Transcript electrostatic-2 (cont`d)

LINE,SURFACE & VOLUME CHARGES
Electric fields due to continuous charge
distributions:
LINE,SURFACE & VOLUME CHARGES (cont’d)
To determine the charge for each distributions:
Line charge
Surface charge
Volume charge
dQ   L dl
dQ   S dS
dQ  V dV
Q    L dl
Q    S dS
Q   V dV
L
S
V
LINE CHARGE
Infinite Length of Line Charge:
To derive the electric field intensity at any
point in space resulting from an infinite length
line of charge placed conveniently along
the z-axis
LINE CHARGE (Cont’d)
Place an amount of charge in
coulombs along the z axis.
The linear charge density is
coulombs of charge per meter
length,
 
L C m
Choose an arbitrary point P
where we want to find the
electric field intensity.
P  ,  , z 
LINE CHARGE (Cont’d)
The electric field intensity is:
E   E a   E a  Eza z
But, the field is only vary with
the radial distance from the
line.
There is no segment of charge
dQ anywhere on the z-axis
that will give us E . So,
E   E a   Eza z
LINE CHARGE (Cont’d)
Consider a dQ segment a
distance z above radial axis,
which will add the field
components for the second
charge element dQ.
The E zcomponents cancel
each other (by symmetry) ,
and the E adds, will give:
E   E a 
LINE CHARGE (Cont’d)
Recall for point charge,
E
Q
4 0 R
2
aR
For continuous charge distribution, the
summation of vector field for each charges
becomes an integral,
E
dQ
4 0 R
2
aR
LINE CHARGE (Cont’d)
The differential charge,
dQ   L dl
  L dz
The vector from source
to test point P,
R  Ra R
 a   z a z
LINE CHARGE (Cont’d)
Which has magnitude, R 
unit vector,
aR 
 2  z 2 and a
a   za z
 2  z2
So, the equation for integral of continuous
charge distribution becomes:
E
4 0

 a   za z

2
2
2
2
2


z

 z
 L dz





LINE CHARGE (Cont’d)
Since there is no
E
az
component,
 L dz

4 0   z
2
2

3
a 
2
L  
dz


4 0   2
  z2


3
a
2
LINE CHARGE (Cont’d)
Hence, the electric field intensity at any point ρ
away from an infinite length is:
L
E
a
2 0 
For any finite length, use the limits on the integral.
EXAMPLE 3
Use Coulomb’s Law to
find electric field
intensity at (0,0,h) for
the ring of charge, of
charge density,  L
centered at the origin in
the x-y plane.
SOLUTION TO EXAMPLE 3
By inspection, the ring
charges delivers only
and
dE 
dE z contribution
to the field.
dE  component will be
cancelled by symmetry.
SOLUTION TO EXAMPLE 3 (cont’d)
Each term need to be
determined:
E
dQ
4 0 R
2
aR
The differential charge,
dQ   L dl
  L ad
SOLUTION TO EXAMPLE 3 (cont’d)
The vector from source to test point,
R  Ra R
  a a   ha z
Which has magnitude, R 
aR 
a 2  h2 and a unit vector,
 aa   ha z
a 2  h2
The integral of continuous charge distribution becomes:
E
E
  aa   ha z

2
2
2
2
2
a

h

a h
 L ad
4 0

 L ad
4 0

a 2  h2


3
ha z
2




SOLUTION TO EXAMPLE 3 (cont’d)
Rearranging,
2
 L ah
E

4 0 a 2  h 2

3
 da z
2  0
Easily solved,
E

 L ah
2 0 a 2  h 2

3
az
2
EXAMPLE 4
An infinite length line of charge  L  4.0
nC
exists at x = 2m
m
and z = 4m. Find the electric field intensity at the origin.
SOLUTION TO EXAMPLE 4
Sketch in three dimensions and the cross section:
SOLUTION TO EXAMPLE 4 (cont’d)
The vector from line charge to the origin:
R  a   2a x  4a z
Which has magnitude, R  20
aR  a 
and a unit vector,
2
4
ax 
az
20
20
Inserting into the infinite line charge equation:
L
E
a
2 0 
4  10 9

2 8.854  10 12

 7.2a x  14.4a z

  2a x  4a z 


20
20 

V
m

SURFACE CHARGE
Infinite Sheet of Surface Charge:
To derive the electric field intensity at point
P at a height h above a charge sheet of
infinite area (x-y plane).
The charge distribution,
 S is in C
m2
SURFACE CHARGE (CONT’D)
SURFACE CHARGE (CONT’D)
Consider a differential charge,
dQ   S dS
  S dd
The vector from surface charge to the origin:
R    a    ha z
Which has magnitude, R 
aR 
 2  h2 and a unit vector,
 a   ha z
 2  h2
Where, for continuous charge distribution:
E

dQ
4 0 R
2
aR
SURFACE CHARGE (CONT’D)
The equation becomes:
E
  a   ha z

2
2
2
2
2
a

h

 h
 S dd
4 0


Since only z components exists,
E
 S dd
4 0

 2  h2

3
ha z
2




SURFACE CHARGE (CONT’D)
 S 2 
E


4 0   0   0

hdd
 h
2


S h
2
2

2    h
4 0
 0

S h 
2
2




h
2 0 
S
E 
az
2 0

1

2
3
2

3
az
2
2 da

 az
0
z
SURFACE CHARGE (CONT’D)
A general expression for the field from a sheet
charge is:
S
E
aN
2 0
Where
a N is the unit vector normal from the
sheet to the test point.
EXAMPLE 5
nC
An infinite extent sheet of charge  S  10 2
m
exists at the plane y = -2m. Find the electric
field intensity at point P (0, 2m, 1m).
SOLUTION TO EXAMPLE 5 (CONT”D)
Sketch the figure:
The unit vector directed away from the
sheet and toward the point P is
ay
S
E
aN
2 0
10  10  9

ay
12
2 8.854  10
V
 565a y
m


VOLUME CHARGE
A volume charge is distributed over a
volume and is characterized by its volume
charge density, V in C 3
m
The total charge in a volume containing a
distribution, V is found by
integrating over the volume: Q   dV
charge

V
V
EXAMPLE 6
Find the total charge
over the volume with
volume charge density,
V  5e
10 5 z
C
3
m
SOLUTION TO EXAMPLE 6
The total charge,
Q   V dV
dV  dddz
Thus,
with volume:
V
Q   V dV

V
0.01 2
 
0.04
  5e
10 5 z
  0   0 z  0.02
 7.854  10 14 C
dddz
VOLUME CHARGE (CONT’D)
To find the electric field intensity resulting from
a volume charge, we use:
V dV
E
a 
a
2 R
2 R
4 0 R
4 0 R
dQ
Since the vector R and
V
will vary over the
volume, this triple integral can be difficult. It can
be much simpler to determine E using Gauss’s
Law.
ELECTRIC FLUX DENSITY, (D)
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.
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.
ELECTRIC FLUX DENSITY (CONT’D)
Electric flux,  psi  extends from the positive
charge and casts about for a negative charge. It
begins at the +Q charge and terminates at the
–Q charge.
The electric flux density, D in C
D

4 0 R
2
a R where
m
2
is:
D  E
ELECTRIC FLUX DENSITY (CONT’D)
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.
We can find the total flux over a surface as:
   D  dS
ELECTRIC FLUX DENSITY (CONT’D)
We could also find the electric flux density, D for:

Infinite line of charge:

L
Where, E 
a
2 0 

Infinite sheet of charge:
S
aN
Where, E 
2 0

L
D
a
2
Volume charge distribution:
V dV
a
Where, E  
2 R
4 0 R
D
S
2
aN
V dV
D
a
2 R
4 R
EXAMPLE 7
Find the amount of electric flux through the
surface at z = 0 with 0  x  5m , 0  y  3m
and

D  3xya x  4 xa z C m
2
SOLUTION TO EXAMPLE 7
The differential surface vector is
We could have chosen
dS  dxdy a z
dS  dxdy  a z  but the positive
differential surface vector is pointing in the same direction as
the flux, which give us a positive answer.
Therefore,
   D  dS
  3 xya x  4 xa z   dxdya z

5
3
  4 xdxdy
x 0 y 0
 150 C
0
…Why?!
EXAMPLE 8
Determine D at (4,0,3) if there is a point charge
(4,0,0) and a line charge
3 m C m
SOLUTION TO EXAMPLE 8
How to visualize ?!
 5 mC
along the y axis.
at
SOLUTION TO EXAMPLE 8 (CONT’D)
Let total flux,

DTOTAL


 DQ  D L
Where DQ is flux densities due to point charge and DL is
flux densities due to line charge.




Q

aR 
Thus, DQ   0 E   0
2
 4 R

0


Q

a
2 R
4 R
Where,

R  4,0,3  4,0,0   0,0,3
 3a z
Which has magnitude, R  3 and a unit vector,
aR 
3a z
 az
3
SOLUTION TO EXAMPLE 8 (CONT’D)

So, DQ 
Q
4 R
2
aR
 5  10 3

az
4 9 
 0.138a m C
z

And D L 
L
a
2
Where, a  
So,
Therefore, total flux:

DTOTAL
4,0,3  0,0,0  4a x  3a z
4,0,3  0,0,0
5

x
z

 DQ  D L
  0.318a z   0.24a x  0.18a z 
 240a x  42a z  C
m2
3  4a x  3a z 
DL 


2 5 
5

 0.24a  0.18a m C

m2
m2
GAUSS’S LAW
If a charge is enclosed, the net flux passing through
the enclosing surface must be equal to the charge
enclosed, Qenc.
Gauss’s law constitutes one of the fundamental laws
of electromagnetism
Gauss’s Law states that:
The total electric flux,  through any closed surface is
equal to the total charge enclosed by that surface
 D  dS  Q
enc
GAUSS’S LAW (Cont’d)
How to derive the Gauss’s Law (which is the first of the four
Maxwell’s equations to be derived ??
It can be rearranged so that we have relation between the Gauss’s
Law and the electric flux.
Thus:
  Qen c
   D  dS  Qenc
Qenc   V dV
V
 Q   D  dS   V dV
S
V
(1)- Integral form
GAUSS’S LAW (Cont’d)
Related to Gauss’s Law, where net flux is evaluated
exiting a closed surface, is the concept of divergence.
By applying divergence theorem
 D  dS     DdV to
V
the middle term in equations Q  D  dS   dV

 V
S
V
Comparing the two volume integrals in above equations,
results in:
(2)-Differential or point
V form of Gauss’s law
D 
Which is the first of the four Maxwell’s equations to be derived.
Equation (2) states that the divergence of the electric flux
density is the same as the volume charge density.
GAUSS’S LAW (Cont’d)
Gauss’s Law is an alternative statement of Coulomb’s Law; proper
application of the divergence theorem to Coulomb’s law results in
Gauss’s Law.
Gauss’s Law is useful in finding the fields for problems that have
high degree of symmetry.
• Determine variables influence D and what components D present
• Select an enclosing surface, called Gaussian Surface, whose
differential surface is directed outward from the enclosed volume
and is everywhere (either tangent or normal to D)
GAUSS’S LAW (Cont’d)
The expression is also called the point form of
Gauss’s Law, since it occurs at some particular
point in space. For instance,
Plunger stationary – no net
movement of molecules
Plunger moves up – net
movement where air molecules
diverging  air is expanding
Plunger pushes in – net flux is
negative and molecules
diverging  air is compressing
GAUSS’S LAW APPLICATION (Cont’d)
Use Gauss’s Law to determine electric
field intensity, (E), for each cases below:

Point Charge

Infinite length of Line Charge

Infinite extent Sheet of Charge
POINT CHARGE
• Point Charge:
It has spherical coordinate
symmetry, where the field
is everywhere directed
radially away from the
origin. Thus,
D  Dr a r
POINT CHARGE (Cont’d)
For a gaussian surface, we could find the
differential surface vector is:
dS  r sin dda r
2
So,
 D  dS   Dr a r  r sin dda r
2
  Dr r sin dd
2
POINT CHARGE (Cont’d)
Since the gaussian surface has a fixed radius, Dr will be constant
and can be taken from integration to yield
 D  dS  Dr r

2
2
 sin dd

 0  0
 4r 2 Dr
By using Gauss’s Law, where:
So, Dr 
 D  dS  Qenc
Dr 4r 2  Q
Q
which leads to expected result:
2
4r
E
Q
4 0 r
2
ar
INFINITE LENGTH LINE CHARGE
• Infinite length line of charge:
Find D and then E at any
point
P  ,  , z 
A Gaussian surface
containing the point P is
placed around a section
of an infinite length line of
charge density L
occupying the z-axis.
INFINITE LENGTH LINE CHARGE (Cont’d)
An element of charge dQ along the line will give
Dρ and Dz. But second element of dQ will result in
cancellation of Dz. Thus,
D  D a 
The flux through the closed surface is:
 D  dS   D  dStop   D  dSbottom
  D  dS side
INFINITE LENGTH LINE CHARGE (Cont’d)
Where,
dStop  dda z ,
dSbottom  dd  a z  , dS side  ddza 
Then, we know that Dρ is constant on the side
of gaussian surface
 D  dS side   D a   ddza 
 D 
2
h
  ddz  2hD
 0 z 0
INFINITE LENGTH LINE CHARGE (Cont’d)
The charge enclosed by the gaussian surface:
h
Qenc    L dz   L h
0
We know that,
So,
 D  dS  2hD
L
D 
2
Thus, as expected:
L
E
a
2 0 
  L h  Qenc
INFINITE EXTENT SHEET OF CHARGE
• Infinite extent sheet of charge:
Determine the field everywhere resulting
from an infinite extent sheet of charge ρS
placed on the x-y plane at z = 0.
Locate a point at which we want to find the
field along the z axis at height h.
SHEET OF CHARGE (Cont’d)
Gaussian surface must contain this point and
surround some portion of the charged sheet.
A rectangular box is
employed as the
Gaussian surface
surrounding a
section of sheet
charge with sides
2x, 2y and 2z
SHEET OF CHARGE (Cont’d)
Only a DZ component will be present, and the
charge enclosed is simply:
x
y
x
y
Q    S dS   S  dx  dy
 4  S xy
No flux through the side of the box, so find
the flux through the top and bottom surface
SHEET OF CHARGE (Cont’d)
  D  dS   D  dStop   D  dSbottom

D a
z
z
 D  a   dxdy a 
 dxdya z 
top
z
z
z
bottom
 24 xyDz
Notice that the answer is independent of the height of the box.
Then we have:
 D  dS  Q
24 xy Dz  4  S xy
 Dz 
S
2
or
And electric field intensity, as expected:
D
S
2
az
S
E
aN
2 0
Example 9

Suppose:
D   2a 
Find the flux through the surface of a cylinder with
and
0 zh
  a by evaluating the left side and the right side of the
divergence theorem.
Solution to Example 9
Remember the divergence theorem?
 D  dS     DdV
V
We can first evaluate the left side of the divergence theorem by
considering:
   D  dS   D  dStop   D  dSbottom   D  dSside
Solution to Example 9 (cont’d)
A sketch of this cylinder is shown with differential vectors.
The integrals over the top
and bottom surfaces are
each zero, since:
a  az  0
Thus,
   D  dS   D  dS side

2

h
2

 a   ddz
 0 z 0
 2ha3
Solution to Example 9 (cont’d)
For evaluation of the right side of the divergence theorem, first find
the divergence in cylindrical coordinate:
1 

D 
 
1  3

  3
 
D
 
Performing a volume
     DdV
V
integration on this divergence,
  3 dddz
3
2
a
 
h
2

 dddz
 0  0 z 0
 2ha3
This is the same!