1
EELE 3331 – Electromagnetic I
Chapter 4
Electrostatic fields
Islamic University of Gaza
Electrical Engineering Department
Dr. Talal Skaik
2012
2
Electrostatic Fields
Electrostatic Fields (time-invariant)
→ produced by static charge distribution.
Electrostatics Applications:
Electric power transmission, X-ray machines, lightning protection,
photocopiers, liquid crystal displays, spray painting…etc.
Two Fundamental Laws:
•
Coulomb’s Law.
•
Gauss Law
3
Electrostatics Application: Photocopiers and laser printers
4
Coulomb’s Law
Deals with the force a point charge exerts on another point charge.
Two polarity of charges may be positive or negative.
→ Like charges repel, unlike charges attract.
Charges measured in Coulombs (C). 1C is approximately
equivalent to 6x1018 electrons.
e=1.602x10-19 C
 Proton has a charge of e.
 Electron has a charge of –e.
 Atom (or group of atoms) that has lost one or more electrons gives
a net positive charge.
 If it gained one or more electrons, it gives net negative charge. 5
Coulomb’s Law
Coulombs law states that the force F between two point charges Q1
and Q2 is:
1) Along the line joining them.
2) Directly Proportional to the product Q1Q2 of charges.
3) Inversely Proportional to the square of the distance R between
them.
KQ1Q2
Q1Q2
=
F =
R2
4πε 0 R 2
−9
10
8.854 × 10−12 
F/m
ε 0 → permitivity of free space, ε 0 =
36π
1
k (constant) → k =
 9 × 109 m/F
6
4πε 0
Coulomb’s Law
F
KQ1Q2
Q1Q2
=
R2
4πε 0 R 2
The force F12 on Q2 due to Q1 is:

Q1Q2 ( r2 − r1 )
Q1Q2
F12 =
a =
3
2 R12
4πε 0 R
4πε 0 r2 − r1


R 12 r2 − r1
where a R12 =
=
, R R 12
R
R
7
Coulomb’s Law

Q1Q2 ( r2 − r1 )
Q1Q2
F12 =
a =
3
2 R12
4πε 0 R
4πε 0 r2 − r1
Notes:
→ The force F21 on Q1 due to Q2 is:

F21 = F12 a=
F12 −a R12
R21


F21 = −F12
or
(
)
→ The distance R must be >> dimensions of charged
bodies Q1 and Q2
→ Q1 and Q2 must be static (at rest)
→ Signs of Q1 and Q2 must be taken into account.
8
Coulomb’s Law
If there are N point charges Q1, Q2,…,QN located at points with
position vectors r1, r2,….,rN, the resultant force F on a charge
Q located at point with position vector r is given by using
principle of Superposition.
 Q Q1 ( r − r1 ) Q Q2 ( r − r2 )
Q QN ( r − rN )
+
+ +
F12 =
3
3
3
4πε 0 r − r1
4πε 0 r − r2
4πε 0 r − rN
→

Q
F12 =
4πε 0
N
∑
k =1
QK ( r − rK )
r − rK
3
9
Electric Field Intensity (E)
The Electric Field Intensity (or electric field strength) E is the
force per unit charge when placed in an electric field.
F
E=
Qt
F=
Q Qt ( r − r')
4πε 0 r − r'
3
Q ( r − r')
F
Q
a (Newton/Coulomb) or (V/m)
→E=
=
=
3
2 R
4πε 0 R
Qt 4πε 0 r − r'
E: Electric field intensity at point r due to a point charge (Q)
Note: If we increase Qt, force F increases by the same factor, and hence
E=F/Qt is the same at the location where E is to be found.
10
Electric Field Intensity (E)
If there are N point charges Q1, Q2,…,QN located at r1, r2,….,rN,
the Electric field intensity at point r is:
E=
or
Q1 ( r − r1 )
4πε 0 r − r1
E=
3
+
Q2 ( r − r2 )
4πε 0 r − r2
1
4πε 0
N
∑
k =1
3
+ +
QN ( r − rN )
4πε 0 r − rN
3
QK ( r − rK )
r − rK
3
11
Electric Field Intensity (E)
12
Example 4.1
Point Charges 1 mC and -2 mC are located at (3,2,-1) and (-1,-1,4),
respectively. Calculate the electric force on a 10 nC charge
located at (0,3,1) and the electric field intensity at that point.
F=
1
4πε 0
∑
k =1,2
Q = 10 nC,
r= ( 0,3,1) ,
F
QQK ( r − rK )
r − rK
3
Q1 = 1 mC,
r1 = ( 3, 2, −1) ,
Q2 = −2 mC
r2 = ( −1, −1, 4 )
−3
−3






10
0,3,1
3,
2,
1
2
10
0,3,1
1,
1,
4
−
−
×
−
−
−
(
)
(
)
(
)
(
)
Q 

−




3
3
4πε 0  ( 0,3,1) − ( 3, 2, −1)
( 0,3,1) − ( −1, −1, 4 )


13
Example 4.1 - continued
2 (1, 4, −3) 
(10 ⋅10−9 )(10−3 )  ( −3,1, 2 )
−
F


−9
3/2
3/2
10
 ( 9 + 1 + 4 )
(1 + 16 + 9 ) 
4π ⋅
36π

−2, −8, 6 ) 
(
−2 ( −3,1, 2 )
F=9 ×10 
+

26 26 
 14 14
F= − 6.507 a x − 3.817 a y − 7.506 a z , mN
−3
10
−6.507 a x − 3.817 a y − 7.506 a z )
(
F
→ At that point, E= = −9
10 ⋅10
Q
→ E= − 650.7 a x − 381.7 a y − 750.6 a z KV/m
14
Example 4.2
Two point charges of equal mass m, and charge Q are suspended at a
common point by two threads of neglible mass and length l. Show
that at equilibrium the inclination angle α of each thread to the
vertical is given by
Q = 16πε 0 mgl sin α tan α
2
2
2
If α is very small, show that
α=3
Q
2
16πε 0 mgl 2
15
Example 4.2 - continued
At A or B, Since the point charges are at equilibrium, net horizontal
and net vertical forces are zero.
Vertical
→ T cosα=mg
Horizontal→ T sinα=Fe
Fe
Q
sin α
1
= =
⋅
cos α mg mg 4πε 0 r 2
2
But=
r 2l sin α →=
r 2 4l 2 sin 2 α
Q cos α = 16πε 0 mgl sin α
2
2
3
Q = 16πε 0 mgl sin α tan α as required
2
2
2
2
3
Q 2 /16πε 0 mgl
For very small α , tanα  α  sin α ⇒ α =
16
Electric Field due to continuous
charge distributions
17
Electric Field due to continuous charge distributions
Line Charge Density ρ L (in C/m)
Surface Charge Density ρ S (in C/m 2 )
Volume Charge Density ρV (in C/m 3 )
To find the charge element dQ :
dQ= ρ L dl → Q=
∫ ρ dl
L
(Line Charge)
L
= ρ S dS → Q
=
dQ
∫ ρ dS
S
( Surface Charge)
S
dQ
= ρV dV → =
Q
∫ρ
V
V
dV
(Volume Charge)
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Electric Field due to continuous charge distributions
E of a point charge is:
E=
Q
4πε 0 R
2
aR
Replace Q by charge elements and integrate:
ρ L dl
a
E= ∫
(Line Charge)
2 R
4πε 0 R
L
ρ S dS
a
E= ∫
2 R
4πε 0 R
S
(Surface Charge)
ρV dV
a
E= ∫
2 R
4πε 0 R
V
(Volume Charge)
19
A.
A line Charge
Consider a line charge with uniform charge density ρL extending from
A to B along the z-axis. Find E(x,y,z).
∗ Charge element
dQ = ρ L dl
∗ Use (x, y , z ) for
the field point.
∗ Use (x ', y ', z ' ) for
the source point.
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A line Charge
=
dQ ρ=
ρ L dz '
L dl
Q=
ZB
∫ ρ dz '
L
ZA
E=
ZB
∫
ZA
ρ L dz '
aR
2
4πε 0 R

R ( x, y , z ) − (0,0, z ')


R = xa x + ya y + ( z − z ')a z → R = ρ a ρ + ( z − z ') a z

ρ a ρ + ( z − z ')a z
ρ a ρ + ( z − z ')a z
ρL
aR
R
=  3=
→ E=
dz '
3/2
3/2
2
∫
R
4πε 0  ρ 2 + ( z − z ') 2  21
 ρ 2 + ( z − z ') 2 
R


A line Charge
ρ 2 + ( z − z ') 2= ρ sec α
z ' 0T − ρ tan α
=
R=
dz ' = − ρ sec 2 α dα
z − z' =
R sin α
→
ρ aρ + ( z − z ')a z
ρL
E=
dz '
3/2
∫
4πε 0  ρ 2 + ( z − z ') 2 

→E
α2
ρL
4πε 0 α∫

R cos α aρ + R sin α a z
R ⋅ R2
1
α2
ρL
→E=
−
4πε 0 α∫
1
( − ρ sec α ) dα
2
ρ sec 2 α cos α aρ + sin α a z 
dα
2
2
ρ sec α
22
A line Charge
α2
ρL
E= −
4πε 0 α∫
1
ρ sec 2 α cos α aρ + sin α a z 
dα
2
2
ρ sec α
α2
ρL
cos α aρ + sin α a z  dα
−
E=
∫
4πε 0 ρ α
1
Hence, For finite Line Charge:
ρL
 − ( sin α 2 − sin α1 ) aρ + ( cos α 2 − cos α1 ) a z 
→
=
E
4πε 0 ρ
23
A line Charge
For Finite Line Charge:
ρL
 − ( sin α 2 − sin α1 ) aρ + ( cos α 2 − cos α1 ) a z 
→=
E
4πε 0 ρ
For Infinite Line Charge: B at (0,0,∞), A at (0,0, − ∞)
→ α1 =
−π / 2, z-componet vanishes,
π / 2, α 2 =
⇒
ρL
aρ
E=
2πε 0 ρ
If the line is not along the z-axis:
ρ is the perpendicular distance from the line to the point
of interest, and aρ is a unit vector along the distance
directed from the line charge to the field point.
24
B.
Surface Charge
Infinite sheet of charge in the xy plane with uniform charge density.
dQ=ρS dS (charge associated with elemental area dS)
25
Surface Charge
E field at P(0,0,h) due to dQ on elemental surface 1 is:
dQ
a
dE=
2 R
4πε 0 R

R = ρ ( −a ρ ) + h a z ,


R
2
2
R =ρ + h , a R =
R
dQ
ρ=
ρ S ρ dφ d ρ
S dS
ρ S ρ dφ d ρ  − ρ a ρ + h a z 
dE=
2
2 3/2
4πε 0  ρ + h 
26
Surface Charge
Due to symmetry of charge distribution, for every element 1,
there is element 2 whose contribution along a ρ cancels that
of element 1.
→ E has only z-component.
E= ∫ dE Z
S
2π
∞
ρS
hρ d ρ dφ
E=
az
3/2
∫
∫
4πε 0 =φ 0=ρ 0  ρ 2 + h 2 


∞
∞
1/2
−
ρSh
ρ dρ
ρSh 

2
2
E=
( 2π ) ∫ 2 2 3/2 a z =
 −  ρ + h 
az
4πε 0
2ε 0 
0 
ρ =0  ρ + h 


27
Surface Charge
∞
−
1/2
ρSh 

2
2
E=
 −  ρ + h 
az
2ε 0 
0 
ρS
→ E = az
2ε 0
(For h>0)
ρS
For (h<0) → E=
( −a z )
2ε 0
⇒ In general, for infinite sheet of charge
ρS
E=
a n , a n is unit vector normal to the sheet.
2ε 0
Integration: Let u = ρ 2 + h 2 → du = 2 ρ d ρ
1 du
1 −3/2
−1/2
⇒∫
=
u
du
=
−
u
2 u 3/2 ∫ 2
28
Surface Charge
ρS
E=
an
2ε 0
Notes:
→ E is independent of the distance between the sheet
and point P.
→ In parallel plate capacitor, the electric field existing
between the two plates having equal and opposite
charges is given by:
ρS
−ρS
ρS
E=
an +
( −a n ) =a n
29
2ε 0
2ε 0
ε0
C.
Volume Charge
A sphere with radius a centred at the origin, with volume charge
density ρV (in C/m3). Find E(0,0,z).
dQ = ρ v dv

ρ v dv
dE=
a , R= R
2 R
4πε 0 R

R
R cos α a z − R sin α a ρ

R
= cos α a z − sin α a ρ
a=
R
R
Eρ components add up to zero.
ρV dv cos α
→ Ez = ∫
R2
4πε 0
30
Volume Charge
ρ v dv cos α
Ez =
4πε 0 ∫ R 2
dv = r '2 sin θ ' dr ' dθ ' dφ '
→ [See integration page 119]
E=
Q
4πε 0 z
2
az
[at point (0,0,z)]
→ Q is the total charge on sphere
4π a 3
ρ v dv ρ=
ρv
Q=∫ =
v ∫ dv
3
v
v
→ Generally, the electric field at P( r, θ , φ ) is:
E=
Q
4πε 0 r
2
ar
[Identical to point charge] (see Gauss's law)
31
Example 4.4
Circular ring of radius a, carries
a uniform ρL C/m.
(a) Show that
E(0,0, h ) =
ρ L ah
2 3/2
2ε 0  h + a 
2
az
(b) What values of h gives Emax?
(c) If the total charge on the ring
is Q, find E as a→0.
32
Example 4.4 - continued
ρ L dl
( a ) E= ∫
a
2 R
4πε 0 R
L

dl adφ , R=a ( −a ρ ) + ha z


R
2
2
R= R= a + h , a R =
R
ρ L ( −a a ρ ) + ha z
a dφ
E=
3/2
∫
4πε 0  a 2 + h 2 


By symmetry, the contributions
along a ρ add up to zero.
E=
2π
ρ L aha z
2 3/2
4πε 0  a 2 + h 
∫ dφ → E= 2ε
0
ρ L ah
2 3/ 2
2

+ h 
a
0
az
33
Example 4.4 - continued
2
2 3/2
2
2 1/2






(1)
(3
/
2)(2
)
(
)
a
h
h
a
h
h
+
−
+
d E  




( b)
=

3
2
2
dh 
 a + h 


d E
For maximum E,
=0
dh
2 1/2
2
2
2



0
→  a + h   a + h − 3h  =
a
2
2
0 or h =
±
a − 2h =
2
2
34
Example 4.4 - continued
( c ) Since the charge is unifromly distributed,
Q
,
ρL =
2π a
ρ Lah
az
 E=
3/2
2ε 0  a 2 + h 2 
Qh
az
→ E=
3/2
4πε 0  a 2 + h 2 
Q
As a → 0, E=
a
2 z
4πε 0h
or in general E=
Q
4πε 0 r
2
a R (same as that of point charge)
35
Example 4.5
The finite sheet 0≤x ≤1, 0 ≤y ≤1 on the z=0 plane has a charge density
ρS = xy(x2+y2+25)3/2 C/m2. Find:
(a) The total charge on the sheet
(b) The electric field at (0,0,5)
(c) The force experienced by -1 mC charge located at (0,0,5).
(a) Q= ∫ ρ=
S dS
1 1
∫ ∫ xy ( x
2
+ y + 25) dxdy nC
2
3/2
0 0
S
1
1
3/2
du
2
2
→ ∫ ∫ y ( u + y + 2 5) ( du / 2)dy
Let x = u, du = 2 x dx, xdx =
2
y 0=
u 0
=
1
1
u + y + 25)
5/2
5/2
(
1
1
2
2

 dy
=
+
−
+
y
dy
y
y
y
26
25
(
)
(
)

2 ∫0
5/2
5 ∫0 
2
5/2 1
0
36
Example 4.5 - continued
1
Q
5/2
5/2
1
2
2

 dy
+
−
+
26
25
y
y
y
(
)
(
)

5 ∫0 
2
Let v y=
, dv 2 y dy , =
Similarly , =
y dy dv / 2
1
1 
5/2
5/2
Q= ∫ ( v + 26 ) − ( v + 25)  dv =
33.15 nC

10 0 
ρ S dS
(b) E= ∫
a ,
2 R
4πε 0 R
S


2
2
R=(0,0,5) - ( x, y ,0) =( − x, − y ,5), R = x + y + 25
1 1
E = ∫∫
0 0
(10 ) xy ( x + y + 25)
−9
2
2
3/2
4πε 0 ( x 2 + y 2
( − xa
+ 25)
− ya y + 5a z )
x
3/2
dxdy
37
Example 4.5
1 1
E = ∫∫
0 0
(10 ) xy ( x + y + 25)
−9
2
2
3/2
( − xa
x
− ya y + 5a z ) dxdy
4πε 0 ( x + y + 25)
2
2
3/2
1 1
1 1
 11 2

2
E=
9  − ∫ ∫ x y dxdy a x − ∫ ∫ xy dxdy a y + 5∫ ∫ xy dxdy a z 
0 0
0 0
 00

  1  1 
 1  1 
 1  1  
E=9  −     a x −     a y + 5     a z 
 2  3
 2  2  
  3 2 
→ E= − 1.5 a x − 1.5 a y + 11.25 a z
(c) F=qE=(1.5,1.5, − 11.25) mN
38
Example 4.6
Planes x=2 and y=-3, respectively, carry charges 10 nC/m2 and
15nC/m2. If the line x=0, z=2 carries charge 10π nC/m, Calculate
E at (1,1,-1) due to the three charge distributions.
Let E=E1 +E 2 +E 3
E1 : due to sheet 1, E 2 : due to sheet 2, E 3 : due to Line
ρs
10 × 10−9
E1 =
( −a x ) =
2ε 0
(2)(10−9 / 36π )
E1 = −180π a x
ρs
15 × 10−9
E2 =
ay
2ε 0
(2)(10−9 / 36π )
E 2 = 270π a y
1
2
39
Example 4.6
ρL
E3 =
aρ
2πε 0 ρ
R=(1,1,-1)-(0,1,2)=(1,0,-3)
R=a x − 3 a z , ρ =
R =
10
a=
ρ
R ax − 3 az
=
=
R
10
E3
10π × 10−9
 ax − 3 az 
=

 18π ( a x − 3 a z )
−9
2π (10 / 36π )( 10) 
10 
1
3
ax −
az
10
10
Total field is E=E1 + E 2 + E 3
−162π a x + 270π a y − 54π a z V/m
E=
40
Electric Flux Density (D)
Electric Flux ψ = ∫ D ⋅ dS [measure of the number of field lines
S
passing through an area]. (in Coulombs)
D → Electric Flux Density (C/m 2 )
D=ε 0 E
(for free space)
ρS
ρS
For an infinite sheet of charge, E=
a n → D= a n
2ε 0
2
ρ v dv
For a volume charge distribution: D= ∫
a
2 R
4π R
v
D is a function of charge and position only, it is independent
of the medium.
41
Example 4.7
Determine D at (4,0,3) if there is a point charge -5π mC at (4,0,0) and
a line charge 3π mC/m along the y-axis.
Let D=DQ +D L
Q ( r − r ')
Q
DQ =ε 0 E=
a
, r −=
0) (0, 0,3)
=
r ' (4, 0,3) − (4, 0,=
3
2 R
4π R
4π r − r '
−5π ×10 −3 (0, 0,3)
2
0.138
a
mC/m
→ DQ =
=
−
z
3
4π (0, 0,3)
DL =
ρL
(4, 0,3) − (0, 0, 0) (4, 0,3)
aρ , aρ =
, ρ = (4, 0,3) − (0, 0, 0) = 5
=
2πρ
(4, 0,3) − (0, 0, 0)
5
3π ×10−3 (4a x + 3a z )
DL =
= 0.24 a x + 0.18 a z mC/m 2
(2π )(5)
5
⇒ D = DQ + D L = 240 a x + 42 a z µ C/m 2
42