Download electrostatic 2014-2015(chap1)

Institute of Technology of
Cambodia
Department of Foundation Year
Electricity
Part I
Electrostatics
Chapter I
Field, Potential, and Energy of Electrostatics
I- Coulomb’s law
 Coulomb’s law explain the interaction of charges like
points
Supposing, there are two point charges at P and M
P
q1

F21

F12 M
 q2
The charge q1 exerces the force F12 on q2, and the charge q2
exerces the force F21 on q1. F12 and F21 are defined by:
II- Charge Distribution
1) Discontinuous Distribution of Charge
A system consists of many point charges. The Total charge of
the system is defined by the sum of each point charge:
n
Q   qi
i 1
2) Continuous Distribution of Charge
There are three kinds of distributions:
a) Volume distribution
Let ρ is the density of charge in a volume, (C/m3)
So that the charge contained within the volume is:
b)Surface Charge distribution
Let σ is a charge density on a surface
So that the charge contained on the surface is:
c) Line charge distribution
Let λ is density of charge (C/m)
So that the charge contained the line is:
III Electrostatics Field
1) Electrostatics field of a point charge
At P we place a charge of q, at M we place charges of q1, q2,
……., qn . They are called test charges.
 If we place q1at M, it experiences a force F1 by q
P
q

F1 M
 q1
 If we place q2at M, it experiences a force F2 by q
P
q

F2 M
q2
 If we place qn at M, it experiences a force Fn by q

Fn M
 qn
P
q
The Force Fn is defined by:

Fn 
1
qq2
4 0 r PM
3

F
1
q
PM  n 
qn 4 0 r PM
3
PM
So that the last expression is called the
“Electrostatics Field, E” of a point charge q
P
q
M


EM
In general a charge q at point P, creates an Electrostatics
field at M. The vector field E is defined by:

EM 
1
q
40 r PM
3
PM
2) Superposition of many point charges
There are n point charges q1, q2, ….., qn are placed at
point P1, P2, ….., Pn which radius vector r1, r2, ….., rn
respectively. M is an other point with radius vector r.
The resultant of electrostatics field at
point M is defined by the superposition
of the fields created by each charge.
The expression of electric field at M is defined by:
3) Electric Field of a Continuous charge distribution
 For volume charge
The element charge dq at point P creates
An element field of dE at point M.
The resultant field at point M is defined by intergral of dE
The same for surface and Line charge distribution
4) Tube line and streamline
 If (C) is a closed contour, the vector field which across
this contour form a tube. We call Tube Line (បំពង់ដែន)
 All point, the vector field is always tangent to the dr
 By solving this system equation we obtain the equation
of streamline(សមីការដសែដែន)
IV Gauss’ theorem
1- Orientation of a surface
- n is called normal vector of the surface (S)
- n is define by the rule of Maxwell or
screw rule (For an open surface)
- For a clossed surface direction of n is from
- inside to out side
2) Flux of a vector Field
The flux of the arbitrary vector field A through an arbitrary
surface is the average normal component of ector A over
the surface multiplied by the area of the surface.
Flux across an element surface of dS is dφ
3) Solid Angle
By definition; the Solid Angle is defined
by normal area divided by square of radius
Solid Angle of a closed surface
In case of observation point O is out side the
surface
We obtain the total solid angle by integral :
 Which means that if the observation point is out side a
closed surface the total solid angle is zero.
In case of the observation point O is inside the
closed surface
Firstly, we would calculate the solid angle of a sphere of
radius R center O and the observation point in coincide to
the center O of the sphere.
By supposing that we have a closed surface(Σ), the pink one.
The observation point is O (inside Σ) then we chose a sphere
center O, Radius R
 For all sharp of any closed surface if the observation
point inside the surface, the solid angle equal 4π
4) Apply Gauss’s Theorem
By supposing that we have a charge which distributes over a volume.
At point P the element charge dq creates an element field dE at M on
an arbitrary enclosed surface
The element of flux across the surface dS is


1 d
d   dE.n ds 
PM
.
n
ds
3
4 0 PM
2
d 2 
1
4 0
dd
Flux of E throught a surface element ds is:
d 
1
40
d d 
v
Q
40
d
Total Flux of E throught a closed surface S is:
Qin

4 0
Qin
Qin
s d  4 0 4   0
Where Qin is total charge continue in a closed surface.
By putting ε0E = D; is called Electric Flux Density Vector,
we obtain:
 
 D.nds  Qin
s
V - Electrostatics work and Electrostatics Potential
1) Expression of elementary work of an electrostatics force .
charge dq at point P created at point M an electrostatics filed dEM

dEM 

dEM 
d
1
40 r P M
i
3
PM ;
d
 
  3 (r  r ' )
40 r r  r '
1
q is placed at M, the force dF applies on q is

dF  q
d
 
  3 (r  r ' )
40 r r  r '
1
The elementary work effected by the elementary force dF is:
 
1
d   
 W  dF .dr  q
  3 (r  r ' ).dr
40 r  r '
2
Or
 2W  q
 
1 d
.
d
r
 r'


2
4 0 r  r '
Or
 1
 W  q
d .d   
4 0
 r  r'
2
1
The Elementary work effected by the force F=qE is:
 1 
W  q 
 .d    d
40
V
 r  r' 
d  1 
W  qd 
    qdV

40  r  r ' 
V
1




2) Electrostatics Potential
a) For a point charge
There is a point charge at point P. The electrostatics potential
at M is defined by:
Note: Potential is a scalar quantity, it can be positive
or negative
b) Superposition of Potential
There are n point charges q1, q2, ….., qn are placed at
point P1, P2, ….., Pn which radius vector r1, r2, ….., rn
respectively. M is an other point with radius vector is r.
The potential at M is defined by the sum of
Potential of each point charge.
V 
1
40
n
qi
i 1
Pi M

c) For a Continuous Charge
VI Relationship between E and V
1) Gradient Field
Let M is a point in free space:





OM  r  xux  yu y  zuz  rur
 Which means that E is a conservation field and rotE = 0
From the above relation we obtain:
Is called electric potential
2) Gradient of V in the Three Coordinate Systems
3)Stock’s Theorem
The circulation of an Electrostatics field along a closed
contour is zero.
VII- Local Equation of electromagnetism
 Divergence of a vector field
By definition the flux of a vector A is defined by:
and when the vector field is the electric flux density:
= div D
Maxwell’s first equation
By Green Ostrogradsky
By Gauss’Theorem:

Qin 1
d
s  E.nds   0   0 
v 
 Divergence Expressions in the Three Coordinate Systems
VIII Potential Energy of interaction
1) Electric work of a point charge in an External Field
The is a point charge is placed in an external electric field E. The charge
experiences an electric force F=qE.
 Work done by the electric force
to move the charge q from A to B is:
By putting UP(A)=qVA; UP(B)=qVB are called potential energies at A and B
 If a man (external force) want to move the charge q from A to B,
he need a force Fex= - F = -qE, and the work done by himself is Wex:
 
 
Wex  Fex .dr  qE.dr
B 
B

Wex  q  E.dr  q  dV  qVB  VA   U P ( B )  U P  A) 
A
A
 This means that UP(a) and UP(B) is the work done by the electric force
to take the charge q from point A and B to infinite, or the work done by
the man (external force) to take the charge q from infinite to point A and B .
 The meaning of potential energy at a point is the
work done by the man (external force) to move the
charge q from infinite to that point disused.
2) Potential Energy of Interaction of a Point Charge
Distribution
Supposing there is a system consists of three point charges q1;
q2; q3 are located with radius vectors r1; r2; r3 respectively. The
system produces potential interaction energy. It equal the work
done by the man to take them from infinite to each point
 Firstly, to bring q1 from infinite to r1
we don’t need work, W(ex1) = 0
 Secondly, to bring q2 from infinite to r1
we need work W2
 
1
q1q2
Wex 2    F12 .dr  
4 0 r  r1
Wex 2   
1
4 0
 
q1q2
.
d
r
 r1
  2
r  r1
  

r
 r1 .dr
3
Wex 2 
W( ex 2 )
Wex 2 
 q1 q 2 

d    
4 0  r  r1 
1
 q1 q 2 

d    

4 0   r  r1 
q1 q 2
1
1 q1 q 2


 

4 0 r  r1
4 0 r12
1
r2
 Thirdly, to bring q3 from infinite to r3
we need work W(ex3)
q2 q3
1  q1q3

 
 
4 0  r3  r1
r3  r2
1  q1q3 q2 q3 
Wex 3  
   
4 0  r13
r23 
Wex 3  




 The total work to bring each charge to each place we
need W(ex)= W(ex1)+ W(ex2)+ W(ex3)
1  q1q2 q1q3 q2 q3 
Wex  
     

4 0  r12
r13
r23 
Wex 
qi q j



40 i 1 j 1 rij
1
3
3
j i
 In case of many point charges can reduce this relation to:
 As we know the potential energy interaction equal to the
work done by the external (man) to bring all charge from
infinite to each point.
We obtain;
IX Electrostatics Energy of a continuous
charge
1) Electrostatics Energy Interm of Potential
We have;
For a continuous charge we just change Σ to ʃ;
to dq; and V(ri) to V(r)
We obtain;
2) Electrostatics Energy Interm of Electric Field
From
1
U e   Vd
2 v
And
 
divE 
0

1
 U e   0 V .divEd
2 v




div (VE )  VdivE  gradV .E  VdivE  E 2
0

0
U e   div (VE )d   E 2d
2 v
2 v


By Osrograsky Green;  div (V .E ).d   V .E.ds
v
s
As;

r  ,V  0  V .E.ds  0
s
We obtain;
1
U e   0  E 2 d
2 all space
X) Boundary Condition
Supposing, there are two deference medium separated by a
surface, and their permittivity are ε1 and ε2 respectively. An
electric field traverses from medium (1) to medium (2). We want
to study about their Normal component and Tangent
components.
1) Tangent Components of Field
We have;








   c    d 
  a   
  E1n  E1t drab   En  Et drbc   E2 n  E2t drcd   En  Et drda  0
b
a
b
c
  c  d  a 
  E1t drab   En drbc   E2t drcd   En drda  0
b
a
b
c
d
d
a 
 

  En drbc   En drda  0
c
b
d
  d  
  E2t drab   E1t drcd  0
b
a
c


 E1t l  E2t l  0
2) Normal Components of a Field
The total flux of field D across a closed cylindrical is zero.

 D.nds   ds
s
s
 
 
  D1n .n1ds1   D2 n .n2 ds2   ds
s
 


s

 
D2 n  D1n n21ds   ds
s
s
s
 D2 n  D1n  
We obtain;
 2 E2n  1E1n   