
Put this value of Q4 in eq. (2)
Q  Q  Q   4  r 2
0  1 2 3 s
a r . . . . . . . . ( 3 )
4  r 2
Multiplying both sides of eq. (3) with 4r2
we get;
0  Q  Q  Q   4  r 2
a
1 2 3 s r
 ( Q 1  Q 2  Q 3 )
4  r 2
s
  Put values & getAns.

7. Application of Gauss’s Law: Differential Volume Element
Consider any point P, located by a
rectangular coordinate system.
The value of D at the point P may be
expressed in rectangular components:
D0Dx0axDy0ay Dz0az
We now choose as our closed surface,
 the small rectangular box, centered at P,
having sides of lengthsΔx,Δy, andΔz,
 apply Gauss’s law we know that
for a surface:
 For volumetric object Gauss’s law can be
applied for a six surfaces as following:
S D dS

8. We will now consider the front surface in
detail.
The surface element is very small, thus D is
essentially constant over this surface (a
portion of the entire closed surface):
front
DfrontSfront
Dx,front =yz
Dfront yzax
Suppose The front face is at a distance of Δx/2 from P, and therefore:
2
x,front x0 x x0
D D 
x
rate ofchangeof D withx D
 xDx
2 x
D  dS  front
 back
 left
 right
 top
bottom
=
=
=

9. We have now, for front surface:
x0
front
 D x Dx  yz
 2 x 
 
 In the same way, the integral over the back surface can be foundas:
back
Dback Sback Dback  (yzax ) = Dx,backyz
x,back x0
D D
 x Dx
2 x
back
D 
x Dx  yz
 x0
2 x 
 
=
=
=
= =

10. If we combine the two integrals over the front and back surface, we
have:
fro n t b ack
 D x
 x  y  z
 x
  
right left
Dy
yxz
y
   top bottm z
  
Dz
zxy
Repeating the same process to the remaining surfaces, we find:
These results may be collected to yield:
 Dx
 Dy
Dz 
xyz
 x y z 

S
D dS

  D x
  D y  Dz 
 v
 x y z


S
D  d S  Q

=
=

11. Chargeenclosedin volume v=
Dx

Dy

Dz
v
 x y z 
 
The previous equation is an approximation, which becomes better as
Δv becomes smaller, and in the following section the volume Δv will
be let to approach zero.
For the moment, we have applied Gauss’s law to the closed surface
surrounding the volume elementΔv.
The result is the approximation stating that:

12. Example: 3.3
Let D = y2z3 ax + 2xyz3 ay + 3xy2z2 az nC/m2 in free space.
(a)Find the total electric flux passing through the surface x = 3,
0 ≤ y ≤ 2, 0 ≤ z ≤ 1 in a direction away from the origin.
(b) Find |E| at P(3,2,1).
(c)Find an approximate value for the total charge contained in an
incremental sphere having a radius of 2 mm centered at P(3,2,1).
ψ  S
S
D dS

(a)
 
1 2
x3
x
z0 y0
y
2xyz3
a z
3xy2
z2
a x
 dydz a
   y2
z3
a
1 2 2
1
1
z4
1
3
y2
z3
dydz 
3
4 3
 2
nC
0 0
0 0
y
  

13. (b)
D=y2
z3
a 2xyz3
a 3xy2
z2
a
x y z
2 3 3 2 2
DP = (2) (1) ax  2(3)(2)(1) ay 3(3)(2) (1) az
2
= 4ax 12ay  36az nC m
2 2 2 2
DP = DP  (4) (12) (36) 38.158nC m
0
P

D P
E 
m 2

3 8 . 1 5 8 n C
8 . 8 5 4  1 0  1 2
 4 .3 1 k V m

14. (c)
Q
 Dx
 Dy
Dz 
v
 x y z 
 
 Dx
 Dy  Dz 
P
Q v
 x y z 
  P
  4
3 2
3
3 3 3 3
y2
z1
x3 nC m  (210 ) m
02xz 6xy z
  4
3 2
3
3 3
02(3)(1) 6(3)(2) (1)  (210 ) nC
2.611015
C
=
=
=
=
=

15. (b)
(c) Q=charge enclosed byvolume
Drill Pb#3.6
D 8xyz4
a
x
 4x2
z4
a
y
 16x2
yz3
a pC / m²
z
(a)
1 0
dxdyaz
3 2
Q   Ds .ds    Ds Put values and solve
D
 0
E 
v  (
Dx
 Dy  Dz
) v
x  y z
Next step-I: Find partial derivative of each term separately at limits
x= 2, y= -1, z= 3 and then add them.
Next step-II: Multiply the result of step-I with the value of ΔV andget
the finalAnswer.
Put values and solve