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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