Electrostatic Fields - Coulomb’ Law - 1

The force between two Q1 Q2

R  r2  r1
F2
point charges Q1 ,Q 2 : F1
aR
kQ1Q 2 1
F2  a , k  r1
4 o
R
R2 r2

R 摋r2  r1 摋 o
 o  free space permittivity,
[ o ]  farad / m  F / m
109 Q1Q 2
o   8.85 1012 F / m, F2  F1  a
36 4o R 2 R

k = 9 109 m / F
Dr Alaa K. Abdelmageed Eng. Math & Phy Dep., Cairo U Elect/1
 Electrostatic Fields - Coulomb’ Law - 2

F2  Force on Q 2 due to Q1
F2  Force directed along R   r 2  r1 

F2 
Q1Q 2
aR R  r 2  r1 
4 o R aR  
r 2  r1
2
R

Q1Q 2R Q1
R  r2  r1
Q2
F2  F2
4o R 3 F1
aR

Q1Q 2  r 2  r1  r1
F2  3 r2
4 o r 2  r1
o
Dr Alaa K. Abdelmageed Eng. Math & Phy Dep., Cairo U Elect/2
 Coulomb’ Law - Superposition Principle

Q Q1 R1 Q Q2 R2
F  3  3 
4  o R1 4  o R 2
QQN RN
  3 
4  o R N
QQ1  r  r1  QQ 2  r  r2 
F  3  3 
4  o r  r1 4  o r  r2
QQ N  r  rN 
  3 
4   o r  rN

Q Q i  r  ri 
F 
N

4  o i 1
r  ri
3

Dr Alaa K. Abdelmageed Eng. Math & Phy Dep., Cairo U Elect/3

 Electric Field intensity - 1
An electric charge produces an electric field everywhere. To
determine the strength of the field created by that charge, we can
measure the force a positive “test charge” experiences at some point.
Accordingly, the electric field is defined as:

The electric field intensity is the force per a unit positive

test charge when introduced into that field.

qo
F 1 qoQ Q R F
E = lim  lim a
qo 0 q qo 0 q 4 R 2 R aR
o o o

Q
E= aR
4 o R 2

Unit:
Dr Alaa K. Abdelmageed Eng. Math & Phy Dep., Cairo U Elect/4
 Electric Field intensity - 2
For single charge:

Q Q  r  r '
E  aR 
4  o R 2
4  o r  r '
3

For multiple charges:

1 Q i  r  ri ' 
E 
N
 R r  r' 
4  o i 1
r  ri ' 3 aR  
R r  r'

Field lines

Dr Alaa K. Abdelmageed Eng. Math & Phy Dep., Cairo U Elect/5

 Electric Field Lines

An electric flux line: is an imaginary line drawn such that its

direction at any point is the direction of electric field at that
point. It shows the direction of force if an infinitesimal positive
charge is placed in that field. It begins at positive charges and
end in negative charges.

The charge on the right is twice the magnitude of the charge on the left (and
opposite in sign), so there are twice as many field lines, and they point towards
the charge rather than away from it.

Dr Alaa K. Abdelmageed Eng. Math & Phy Dep., Cairo U Elect/6

 Continuous Charge Distribution

R R

r '   x, y, z  r '   x, y, z   R

d
r '   x, y , z  dl dS´
S dV´
dQ  s  r ' dS 
dQ  l  r ' dl  dQ   v  r ' dV 

dQ  l dl ' dQ   S dS '
dQ  v dv '
Q   l dl ' Q   v dv
'
Q    S dS '

[ l ]  C / m [s ]  C / m 2 [  v ]  C / m3

l  r ' ,  s  r ' , v  r '  mean that l ,  s , v are fn. of r '

Dr Alaa K. Abdelmageed Eng. Math & Phy Dep., Cairo U Elect/7
 Example 2.1: Find the total charge within a sphere of radius
if it has a nonuniform charge density .

Solution:
a 2 
Q   v dV     o r 2 r 2 sin  d  d dr
V 0 0 0

2  z
 
 o  r 2 r 2   
sin  d  d  
a
dr
0
 
0 0

a
 4  a

 4o  r 2 r 2 dr y
r
0

a5
x
 4o
5
Dr Alaa K. Abdelmageed Eng. Math & Phy Dep., Cairo U Elect/8
 Field from Line Charge
z

dQ
R dE  a
4 0 R 2 R

l  x, y, z   dl '

 aR
4 0 R 2

dl
y
r '   x, y , z  l  r ' 
E (r )   a dl '
x l
4 0 R 2 R

dQ  l  r ' dl 
l  r '  R
 dl '
R  r r' 4 0 R 3
l
( x  x ') a x  ( y  y ') a y  ( z  z ') a z
l  r ' (r - r' )
R r r'  dl '
aR   l
4 0 摋 r - r' 摋 3

R 摋 r  r '摋
Dr Alaa K. Abdelmageed Eng. Math & Phy Dep., Cairo U Elect/9
 Field from Surface Charge
z

dQ
R dE  a
4 0 R 2 R

 s  x, y , z   dS '
 aR
4 0 R 2

dSdS´ y
 s  r '
E (r )   a dS '
r '   x, y , z  4 0 R 2 R
dQ   s  r ' dS  S
x  s  r ' R
R  r r'  dS '
S
4 0 R 3

( x  x ') a x  ( y  y ') a y  ( z  z ') a z

 s  r ' (r - r' )
R r r'  dS '
aR   4 0 摋 r - r' 摋 3
R 摋 r  r '摋 S

Dr Alaa K. Abdelmageed Eng. Math & Phy Dep., Cairo U Elect/10

 Field from Volume Charge
z

dQ
dE  a
R 4 0 R 2 R

r '   x, y, z  

v  x, y, z   dV 
dV´  aR
4 0 R 2

dQ   v  r ' dV 
v  r '
y E (r )   a dV 
V
4 0 R 2 R

x R  r r' v  r ' R
E (r )   dV 
( x  x ') a x  ( y  y ') a y  ( z  z ') a z V
4 0 R 3

aR 
R

r r' v  r ' (r - r' )
 dV 
R 摋 r  r '摋 4 0 摋 r - r' 摋 3
V

Dr Alaa K. Abdelmageed Eng. Math & Phy Dep., Cairo U Elect/11

 Position Vector in Coordinate Systems

Rectangular Coordinates:

r  ax x  a y y  az z

Cylindrical Coordinates:

r  a   az z

Spherical Coordinates:

r  ar r

Dr Alaa K. Abdelmageed Eng. Math & Phy Dep., Cairo U Elect/12

 Example 2.2: A charged rectangle
(2 ) has a charge density:

 s  2( x 2  y 2  9)3/2 nC / m2

Find: (I) The electric field at ,

(II) The force on a mC charge at

E
1 b dy ' a dx ' s   x 'ax  y 'a y  3a z 
Solution 4 o b a 3
x'  y'  9
2 2

r  3a z , r '  x 'ax  y 'a y

4 o b a  z

2 b dy ' a dx '  x 'a  y 'a  3a
x y

r  r '  3a z  x ' a x  y ' a y

1 b dy ' a dx ' 3a  3  2a  2b
2 o b a  z  2 o
E az
r  r '  x '2  y '2  9 6ab a nV/m
E
 o z
s  r  r ' 
E   dS F = qE = 
12ab  1012 a N
4 o r  r '
3
 o z

Dr Alaa K. Abdelmageed Eng. Math & Phy Dep., Cairo U Elect/13

Example 2.3: Annulus of radii and uniform surface
charge density ௦௢ is located on the xy-plane. Find the electric field
at a point ௢ on the axis of the annulus .

Solution: z
௢
r  a z zo
r  zo a z R
r'  a    s = s0 [C/m2]

R  r r' a
y
 a         a z zo b
a
x r'    a  
dS     d   d 

Dr Alaa K. Abdelmageed Eng. Math & Phy Dep., Cairo U Elect/14

 Example 2.3: ctd
y

R  r  r '  a        az z o d 

dS 
R  r  r '    z o 2 2
x

E field equation:
 s  r  r '
E   dS ' 3

4 0 r  r '
 
2 b
  s 0   a         a z zo 
E  a  4 0   3    d   d 
0      zo 
2 2 2

 
Dr Alaa K. Abdelmageed Eng. Math & Phy Dep., Cairo U Elect/15
 Example 2.3: ctd
function of  
 
 a         a z zo
2 b
 s0  
E       d   d 
 4 0  0 a      zo 
2 2 2
3

 

2
Note that  a
0
 .... d   0 due to symmetry

2
 
 s0  b
 a z zo 
E       d   d 
 4 0  0 a
3
    2  zo 2  2 
 

Dr Alaa K. Abdelmageed Eng. Math & Phy Dep., Cairo U Elect/16

 Example 2.3: ctd

2
 
 s 0  b
 a z zo 
E       d   d 
 4 0  0 a      zo 
2
3
2 2

 

Note: We have a separable integrand with fixed limits

(we can split this double integral into the product of two one-dimensional integrals).

  2
 s0  b
 1 
E   a z zo      d    d 
 4 0 
3
a      zo 
2 2 2
 0
 

 s 0  b

E   a z zo  d   2 
 4 0 
3
a
  2  z 2  2
Dr Alaa K. Abdelmageed Eng. Math & Phy Dep., Cairo U Elect/17
 Example 2.3: ctd

 s0  b

E  2   a z zo  d 
 4 0 
3
a
  2  z o 
2 2

b
  1 
E  s 0 a z zo  
2 0   2  zo 2 
a

 s0   1 1 
E   a z zo     V/m 
 2 0   a  zo b 2  zo 2 
2 2

Dr Alaa K. Abdelmageed Eng. Math & Phy Dep., Cairo U Elect/18

 Example 2.3: ctd
Limiting case: a  0, b   (infinite sheet of charge)

 s0   1 1 
E   z o
a z     V/m 
  a 2  zo 2 b 2  zo 2 
2
 0

z   S 0  zo
E  az  
 2 0  zo
E

 S 0 
E   az    V/m
s0
y  2 0 
+ for
E ௢
x - for ௢
Dr Alaa K. Abdelmageed Eng. Math & Phy Dep., Cairo U Elect/19
 z
Example2.4: Find at arbitrary point dQ  l dz '
due to a finite wire along z (0, 0, z )
axis extending from to R =摋 r  r' 摋

r
Solution:  z y

r   a  z az r '  z 'az x a

  r  r ' 
l
E   dl  
L b L
3 ( z  z ') a z
4 o r  r '  dz '
4 o a 2 2 3/ 2

4 

  ( z  z ') o

E 
L b   a   ( z  z ') a z  [ 
1 ( z  z ')
a 
1
a z ]ba
4 o  dz ' 2  2  ( z  z ') 2  2  ( z  z ') 2
2 3/ 2
a  2  ( z  z ')
if a   , b  
L b  a
E  dz '   2  L
4 o E =   a    2  a 
L
2 3/ 2
a  2  ( z  z ') 4 o   o
Dr Alaa K. Abdelmageed Eng. Math & Phy Dep., Cairo U Elect/20
 Electric Field for infinite Line/Sheet of charge
z
infinite sheet of charge
E

 S 
E   az  
2
 0 y

E
x

Infinite Line of charge:

L E
E= a
2 o 

Dr Alaa K. Abdelmageed Eng. Math & Phy Dep., Cairo U Elect/21