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Tutorial 4
Gauss’s Law
PHY 1202
Lecture Outline
 Chapter 23, D Halliday, R Resnick, and J Walker,
“Fundamentals of Physics” 9th Edition, Wiley (2005).
 Gauss’s Law
– Electric flux
– Gauss’s law and Coulomb’s law
– Applications of Gauss’ law
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Lecture 04 – Review
Electric Flux
 In Lecture 04, we introduced the concept of electric flux F, which is a
measure of the flow of the electric field through a given area.
 In order to determine the electric flux, we need to represent the
given area by its area vector 𝐴, whose direction is perpendicular
(normal) to the face of the area surface and magnitude equals to the
area of the surface.
 With the definition of area vector, the electric flux is simply
Φ = 𝐸 ∙ 𝐴 = 𝐸𝐴cos𝜃
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23.2: Flux
Fig. 23-2 (a) A uniform airstream of velocity is perpendicular to the plane of a square loop of area A.
(b) The component of perpendicular to the plane of the loop is v cos q, where q is the angle between v
and a normal to the plane. (c) The area vector A is perpendicular to the plane of the loop and makes an
angle q with v. (d) The velocity field intercepted by the area of the loop. The rate of volume flow
through the loop is F= (v cos q)A.
This rate of flow through an area is an example of a flux—a volume flux in this
situation.
#The actual physical quantity referred to by the term ‘flux’ can be different in
different occasions.
PHY 1202
Lecture 04 – Review
Gauss’s Law
 A Gaussian surface is a closed surface in 3D
space.
 Gauss’ law states that the net charge
enclosed in a volume is equal to the
product e0 and F through the Gaussian
surface.
FE = E·DA
e 0 F  qenc
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or
𝜀0
𝐸 ∙ 𝑑𝐴 = 𝑞𝑒𝑛𝑐
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Lecture 04 – Review
Gauss’s Law and Coulomb’s Law
 We shown that one may derive Coulomb’s law from Gauss’ law as the
two laws are equivalent.
Gauss’ Law for point charge q1
E
1
q1
4e 0 r 2
Coulomb’s Law for point charges q1 & q2
F  q2 E 
𝜀0
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1 q1q2
4e 0 r 2
𝐸 ∙ 𝑑𝐴 = 𝑞
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Lecture 04 – Review
E field vanishes inside a conductor
 Using Gauss’ law, we argue that when
an excess charge is placed on an isolated
conductor, the charge must remains on
the outer surface of the conductor as
the static electric field inside the
conductor must be zero.
 In a conductor, the charge is free to
move. If the E field inside the conductor
is non-zero, the charges inside the
conductor will move. They will only stop
moving,
when
they
rearrange
themselves in such a way that the E field
inside
the
conductor
vanishes.
Therefore, these charges must reside on
the conductor surface.
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Lecture 04 – Review
Distribution of charge on a conducting spherical shell
 The charge inside the conducting shell will induce the opposite charges on
the inside surface of the shell. The distribution of the induced charges will be
non-uniform, if the charge is positioned off the center of the shell. However,
the distribution of the charges on the outer surface of the sphere will be
uniform, since these charges are shielded from (the charges on the outer
surface are not affected by those in the inner surface or the original charge)
the non-uniform charge distribution inside the shell, because the E field in
the conductor is zero.
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Lecture 04 – Review
E Field on a conducting surface
 If E=0 inside the conductor, the
electric field on the conducting
surface can be determined by
constructing a small Gaussian surface
that overlaps both inside and outside
the surface and show the field is
normal to the surface and
proportional to the surface charge
density s
E
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s
e0
(conducting surface)
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Lecture 04 – Review
E field from a non-conducting sheet
 In a non-conducting sheet, with non
zero electric field on both sides of the
surface, the magnitude of the electric
field becomes, which is half of the field
from the conducting surface with the
same amount of surface charge density
E
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s
2e 0
(sheet of charge)
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Lecture 04 – Review
E field from a line of charge
 When we apply the Gauss’s law on a
line of charge with charge density l, we
found that
E
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l
2e 0 r
(line of charge)
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Lecture 04 – Review
E field from a charged sphere
 In a spherically symmetric
distribution of charges of radius
R with volume charge density r.
The electric field inside the
sphere is proportional to r, but
proportional to 1 𝑟 2 for r>R.
r
r  R; E 
r,
3e 0
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R3r
r  R; E 
3e 0 r 2
4r 3
r  R ; q' 
r
3
4R 3
r  R ; q' 
r
3
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Practice Questions: Gauss’s Law
Chapter 23
Questions: **3,*7,**14,*24,*35,**40,**49
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**3
.
2
(a) F   6.00 N C  ˆi  1.40 m  ˆj  0.
(b) F   2.00 N C  ˆj  1.40 m 2 ˆj
 3.92 N  m 2 C.
Electric flux:
 
F E A
Surface vector of the right face:
2
A  Aˆj  1.40m  ˆj
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(c) F   3.00 N C  ˆi   400 N C  kˆ   1.40 m  ˆj
0
(d) The total flux of a uniform field through a
closed surface is always zero.
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*7
Gauss’ law:
e 0F  q
1.8 106 C
5
2
F 

2.0

10
N

m
C.
e 0 8.85 1012 C2 N  m 2
q
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**14
(b) The next value that F takes is F  4.0  105 N  m 2 /C,
6
which implies that qenc  3.54 10 C.
But we have already accounted for some of that charge
in part (a), so the result for part (b) is
qA = qenc – qcentral = – 5.3 ×106 C.
r=A
r=B
A
B
(c) Finally, the large r value for F is F  6.0 105 N  m 2 /C,
6
which implies that qtotal enc  5.3110 C. Considering
what we have already found, then the result is
qtotal enc  q A  qcentral  8.9  C.
(a) The value F  2.0 105 N  m2 /C for small r leads to
qcentral  e 0 F  (8.85  1012 C2 /N  m 2 )(2.0 105 N  m 2 /C)
 1.77 106 C  1.8 106 C
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*24

A
E  dA  2rE 
qenc
e0
.
(a) For r < R, qenc = 0, so E = 0.
(b) For r > R, qenc = l, so E (r )  l / 2 re 0 .
With l  2.00 108 C/m and r = 2.00R =0.0500 m,
Cylindrical Gaussian Surface
we obtain
 2.0 10 C/m 
E
2  0.0500 m   8.85 10 C
L
8
12
r
2 rL
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2
/ N  m2

 7.19 103 N/C.
 r2
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*24
(c) The plot of E vs. r is shown below.
Here, the maximum value is
Emax


2.0 108 C/m
l


 1.4 104 N/C.
12
2
2
2 re 0 2  0.025 m  8.85 10 C / N  m
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

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*35
 E1  E2  E3  0
E1
E2
E3
A
B
C
D
(1)
E1  E2  E3  2  105
(2)
E1  E2  E3  6  105
(3)
E1  E2  E3  0
(4)
(1)  (2)
2 E1  2  105  E1  1 105
(2)  (3)
2 E2  4 105  E2  2 105
Sub E1 , E2into (4)
1105  2 105   E3  E3  3  105
E
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s
 E s
e0
s 3 E3
3 105


 1.5
5
s 2 E2
2 10
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**40
If d = 0.20 m (which is less than the magnitude of r
found above), then neither of the points (x  ±
0.691 m) is in the “forbidden region” between the
particle and the sheet. Thus, both values are
allowed. Thus, we have
(a) x = 0.691 m on the positive axis, and
The point where the individual fields cancel cannot be in the
region between the sheet and the particle (d < x < 0) since
the sheet and the particle have opposite-signed charges. The
point(s) could be in the region to the right of the particle (x >
0) and in the region to the left of the sheet (x < d); this is
where the condition
|s |
Q

2e 0 4e 0 r 2
(b) x =  0.691 m on the negative axis.
(c) If, however, d = 0.80 m (greater than the
magnitude of r found above), then one of the points
(x  0.691 m) is in the “forbidden region” between
the particle and the sheet and is disallowed. In this
part, the fields cancel only at the point x  +0.691
m.
must hold. Solving this with the given values, we find
r = x = ±  ± 0.691 m.
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**49
For r < a, the charge enclosed by the Gaussian surface is
q1(r/a)3. Gauss’ law yields
 q  r 
qr
4 r E   1     E  1 3 .
4e 0 a
 e0   a 
3
2
(a) For r = 0, the above equation implies E = 0.
(b) For r = a/2, we have
E
At all points where there is an electric field, it is radially
outward. For each part of the problem, use a Gaussian
surface in the form of a sphere that is concentric with the
sphere of charge and passes through the point where the
electric field is to be found. The field is uniform on the
surface, so
2
 E  dA  4r
E
(8.99 109 N  m 2 /C 2 )(5.00 1015 C)

2(2.00 102 m) 2
 5.62 102 N/C.
(c) For r = a, we have
E
where r is the radius of the Gaussian surface.

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q1 (a / 2)
4e 0 a 3
q1
4e 0 a 2
(8.99 109 N  m 2 /C 2 )(5.00 10 15 C)
 0.112 N/C.
(2.00 102 m) 2
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**49
In the case where a < r < b, the charge enclosed by the
Gaussian surface is q1, so Gauss’ law leads to
4 r 2 E 
q1
e0

E
q1
4e 0 r 2
.
(d) For r = 1.50a, we have
E
q1
4e 0 r 2

(8.99 109 N  m2 /C2 )(5.00 1015 C)
 0.0499 N/C.
(1.50  2.00 102 m) 2
(e) In the region b < r < c, since the shell is conducting, the
electric field is zero. Thus, for r = 2.30a, we have E = 0.
(f) For r > c, the charge enclosed by the Gaussian surface is
zero. Gauss’ law yields
(g) Consider a Gaussian surface that lies completely
within the conducting shell. Since the electric field is
everywhere zero on the surface,
z
 
E  dA  0
and, according to Gauss’ law, the net charge enclosed
by the surface is zero. If Qi is the charge on the inner
surface of the shell, then q1 + Qi = 0 and Qi = –q1 = –
5.00 fC.
(h) Let Qo be the charge on the outer surface of the
shell. Since the net charge on the shell is –q, Qi + Qo
= –q1. This means
Qo = –q1 – Qi = –q1 –(–q1) = 0.
4 r 2 E  0  E  0.
Thus, E = 0 at r = 3.50a.
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