Lecture 8
Ch23. Finding the Electric Field – II
University Physics: Waves and Electricity
Dr.-Ing. Erwin Sitompul
http://zitompul.wordpress.com
2013
Homework 6: Three Particles
Three particles are fixed in place and have charges q1 = q2 =
+p and q3 = +2p. Distance a = 6 μm.
What are the magnitude and direction of the net electric field
at point P due to the particles?
p  1.602 1019 C
e  1.602  1019 C
Erwin Sitompul
University Physics: Wave and Electricity
8/2
Solution of Homework 6: Three Particles
EP,net  E1  E2  E3
E2

E3
E1
E1  E2  0
 EP,net  E3  k
EP ,net
r3P  12 aˆi  12 aˆj
r3 P  12 2a
r
rˆ3P  3 P  12 2iˆ  12 2jˆ
r3 P
 cos ˆi  sin  ˆj
   45
Erwin Sitompul
• Both fields cancel
one another
q3
r3P
rˆ
2 3P
(2 1.602 1019 )
 8.99 10 1
( 2 2(6 106 ))2
9
 160 N C • Magnitude
EP,net    45
• Direction
University Physics: Wave and Electricity
8/3
The Electric Field
→
 The calculation of the electric field E can be simplified by
using→symmetry to discard the perpendicular components of
the dE vectors.
 For certain charge distributions
involving symmetry, we can simplify
even more by using a law called
Gauss’ law, developed by German
mathematician and physicist Carl
Friedrich Gauss (1777–1855).
→
 Instead of considering dE in a given charge
distribution, Gauss’ law considers a
hypothetical (imaginary) closed surface
enclosing the charge distribution.
 Gauss’ law relates the electric fields at points
on a closed Gaussian surface to the net
charge enclosed by that surface.
Erwin Sitompul
University Physics: Wave and Electricity
8/4
Flux
→
 Suppose that a wide airstream flows with uniform velocity v
flows through a small square loop of area A.
 Let Φ represent the volume flow rate (volume per unit time) at
which air flows through the loop.
→
 Φ depends on the angle θ between v and the plane of the
loop.
A  A rˆN
Erwin Sitompul
• Unit vector pointing to the normal
direction of the plane
University Physics: Wave and Electricity
8/5
Flux
→
 If v is perpendicular to the plane (or parallel to the plane’s
direction), the rate Φ is equal to vA.
→
 If v is parallel to the plane (or perpendicular to the plane’s
direction), no air moves through the loop, so Φ is zero.
 For an intermediate angle θ, the rate of volume flow through
the loop is:
  (v cos  ) A  v  A
 This rate of flow through an area
is an example of a flux.
 The flux can be interpreted as
the flow of the velocity field
through the loop.
Erwin Sitompul
University Physics: Wave and Electricity
8/6
Flashback: Multiplying Vectors
The Scalar Product
→
→
→→
 The scalar product of the vector a and b is written as a·b
and defined to be:
a  b  ab cos 
→→
 Because of the notation, a·b is also known as the dot
product and is spoken as “a dot b.”
→
→
 If a is perpendicular to b, means Φ = 90°, then the
dot product is equal to zero.
→
→
 If a is parallel to b, means Φ = 0, then the dot product
is equal to ab.
Erwin Sitompul
University Physics: Wave and Electricity
8/7
Flashback: Multiplying Vectors
 The dot product can be regarded as the product of the
magnitude of the first vector and the projection magnitude
of the second vector on the first vector
a  b  ab cos 
 (a cos  )(b)
 (a)(b cos  )
Erwin Sitompul
University Physics: Wave and Electricity
8/8
Flashback: Multiplying Vectors
 When two vectors are in unit vector notation, their dot
product can be written as
ˆ  (b ˆi  b ˆj  b k)
ˆ
a  b  (ax ˆi  ay ˆj  az k)
x
y
z
 axbx  ayby  azbz
 ˆi ˆj kˆ
ˆi 1 0 0
ˆj 0 1 0
kˆ 0 0 1
Erwin Sitompul
University Physics: Wave and Electricity
8/9
Flashback: Multiplying Vectors
→
^
→
^
^
^
What is the angle Φ between a = 3i – 4j and b = –2i + 3k ?
z
Solution:
a  b  ab cos 
3
a  3  (4)  5
2
2

–4
b
–2
y
a
b  (2) 2  32  3.606
3
ˆ  (2iˆ  3k)
ˆ
a  b  (3iˆ  4j)
ˆ 2i)
ˆ
 (3i)(
x
 6
 ˆi ˆj kˆ
ˆi 1 0 0
ˆj 0 1 0
kˆ 0 0 1
6  (5)(3.606) cos 
6
1
  cos
 109.438
(5)(3.606)
Erwin Sitompul
University Physics: Wave and Electricity
8/10
Flux of an Electric Field
 The next figure shows an arbitrary
Gaussian surface immersed in a
nonuniform electric field.
 The surface is divided into small squares
of area ΔA, each being very small to
permit us to consider the individual
square to be flat.
→
 The electric field E may now be taken as
constant over any given square.
 The flux of the electric field for the given
Gaussian surface is:
   E  A
• Φ can be positive, negative,
or zero, depending
→ on the
→
angle θ between E and ΔA
Erwin Sitompul
University Physics: Wave and Electricity
8/11
Flux of an Electric Field
 The exact solution of the flux of electric
field through a closed surface is:

 E  dA
S
 The flux is a scalar, and its Si unit is
Nm2/C.
• The electric flux through a Gaussian
surface is proportional to the net
number of field lines passing through
that surface
• Without any source of electric field
inside the surface as in this case, the
total flux through this surface is in fact
equal to zero
Erwin Sitompul
University Physics: Wave and Electricity
8/12
Checkpoint
The figure below shows a Gaussian cube of face area A
immersed in a uniform electric field E that has the positive
direction of the z axis.
In terms of E and A, determine the flux flowing through:
(a) the front face (xy plane) Φ = +EA
Φ = –EA
(b) the rear face
(c) the top face
Φ=0
(d) the whole cube
Φ=0
Erwin Sitompul
University Physics: Wave and Electricity
8/13
Gaussian Surface: Cylinder
The next figure shows a Gaussian
surface in the form of a cylinder of
radius R immersed
in a uniform
→
electric field E, with the cylinder
axis parallel to the field. What is the
flux Φ of the electric field through
this closed surface?
• left cap
 E  dA   ( E )(dA)(cos180)
a
a
  E  dA   EA
a
• right cap
 E  dA   ( E )(dA)(cos 0)
 E  dA  EA
c
c
c
• cylindrical surface

 E  dA   E  dA   E  dA   E  dA
S
a
b
 EA  0  EA  0
c
• zero net flux
Erwin Sitompul
 E  dA   ( E )(dA)(cos 90)
b
b
0
University Physics: Wave and Electricity
8/14
Gaussian Surface: Cube
A
electric field given by
→nonuniform
^
^
E = 3xi + 4j N/C pierces the Gaussian
cube shown here (x in meters). What
is the electric flux through the right
face, the left face, and the top face?
r   E  dAr
 l   E  dAl
• right face
l
r
ˆ  (dA ˆi)
  (3xˆi  4j)
r
r
2

2
 

l
2

  3x dydz
2

 
 
2
(3)(3)dydz  36 N  m 2 C
z 0 y 0
Erwin Sitompul
(3xˆi  4ˆj)  (dydzˆi)

x 1
  3x dydz
z 0 y 0
x 3
2
2
z 0 y 0
2 2
x 3
z 0 y 0

  (3 xˆi  4ˆj)  (  dAl ˆi)
(3xˆi  4ˆj)  (dydzˆi)
z 0 y 0
2 2
• left face
x 1
2
 
(3)(1)dydz  12 N  m 2 C
z 0 y 0
University Physics: Wave and Electricity
8/15
Gaussian Surface: Cube
A
electric field given by
→nonuniform
^
^
E = 3xi + 4j N/C pierces the Gaussian
cube shown here (x in meters). What
is the electric flux through the right
face, the left face, and the top face?
 t   E  dAt • top face
t
  (3 xˆi  4ˆj)  ( dAt ˆj)
t
2

3
 
(3xˆi  4ˆj)  (dxdzˆj)
 
4 dxdz
z  0 x 1
2 3

y 2
z  0 x 1
Erwin Sitompul
 16 N  m 2 C
y 2
University Physics: Wave and Electricity
8/16
Example: Flux of an Electric Field
In a three-dimensional space, a homogenous electric field of
10 N/C is directed down to the negative z direction.
Calculate the flux flowing through:
(a) the square ABCD (xy plane)
z
E
(b) the rectangular AEFG (xz plane)
3
(a)
E  10kˆ V m
AABCD  4kˆ m2
ABCD  E  AABCD
ˆ  (4k)
ˆ
 (10k)
 40 N  m 2 C
2
1
F
A
2
(b) AAEFG  6jˆ m
AEFG  E  AAEFG
ˆ  (6j)
ˆ
 (10k)
0
Erwin Sitompul
G
1
0
B
1
2
3
y
2
3
D
C
x E
University Physics: Wave and Electricity
8/17
Gauss’ Law
 Gauss’ law relates the net flux Φ of an electric field through a
closed surface (a Gaussian surface) to the net charge qenc
that is enclosed by that surface.
 0  qenc
 0  E  dA  qenc
• in vacuum
S
 If you know how much electric field
is intercepted by the closed
surface, you can calculate how
much net charge is inside in the
volume of that enclosed surface.
Erwin Sitompul
University Physics: Wave and Electricity
8/18
Gauss’ Law
• S1: Electric field is outward for all points
 flux is positive
 enclosed charge is positive
• S2: Electric field is inward for all points
 flux is negative
 enclosed charge is negative
• S3: No charge enclosed
• S4: Net charge enclosed is equal to zero,
the field lines leaving the surface are as
many as the field lines entering it
Erwin Sitompul
University Physics: Wave and Electricity
8/19
Gauss’ Law: Net Enclosed Charge
What is the net charge enclosed by
the Gaussian cube of the previous
example?
 b   E  dAb • bottom face
b
  (3 xˆi  4ˆj)  ( dAb ˆj)
 f   E  dA f • front face
b
f
2

3
 
z  0 x 1
2

ˆ 0
  (3 xˆi  4ˆj)  ( dA f k)
(3xˆi  4ˆj)  (dxdzˆj)
f
y 0
3
 
4 dxdz
z 0 x 1
 16 N  m 2 C
y 0
 back 

E  dAback • back face
back


ˆ 0
(3 xˆi  4ˆj)  ( dAback k)
back
Erwin Sitompul
University Physics: Wave and Electricity
8/20
Gauss’ Law: Net Enclosed Charge
What is the net charge enclosed by
the Gaussian cube of the previous
example?
  r  l  t  b   f  back
 36  12  16  16  0  0  24 N  m2 C
qenc   0  E  dA
S
  0
 (8.854 1012 )(24)
 2.125 1010 C
Erwin Sitompul
University Physics: Wave and Electricity
8/21
Applying Gauss’ Law: Cylindrical Symmetry
 The next figure shows a section of an
infinitely long cylindrical plastic rod with a
uniform positive linear charge density λ, in
C/m.
 We need an expression
for the magnitude
→
of the electric field E at a distance r from
the axis of the rod.
 Due to symmetry, the component of the
resultant electric field will be only directed
radially outward.
  EA cos   E (2 rh) cos 0  E(2 rh)
 But,
 0  qenc
 0 E(2 rh)  h
Erwin Sitompul
• What about top cap
and bottom cap

E
2 0 r
University Physics: Wave and Electricity
8/22
Applying Gauss’ Law: Planar Symmetry
 Imagine a thin, infinite sheet with a
uniform (positive) surface charge
density σ, in →
C/m2. Now, let us find the
electric field E a distance r in front of
the sheet.
 A useful Gaussian surface in this case
is a closed cylinder with end caps of
area A, arranged to pierce the sheet
perpendicularly, as shown.
 0  E  dA  qenc
S
 0 ( EA  EA)   A
• What is the direction
of the electric field?
Erwin Sitompul

E
2 0
University Physics: Wave and Electricity
8/23
Example: Single Thin Plate
Charge of 10–6 C is given to a 2-m2 thin plate. Afterwards, an
electron with the mass 9.109×10–31 kg with the charge
1.602×10–19 C is held in a distance 10 cm from the plate.
(a) Determine the force acting on the electron.
(b) If the electron is released, determine the speed of the
electron when it hits the plate.
q 106
 5 107 C m 2
(a)   
A
2
5 107

4


2.824

10
NC
E
12
2 0 2(8.854 10 ) away from the plate
15
F  qE  (1.602 1019 )(2.824 104 )  4.524 10 N
toward the plate
Erwin Sitompul
University Physics: Wave and Electricity
8/24
Example: Single Thin Plate
Charge of 10–6 C is given to a 2-m2 thin plate. Afterwards, an
electron with the mass 9.109×10–31 kg with the charge
1.602×10–19 C is held in a distance 10 cm from the plate.
(a) Determine the force acting on the electron.
(b) If the electron is released, determine the speed of the
electron when it hits the plate.
F
4.524 1015
15
2

4.967

10
m
s
(b) a  
m 9.109 1031
v2  v02  2a( x  x0 )
 (0)2  2(4.967 1015 )(0  0.1)
 9.934 1014
v  3.152 107 m s
toward the plate
Erwin Sitompul
University Physics: Wave and Electricity
8/25
Example: Double Parallel Plates
Two large, parallel plates, each with a fixed
uniform charge on one side, is shown below. The
magnitudes of the surface charge densities are
2
2
σ(+) = 6.8 μC/m
→ and σ(–) = 4.3 μC/m . Find the
electric field E
(a) to the left (L) of the plates
(b) between (B) the plates
(c) to the right (R) of the plates
E(  )
 ()
6.8 106
5



3.840

10
NC
12
2 0 (2)(8.854 10 )
away from the (+) plate
E(  )
 ( )
4.3 106
5



2.428

10
NC
12
2 0 (2)(8.854 10 )
toward the (–) plate
Erwin Sitompul
University Physics: Wave and Electricity
8/26
Example: Double Parallel Plates
(a) EL  E()  E( )
 3.840 105  2.428 105
 1.412 105 N C
away from the (+) plate
(b) EB  E()  E( )
 3.840 105  2.428 105
 6.268  105 N C
away from the (+) plate
(c) ER  E()  E( )
 3.840 105  2.428 105
 1.412 105 N C
away from the (–) plate
Erwin Sitompul
University Physics: Wave and Electricity
8/27
Applying Gauss’ Law: Spherical Symmetry
 A spherical shell of uniform charge
attracts or repels a charged particle that
is outside the shell as if all the shell’s
charge were concentrated at the center of
the shell.
1
q
E
,
2
4 0 r
rR
 If a charged particle is located inside a
spherical shell of uniform charge, there is
no electrostatic force on the particle from
the shell.
E  0,
Erwin Sitompul
rR
University Physics: Wave and Electricity
8/28
Applying Gauss’ Law: Spherical Symmetry
 Any spherically symmetric charge
distribution, such as on the figure, can be
constructed with a nest of concentric
spherical shells.
 If the entire charge lies within a Gaussian
surface, r > R, the charge produces an
electric field on the Gaussian surface as if
the charge were a point charge located at
the center.
 If only a portion of the charge lies within a
Gaussian surface, r < R, then the charge
enclosed q’ is proportional to q.
E
q
4 0 R
Erwin Sitompul
3
r,
rR
University Physics: Wave and Electricity
8/29
Checkpoint
The figure shows two large, parallel sheets with identical
(positive) uniform surface charge densities, and a sphere with a
uniform (positive) volume charge density.
Rank the four numbered points according to the magnitude of
the net electric field there, greatest first.
3 and 4 tie,
then 2, 1.
• The electric field contributed by the two parallel
sheets is identical at all numbered points.
• The closer to the sphere, the greater the electric field
contributed by it.
Erwin Sitompul
University Physics: Wave and Electricity
8/30
Email Quiz
A long thin cord has a positive a linear charge
density of 3.1 nC/m. The wire is to be enclosed by
a coaxial, thin cylindrical shell of radius 1.8 cm.
The shell is to have negative charge on its surface
with a surface charge density that makes the net
external electric field zero. Calculate the surface
charge density σ of the cylindrical shell.
E 0
Erwin Sitompul
University Physics: Wave and Electricity
8/31
Homework 7
(a) The rectangle ABCD is defined by its corner points of
A(2,0,0), B(0,3,0), C(0,3,2.5), and D(2,0,2.5). Draw a sketch
of the rectangular.
→
^
^
(b) Given an electric field of E = –2i + 6j N/C, draw the electric
field on the sketch from part (a).
(c) Determine the number of flux crossing the area of the
rectangular ABCD.
Erwin Sitompul
University Physics: Wave and Electricity
8/32
Homework 7A
1. (a) The triangle FGH is defined by its corner points of F(2,0,0), G(0,3,0),
and H(0,0,4). Draw a sketch
of the triangle.
→
^
^
(b) Given an electric field of E = –2i + 6j N/C, draw the electric field on the
sketch from part (a).
(c) Determine the number of flux crossing the area of the triangle FGH.
→
^
^
2. A rectangle is under the influence of electric field of E = 2xyi + 4zk N/C.
The dimension of the rectangle is 1 m × 2 m × 3 m, with x1 = 5 m and y1 = 4
m. What is the electric flux flowing through the front face and top face?
Erwin Sitompul
University Physics: Wave and Electricity
8/33