# Electrostatic fields by 56K74R

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```									Electrostatic fields
Sandra Cruz-Pol, Ph. D.
INEL 4151
ECE UPRM
Mayagüez, PR
Some applications
   Power transmission, X rays, lightning protection
   Solid-state Electronics: resistors, capacitors,
FET
   Computer peripherals: touch pads, LCD, CRT
   Medicine: electrocardiograms,
electroencephalograms, monitoring eye activity
   Agriculture: seed sorting, moisture content
monitoring, spinning cotton, …
   Art: spray painting
   …
We will study Electric charges:

 Coulomb's Law
 Gauss’s Law
Coulomb’s Law (1785)
   Force one charge exerts on another

kQ1Q2
F    2
R                         Point
charges
+            R
+
where k= 9 x 109
or k = 1/4peo                    *Superposition
applies
Force with direction

Q1Q2
F12           aˆ
4pe o R 2 12
Example
Example: Point charges 5nC and -2nC are located at
r1=(2,0,4) and r2=(-3,0,5), respectively.
a) Find the force on a 1nC point charge, Qx, located at
(1,-3,7)

Apply superposition:
1  Q1Qx rx  r1  Qx Q2 rx  r2 
F                                      
4pe o  rx  r1

3
rx  r2
3


  5,15,15 8,6,4 
F  9                          1.004,1.285,1.3998
 82.81           156.2 
Electric field intensity
   Is the force per unit charge
when placed in the E field                  F
E
Q
Example: Point charges 5nC and -
Q
2nC are located at (2,0,4) and (-
E                 ˆ
aR
3,0,5), respectively.
4pe o R   2

b) Find the E field at rx=(1,-3,7).

1  Q1 rx  r1  Q2 rx  r2 
E                                
4pe o  rx  r1

3
rx  r2 
3

If we have many charges
Line charge density,             C/m
L
Surface charge density            C/m2
S
Volume charge density             C/m3
v

Q    L dl       Q    S dS         Q    v dv
L                   S                   v
The total E-field intensity is

 L dl
E           ˆ
aR
4pe o R 2

 S dS
E           ˆ
aR
4pe o R 2

 v dv
E           ˆ
aR
4pe o R 2
Find E from LINE charge
B
Q    L dl
    Line charge w/uniform                 A              dl  dz'
charge density, L
z
(x,y,z)                        L dz'
T                                dE           E          ˆ
a
4pe o R 2 R
a
B
z '  OT   tan a
R

(0,0,z’)                    R   seca
dl               
R  R cosa a   R sin a a z
ˆ             ˆ

A                                   R
ˆ R   cos a a   sin a a z
a               ˆ         ˆ
R
x
0
LINE charge                                   z '  OT   tan a
   Substituting in:               L dz'            dz'  [0   sec2 a ]da
E          ˆ
a
4pe o R 2 R
R   seca
z
a R  cos a a   sin a a z
ˆ           ˆ           ˆ
(x,y,z)
T                                     dE
a [  2 sec2 a ]da
E L                   [cosa a   sin a a z ]
ˆ           ˆ
B
4pe o  sec a
2   2
R
finite Line Charge :
(0,0,z’)
dl
L
E         [(sin a 2  sin a1 ) a   (cosa 2  cosa1 ) a z ]
ˆ                       ˆ
4pe o 
A
infinite Line Charge (a1,2  90o )
x
0                        L
E         ˆ
a
2pe o 
More Charge distributions
   Point charge
   Line charge
   Surface charge
   Volume charge
Find E from Surface charge
 S dS
dE           aˆ
   Sheet of charge      dQ   S dS                 4pe o R 2 R

w/uniform density S
z                        dS  dd
R   (  a  )  ha z
ˆ        ˆ
y

R
aR 
ˆ
R

 S  d d   a ρ  ha z 
ˆ      ˆ
dE 

4pe o   h2      2
   3
2
SURFACE charge
 Due to SYMMETRY                  S      2p                      h d
the  component cancels
Ez 
4pe o      0
d 
 0
   2
h   2
   3
2

out.


S
 h 
Ez          2p        
4pe o
  h 

2  2
            0

infinite SurfaceCharge :
S
E      ˆ
an
2e o
More Charge distributions
   Point charge
   Line charge
   Surface charge
   Volume charge
Find E from Volume charge
 v dv
   sphere of charge      dQ   v dv                            dE           aˆ
4pe o R 2 R
w/uniform density, v
dv  r ' sin  ' d ' d ' dr '
2
dE         P(0,0,z)
Law of cosines :
a
(Eq. *)   R 2  z 2  r '2 2 zr ' cos '
(r’,’,’
r '2  z 2  R 2  2 zR cosa
’
v                                  Differentiating (Eq. *)                 RdR
sin  ' d ' 
’                                                                     zr '
Due to symmetry only
x
dEz  dE cos a
survives.
Find E from Volume charge
 v dv
   Substituting…                                            dEz            cos a a z
ˆ
4pe o R 2

P(0,0,z)
dv  r '2 sin  ' d ' d ' dr '
dE

RdR
(r’,’,’                 sin  ' d ' 
zr '
’
v                  2p                z r '
v                      a
RdR    z 2  R 2  r '2 1
0d 'r '0 Rrr '' zr ' dr' 2 zR R 2
’
Ez                               2

4pe o             z
x
3
a                     Q
E  v                    az 
ˆ                    ˆ
ar
De donde salen los
limites de R?                                     3e o r     2
4pe o r   2
P.E. 4.5
   A square plate at plane z=0 and x  2, y  2
carries a charge 12 y mC/m2 . Find the total
charge on the plate and the electric field intensity at
(0,0,10).
2     2              2     2                           2 2
y
Q   dx  12 y dy   dx  12(2) ydy  4 12(2)     192 mC
x  2 y  2    x  2 y  0                 2 0
 
s                       s dS r  r '
E              ˆ
dSar                 
4pe o r 2                 4pe o r  r ' 3
 
r  r '  (0,0,10)  ( x' , y' ,0)  ( x' , y' ,10)
z

Cont…sheet of
charge
y=2

                                                                 x=2
( x, y,10)
2   2
12 y dxdy
E     
 2 y  2
4pe o     x   2
 y  100
2
   3/ 2

2      2
 xdxdya x
ˆ          2 2
 y y dxdya y
ˆ        2  2
10 y dxdya zˆ
 108 10   y 
6
                             y 
 2 x  2 x  y  100                   x  y  100  2 x2 x 2  y 2  100 3/ 2
2    2       3/ 2               2    2       3/ 2
                                 2 x  2

Due to symmetry only Ez survives:
               2 2 10 ydxdyaz 
ˆ
E  108 10  2  
6
3/ 2 
          
2 0 x  y  100 
2   2
     
 16.5az MV / m
ˆ
Electric Flux Density
D is independent of the medium in which the
charge is placed.

            v dv
D  eoE           ˆ
aR [C / m 2 ]
4pR   2

Then the electric flux is :
 
   D  dS [C ]
Gauss’s Law
 
   D  dS  Qenc
S
 
Qenc    v dv   D  dS
S
             
 D  dS     D dv
S          v


v    D
Gauss’s Law
   The total electric flux ,
through any closed
            v dv
D  eoE           ˆ
a
surface is equal to the                   4pR  2 R
 
total charge enclosed by       Qenc   D  dS    v dv
that surface.                            S          v
Some examples: Finding D at point
P from the charges:
D      Point Charge is at the origin.
 
P                   Q   D  dS
r                             S

charge             Choose a spherical dS
   Note where D is perpendicular
to this surface.
Q  Dr  dS  Dr 4pr 2
S

Q
D       ˆ
ar
4pr 2
Some examples: Finding D at point P
from the charges:
   Infinite Line Charge
 
l dl  Q   D  dS
S
Line                D
charge
              Choose a cylindrical dS
P
   Note that integral =0 at top and
bottom surfaces of cylinder

Q  D  dS  D 2p l
 L
D     ˆ                       S
a
2p
Some examples: Find D at point P
from the charges:
   Infinite Sheet of charge
D                           
 s  dS  Q   D  dS
sheet of                                     S
charge
   Choose a cylindrical box
cutting the sheet
Area A                                        
D
 S A  Q  Ds   dS   dS 
top
       bottom 

 S              Note that D is parallel to the
D   ˆ
az            sides of the box.
2                     S A  Ds A  A
P.E.   A point charge of 30nC is located at the origin, while
plane y=3 carries charge 10nC/m2.
4.7   Find D at (0, 4, 3)

            Q        s
D  DQ  D        ar  an
ˆ      ˆ
4pr 2
2
         30 109
D                     (0,4,3)  (0,0,0)  a y
10n
ˆ
4p      4 3
2   2

3
2

 30 10 9
D          (0,4,3)  5na y
ˆ
4p 5
3

 5.08 a y  0.057 a z nC/m 2
ˆ           ˆ

P.E. 4.8
       
If D  2 y 2  z ax  4xyay  xaz C/m2 . Find :
ˆ       ˆ     ˆ
  volume charge density at (-1,0,3)

v (1,0,3)    D  4x  4C/m3

   Flux thru the cube defined by 0  x  1, 0  y  1,0  z  1
1 1 1
  Qenc    v dv     4 x dxdy dz
v           0 0 0
   Total charge enclosed by the cube

Q    2C
Review
Point charge or              Q
volume                   D       ˆ
ar
Charge distribution         4pr 2

Line charge            L
distribution          D     ˆ
a
2p

Sheet charge
distribution           S
D   ˆ
an
2
We will study Electric charges:

 Coulomb's Law (general cases)
 Gauss’s Law (symmetrical cases)

 Electric Potential (uses scalar, not vectors)
Electric Potential, V
   The work done to move a charge Q from A to B is

dW   F  dl

ˆ
ay                        QE  dl
   The (-) means the work is done by an external force.
   The total work= potential energy required in moving Q:
 
B
W  Q  E  dl
A
   The energy per unit charge= potential difference between the 2 points:
B        J
 C   V
W
VAB      E  dl
Q   A           

V is independent of the path taken.
The Potential at any point is the potential difference
between that point and a chosen reference point at
which the potential is zero. (choosing infinity):
r          r                                   r

V (r )   E  dl  
Q
a  dr 'ar 
ˆ      ˆ
Q 1

Q
V
           
4pe o r ' 2 r
4pe o r '  4pe o r
For many Point charges at rk:                                         n

(apply superposition)           V (r ) 
1

Qk
V
4pe                   o k 1         r  rk
ˆ ˆ

1                 L r 'dl '
ˆ
For Line Charges:                  V (r ) 
ˆ
4pe o         L
r  r'
ˆ ˆ
For Surface charges:
1              s r 'dS '
ˆ
V (r ) 
ˆ
4pe o     S
r  r'
ˆ ˆ
For Volume charges:                             1                    v r 'dv'
ˆ
V (r ) 
ˆ
4pe o          
v
r  r'
ˆ ˆ
A point charge of -4mC is located at (2,-1,3)
P.E. 4.10      A point charge of 5mC is located at (0,4,-2)
A point charge of 3mC is located at the origin

Assume V(∞)=0 and Find the potential at (-1, 5, 2)
3
Qk
V (r )                   C
k 1 4pe o r  rk

r  r1  (1,5,2)  (2,1,3)  46
r  r2  (1,5,2)  (0,4,2)  18
r  r3  (1,5,2)  (0,0,0)  30
106   4    5    3 
V (1,5,2)           9 
         =10.23 kV
1 / 9 10  46   18   30 
Example
A line charge of 5nC/m is located on line x=10, y=20
Assume V(0,0,0)=0 and Find the potential at A(3, 0, 5)

              L
V (r )    E  dl   
ˆ                              a  d a
ˆ       ˆ
2pe o 
L
V (r )  
ˆ            ln   C
2pe o
L
Vorigin  VA         ln  o  ln  A 
2pe o
0  VA  4.8               VA=+4.8V
0=|(0,0,0)-(10,20,0)|=22.36 and A=|(3,0,5)-(10,20,0)|= 21.2
A point charge of 5nC is located at the origin
P.E. 4.11        V(0,6,-8)=2V and Find the potential at A(-3, 2, 6)
Find the potential at B(1,5,7), the potential difference VAB

r  (0,0,0)  (0,6,8)  10
Q
V               C             5n                    C  2.5
4pe o r          2
4pe o 10
C

5n
VA                            C  3.93V
4pe o (3,2,6)  (0,0,0)
5n
VB                           2.5  2.696 V
4pe o (1,5,7)  (0,0,0)

VAB  VB  VA  1.233V
Relation between E and V
V is independent of the path taken.
B
V AB  VBA

V AB  VBA          E dl  0

Esto aplica sólo a campos estáticos.
Significa que no hay trabajo NETO en mover una carga
en un paso cerrado donde haya un campo estático E.

       
             
 E dl     E dS  0
S
A
Static E satisfies:

 E  0                     B
Condition for Conservative field = independent
of path of integration

dV   E  dl
  Ex dx  E y dy  Ez dz

V      V      V
dV       dx     dy     dz
x      y      z

A
E  V
Given that E=(3x2+y)ax +x ay kV/m, find the
P.E. 4.12       work done in moving a -2mC charge from
(0,5,0) to (2,-1,0) by taking the straight-line
path.
a) (0,5,0)→(2,5,0) →(2,-1,0)  W
Q

  E  dl   3x  y dx  xdy
2
        
1
W
       
2
  3x 2  y dx  xdy
Q   0             5
W  (Q)(18  12)                W  6(2m )12mJ
b) y = 5-3x
dy  3dx
W
Q
                  
  E  dl   3x 2  5  3x dx  x(3dx)                 
W                         W
  3x  6 x  5dx 
2
2                  8  12  10  6 W  12mJ
Q    0                    Q
Given the potential V  10 sin  cos
Example                                                   2
r
 p 
 2, ,0 
Find D at     2 .
       
D  e o E  e o  V 
In spherical coordinates:
     V      1 V          1 V 
E      ˆr 
a          ˆ 
a              ˆ
a 
 r      r        r sin   
     20                10               10     V 
E   3 sin  cos ar  3 cos cosa  3 sin 
ˆ                 ˆ            ˆ
a 
r                  r                r       
                                      20              
D  eo E                          e o  ar  0a  0a 
ˆ   ˆ     ˆ
( 2 ,p / 2 , 0 )
 8               

D  22 .1a
ˆr C / m 2
Electric Dipole
   Is formed when 2 point charges of equal but opposite
sign are separated by a small distance.

Q 1 1         Q  r2  r1 
z
P
V                         
4pe o  r1 r2  4pe o  r1r2 
r1
Q+                         For far away observation points (r>>d):
r
r2                  Q d cos 
V
d
y         4pe o r 2

Q-
Energy Density in Electrostatic fields
   It can be shown that the
total electric work done
is:

1         eo
WE   D  Edv   E 2 dv
2v          2 v

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