Chapter 24 electric flux
24.1 Electric Flux
24.2 Gauss’s Law
24.3 Application of Gauss’s Law to Various Charge
Distributions
A Spherically Symmetric Charge Distribution (An insulating
solid sphere )
The Electric Field Due to a Point Charge
The Electric Field Due to a Thin Spherical Shell
The Electric Field Due to A Cylindrically Symmetric Charge Distribution
the electric field due to an infinite plane
24.4 Conductors in Electrostatic Equilibrium
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We will use a “charge density” to describe the distribution of charge.
This charge density will be different depending on the geometry
Symbol



Name
Charge per length
Charge per area
Charge per volume
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Unit
C/m
C/m 2
C/m 3
Definitions
 Symmetry—The balanced structure of an object, the halves
of which are alike
Open and Closed Surfaces
A rectangle is an open surface — it does
NOT contain a volume
A sphere is a closed surface — it DOES
contain a volume
 Closed surface—A surface that divides space into an inside
and outside region, so one can’t move from one region to
another without crossing the surface
 Gaussian surface—A hypothetical closed surface that has
the same symmetry as the problem we are working on—
note this is not a real surface it is just an mathematical one
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24-1
Electric Flux
• Like the flow of water, or light energy, we can think of the
electric field as flowing through a surface (although in this
case nothing is actually moving).
• We represent the flux of electric field as F (greek letter
phi),
The units for electric flux are Nm2/C.
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24.1Electric Flux
 The number of lines per unit area is proportional to the
magnitude of the electric field.
 This means that the total number of lines that penetrate a
given area is proportional to the magnitude of the electric
field times the area which is being penetrated. E x A
 The product of the electric field (E) and surface area (A)
which is perpendicular to the field is called the electric
flux (ΦE).
 Consider a uniform electric field, and
a surface area through which the
electric field is passing.
 We define the angle of the given area,
relative to the direction of the electric
field, by a normal vector which is
perpendicular to the surface.
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The electric flux passing through a surface is the
number of electric field lines that pass through it.
Because electric field lines are drawn
arbitrarily, we quantify electric flux like
this: FE=EA, except that…
We define A to be a vector having a
magnitude equal to the area of the surface,
in a direction normal to the surface.
The “amount of surface” perpendicular to
the electric field is A cos.
Because A is perpendicular to the surface, the
amount of A parallel to the electric field is A cos .
A = A cos  so FE = EA = EA cos .
Remember the dot product? FE  E  A
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A
E
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• If the electric field is not uniform, or the
surface is not flat…
• divide the surface into infinitesimal surface
elements and add the flux through each…
• so the flux of the electric field through an
element of area DA is
 
DF  E  DA  E DA cos
• When  < 90˚, the flux is positive (out of the
surface), and when  > 90˚, the flux is
negative.
• When we have a complicated surface, we can
divide it up into tiny elemental areas:
 
dF  E  dA  E dA cos
F E  lim
DA i  0
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E
i
F
 DA i
i
8

E

  E .dA
Electric Flux
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• We are going to be most interested in closed
surfaces, in which case the outward direction
becomes self-evident.
• We can ask, what is the electric flux out of such
a closed surface? Just integrate over the closed
 
surface:
 dF  E  dA Flux positive => out
F 
E

Flux negative => in
• The  symbol has a little circle to indicate
that the integral is over a closed surface.
• The closed surface is called a gaussian surface,
because such surfaces are used by Gauss’ Law,
which states that:
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Spherical Gaussian surfaces around
(a) positive and (b) negative point charge.
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 The flux is a maximum when the surface is perpendicular
to the field Φ = EA
 The flux is zero when the surface is parallel to the field
 If the field varies over the surface, Φ = EA cos θ is valid
for only a small element of the area
 In general, the value of the flux will depend both on the
field pattern and on the surface
 For a uniform electric field perpendicular to a rectangular
surface, ΦE the electric flux is ΦE = EA (Nm^2/C)
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Flux of Electric Field
1. Which of the following figures correctly shows a positive
electric flux out of a surface element?
A. I.
DA
B. II.
I.

C. III.
E
D. IV.
E. I and III.
III.
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
DA
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E
II.
E

DA
IV.
DA
E

Example 1: flux through a cube of a uniform electric field
 The field lines pass




through two surfaces
perpendicularly and are
parallel to the other
four surfaces
For side 1, ΦE = -El 2
For side 2, ΦE = El 2
For the other sides,
ΦE = 0
Therefore, Φtotal = 0
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24-2Gauss’s Law
Gauss’s Law relates the electric flux through a closed
surface with the charge Qin inside that surface.
This is a useful tool for simply determining the electric
field, but only for certain situations where the charge
distribution is either rather simple or possesses a high
degree of symmetry.
 More general and elegant form of Coulomb’s law.
 The electric field by the distribution of charges can be
obtained using Coulomb’s law by summing (or integrating)
over the charge distributions.
 Gauss’ law, however, gives an additional insight into the
nature of electrostatic field and a more general
relationship between the charge and the field
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Electric lines of flux and
Derivation of Gauss’ Law using Coulombs law
 Consider a sphere drawn around a positive point charge.
Evaluate the net flux through the closed surface.
Net Flux =
F
For a Point charge
F


EdA 
 E  dA   E cosdA   EdA
E=kq/r2
2
kq/r
  dA
F  kq/r 2  dA  kq/r 2 (4r 2 )
F  4kq
4k  1/0 where 0  8.85x10
12
Fnet 
qenc
0
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C2
Nm 2
Gauss’ Law
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Gauss’ Law
The flux of electric field through
a closed surface is proportional
to the charge enclosed.
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The Gaussian Surface and Gauss’s Law
 Closed surfaces of various shapes can surround the charge



Only S1 is spherical
The flux through all other surfaces (S2 and S3) are the
same.
These surfaces are all called the Gaussian Surface.
 Gauss’s Law

The net flux through any closed surface
surrounding a charge q is given by q/εo and is
independent of the shape of that surface
 
q
F E   E  dA 
0
The net electric flux through a closed surface that
surrounds no charge is zero
 Since the electric field due to many charges is the vector
sum of the electric fields produced by the individual
charges, the flux through any closed surface can be
expressed as
 q1  q2  ...
 


F E   E  dA   (E1  E 2  ...)  dA 
•
0
Gauss’s Law connects electric field with its source charge
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 in
q

out
Flux lines & Flux
Just what we would expect because the number of
field lines passing through each sphere is the same
N F
and number of lines passing through
each sphere is the same
F1
out
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F S  F 2  F1 
In fact the number of flux lines
passing through any surface
surrounding this charge is the
same
F2
Fs
18
in
out
F N
Q
0
even when a line
passes in and out
of the surface it
crosses out once
more than in
Flux through a sphere from a
point charge
The electric field around a
point charge
1
Q
| E |
2
40 | r1 |
E
r1
Thus the flux
on a sphere
is E × Area
Cancelling we
get
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1
Area
Q
2
F

4

|
r
|
1
2
40 | r1 |
F
Q
0
Flux through a sphere from a
point charge
Now we change the
radius of sphere
The electric field
around a point charge
r1
| E |
1
Q
4 0 | r1 |2
E
Area
Thus the
1
Q
flux on a
F
 4 | r1 |2
2
4

|
r
|
0
1
sphere is E
× Area
Cancelling
we get
F
Q
0
1
Q
| E |
2
40 | r2 |
r2
1
Q
2
F2 

4

|
r
|
2
40 | r2 |2
The flux is
Q
Q
the same as
F2 
F 2  F1 
before
0
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0
Principle of superposition:
What is the flux from two charges?
Since the flux is related to the number of
field lines passing through a surface the
total flux is the total from each charge
FS 
Q1
0

Q2
0
In general
FS  
Q1
Q2
Qi

Fs
Gauss’s Law
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For any
0 surface
The total flux through the below spherical surface is
1.
2.
3.
4.
positive (net outward flux)
negative (net inward flux)
zero.
I don’t know
+q
No enclosed charge  no net flux.
Flux in on left cancelled by flux out on right
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Example
The net outward flux through the surface of a black box is 8
×103 Nm2C–1.
(a) What is the net charge inside the box?
(b) If the net outward flux were zero, could you say
that there were no charges inside the box?
(a) Ф = 8 ×103 Nm2C–1
If the net charge inside is q, then,
F
q
in
q

in
0
 F  0  8 103  8.85  1012  7  108 C
(b) Not necessarily. We can only say that the net
charge inside the box is zero.
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Example:
A surface is so constructed that, at all points on the
surface, the E vector points inward. Therefore, it
can be said that
A.the surface encloses a net positive
charge.
B.the surface encloses a net negative
charge.
C.the surface encloses no net charge.
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Example:
The electric field E in Gauss’s Law is
A. only that part of the electric field due to the charges
inside the surface.
B. only that part of the electric field due to the charges
outside the surface.
C. the total electric field due to all the charges both inside
and outside the surface.
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Example:
The figure shows a surface enclosing the charges q
and –q. The net flux through the surface
surrounding the two charges is
A. q/0
B. 2q/0
C. –q/0
D. zero
E. –2q/0
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Example:
The figure shows a surface enclosing the charges 2q and –
q. The net flux through the surface surrounding the two
charges is
q
A.
0
B.

2q
ε0
C.

q
D.
zero
3q
E.
0
0
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Example:
The figure shows a surface, S, with two charges q and
–2q. The net flux through the surface is
q
A.
0
B.

C. 
D.
E.
2q
ε0
s
q
0
+q
zero
3q
0
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2q
Example:
These are two-dimensional cross sections through threedimensional closed spheres and a cube. Rank order,
from largest to smallest, the electric fluxes Фa to Фe
through surfaces a to e.
1.
2.
3.
4.
5.
Φa > Φc > Φb > Φd > Φe
Φb = Φe > Φa = Φc = Φd
Φe > Φd > Φb > Φc > Φa
Φb > Φa > Φc > Φe > Φd
Φd = Φe > Φc > Φa = Φb
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Sample Problem
Four closed surfaces, S1
through S4, together
with the charges -2Q,
+Q, and –Q are
sketched in the figure.
Find the electric flux
through each surface.
S1
-2Q
+Q
-Q
S3
S4
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S2
Example:
a point charge q1 = 4.00 nC is located on the x-axis at x = 2.00
m, and a second point charge
q2 = -6.00 nC is on the y-axis
at y = 1.00 m. What is the total electric flux due to these two
point charges through a spherical surface centered at the origin if
the radius is (a) 0.500 m? (b) 1.50 m? (c) 2.50 m?
y
q2
1m
x
2m
(a) FE  0
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q1
6nC
(b) F E 
0
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2nC
(c ) F E 
0
Example:
At what angle will the flux be zero and what
angle will it be a maximum.
1. π/4 ; π/2
2. π/2 ; 0
3. 0 ; π/2
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Example
An electric field with a magnitude of 3.50 kN/C is applied
along the x axis. Calculate the electric flux through a
rectangular plane 0.350 m wide and 0.700 m long
assuming that
(a) the plane is parallel to the yz plane;
(b) the plane is parallel to the xy plane;
(c) the plane contains the y axis, and its normal makes an
angle of 40.0° with the x axis.


FE  EA cos  3.50  103  0.350  0.700 cos0  858 N  m 2 C
  90.0

FE  0

FE  3.50  103  0.350  0.700 cos40.0  657 N  m
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2
C
example
What is the net electric
flux through the surface
if Q1=q4=+3.1nC,
q2=q5=-5.9nC,
and q3=-3.1nC?
F
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qenc
0

q1  q2  q3
0
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 670 N  m / C
2
Problem
What is the flux Φ of The electric
field through This closed surface?
Step one:
 
F  EA






 E  dA   E  dA   E  dA
a
Step two:
b
c
 
0
E

d
A

E
cos
180
dA   E  dA   EA



a


 E  dA 
c




E  dA 
 E cos 0 dA  EA
 

E cos 90 0 dA  0
b
Step three:
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F  EA  0  EA  0
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Remember – electric field lines must start and must end on charges!
If no charge is enclosed within Gaussian surface – flux is zero!
Electric flux is proportional to the algebraic number of lines leaving
the surface, outgoing lines have positive sign, incoming - negative
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Example
The following charges are located inside a submarine: 5.00 μC, –9.00 μC,
27.0 μC, and –84.0 μC.
(a) Calculate the net electric flux through the hull of the submarine.
(b) Is the number of electric field lines leaving the submarine greater
than, equal to, or less than the number entering it?
qin  5.00 C  9.00 C  27.0 C  84.0 C 
6
2
2
FE 



6.
89

10
N

m
C
0
8.85  1012 C 2 N  m 2
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Example:
Consider a cube with each edge = 55cm. There is a 1.8 C charge
In the center of the cube. Calculate the total flux exiting the cube.
6
1.8 10
5
2
F 

2
.
03

10
Nm
/C
12
 0 8.85 10
q
NOTE: INDEPENDENT OF THE SHAPE OF THE SURFACE!
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Example
A point charge Q = 5.00 μC is located at the center of a cube of edge L =
0.100 m. In addition, six other identical point charges having q = –1.00
μC are positioned symmetrically around Q as shown in Figure P24.17.
Determine the electric flux through one face of the cube.
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Problem solving guide for Gauss’ law
 Use the symmetry of the charge distribution to
determine the pattern of the field lines.
 Choose a Gaussian surface for which E is either
parallel to or perpendicular to dA.
 If E is parallel to dA, then the magnitude of E
should be constant over this part of the surface. The
integral then reduces to a sum over area elements.
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24-3 Application of gauss s law to various charge distributions
Gauss’s law is useful in determining electric fields when the charge
distribution is characterized by a high degree of symmetry. The following
examples demonstrate ways of choosing the Gaussianqsurface over which
the surface integral given by Equation    E d A   can be
simplified and the electric field determined. In choosing the surface, we
should always take advantage of the symmetry of the charge distribution
so that we can remove E from the integral and solve for it. The goal in
this type of calculation is to determine a surface that satisfies one or
more of the following conditions:
1. The value of the electric field can be argued by symmetry to be
constant over the surface.
2. The dot product in the Equation can be expressed as a simple
algebraic product E dA because E and dA are parallel.
3. The dot product in the Equation is zero because E and d A are
perpendicular.
4. The field can be argued to be zero over the surface.
All four of these conditions are used in examples throughout the
remainder of this chapter.


in
E
0
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Gauss’ Law and Coulomb law
Example 24.4 The Electric Field Due to a Point Charge
Starting with Gauss’s law, calculate the electric field due to an isolated point
charge q.
 The field lines are directed radially outwards and are
perpendicular to the surface at every point, so
The angle  between E and dA is zero at any point
on the surface, we can re-write Gauss’ Law as
 
F E   E  dA   E n dA   EdA  E  dA  E  4r 2
FE 


 E .dA 
Qin

0
E has the has same value
at all points on the surface
E is can be moved out
A spherical Gaussian surface
centered on a point charge q
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Integral is the sum of
surface area
Coulomb’s Law
42
Example 24.5 A Spherically Symmetric Charge Distribution
An insulating solid sphere of radius a has a uniform volume
charge density ρ and carries total charge Q.
(A) Find the magnitude of the E-field at a point outside the
sphere
E
(B) Find the magnitude of the E-field at a point inside the
sphere
Since the charge distribution is spherically symmetric
we select a spherical Gaussian surface of radius r >
a
a centered on the charged sphere. Since the
charged sphere has a positive charge, the field will
be directed radially outward. On the Gaussian
sphere E is always parallel to dA, and is constant.


Qin
E
.
d
A


r
dA
Q

Right side:
Qin
0

Q
0
2
E
d
A

E
d
A

E
(
4

)
r


E  4 r 2  
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0
43
Q
0
or E 
1
Q
Q

k
e
4 0 r 2
r2
Example 3
Find the electric field at a point inside the sphere.
Now we select a spherical Gaussian surface with
radius r < a. Again the symmetry of the charge
distribution allows us to simply evaluate the left side
of Gauss’s law just as before.
r
a

The charge inside the Gaussian sphere is no longer
Q. If we call the Gaussian sphere volume V’ then“V
Q


2
E
d
A

E
d
A

E
(
4

r )


Volume charge density“: ρ = charge / unit volume is used to characterize the charge
distribution
4 3 
Qin
4  r 3
2
Right side: Qin   V     r
E  4 r  

0
3 0
3
4  r 3

Q
1 Q
Q
E

r but  
so E 
r  ke 3 r
3
2
4
3 0
4 0 a
a
3 0 4 r
 a3
3

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
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The electric field of a uniformly charged INSULATING sphere.
 
E (r ) =
outside
inside
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Q
4o a
3
rrˆ
Q
We found for r > a , E  ke 2
r
ke Q
and for r < a , E  3 r
a
45
same as
a point
charge!
Example 24.6 The Electric Field Due to a Thin
Spherical Shell
A thin spherical shell of radius a has a total charge Q distributed
uniformly over its surface Find the electric field at points
(A) outside and
(B) inside the shell.
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 Let’s start with the Gaussian surface outside the sphere of
charge, r > a
 We know from symmetry arguments that the electric field will
be radial outside the charged sphere

If we rotate the sphere, the electric field cannot change
 Spherical
symmetry
 Thus we can apply Gauss’ Law and get
Flux   E  dA  E  4π r 2 
 q / 0
(Gauss)
 … so the electric field is
E
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1
4 0
47
q
r2
 Let’s let’s take the Gaussian surface inside the
sphere of charge, r < a
 We know that the enclosed charge is zero so
Flux  F E  EA  0
 We find that the electric field is zero
everywhere inside spherical shell of charge
E 0
 Thus we obtain two results


The electric field outside a spherical shell of
charge is the same as that of a point charge.
The electric field inside a spherical shell of
charge is zero.
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48
 Let’s let’s take the Gaussian surface inside the
sphere of charge, r < a
 We know that the enclosed charge is zero so
Flux  F E  EA  0
 We find that the electric field is zero
everywhere inside spherical shell of charge
E 0
 Thus we obtain two results


The electric field outside a spherical shell of
charge is the same as that of a point charge.
The electric field inside a spherical shell of
charge is zero.
Norah Ali Al moneef
49
24-6 The Electric Field Due to A Cylindrically Symmetric
Charge Distribution
Find the E-field a distance r from a line of
positive charge of infinite length and
constant charge per unit length λ.

Symmetry  E field must be ^ to line and can only
depend on distance from line
Construct a “Gaussian Surface” that reflects the
symmetry of the charge - cylindrical in this case,
then evaluate
Norah Ali Al moneef
50
q

L


 E .dA 
Qin
Gauss s law

L
E
d
A

E
d
A

0

E
(
2

r
L
)




0
0
E

2 0 r

2k
r

E
20r
2k 
E
r̂
r

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51
Line of Charge
Example 24.8 A Plane of Charge
•Find the electric field due to an infinite plane of positive charge
with uniform surface charge density σ.
 Assume that we have a thin, infinite non-conducting sheet
of positive charge
 The charge density in this case is the charge per unit area,
 From symmetry, we can see that the electric field will be
perpendicular to the surface of the sheet
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52
Planar Symmetry (2)
 To calculate the electric field using Gauss’ Law, we assume a Gaussian
surface in the form of a right cylinder with cross sectional area A and height
2r, chosen to cut through the plane perpendicularly.
 Because the electric field is perpendicular to the plane
everywhere, the electric field will be parallel to the walls of the cylinder and
perpendicular to the ends of the cylinder.
 Using Gauss’ Law we get
Flux  F E   E  dA  EA  EA
 q /  0  A /  0
(Gauss)
 … so the electric field from an infinite
non-conducting sheet with charge density 
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53
Example
A conducting spherical shell of inner radius a and outer radius b with a
net charge -Q is centered on point charge +2Q. Use Gauss’s law to
find the electric field everywhere, and to determine the charge
distribution on the spherical shell.
First find the field for 0 < r < a
This is the same as Ex. 2 and is the field due to a
point charge with charge +2Q.
-Q
a
+2Q
b
2Q
E  ke 2
r
Now find the field for a < r < b
The field must be zero inside a conductor in equilibrium. Thus from
Gauss’s law Qin is zero. There is a + 2Q from the point charge so we
must have Qa = -2Q on the inner surface of the spherical shell. Since the
net charge on the shell is -Q we can get the charge on the outer surface
from Qnet = Qa + Qb.
Qb= Qnet - Qa = -Q - (-2Q) = + Q.
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54
Example
Find the field for r > b
From the symmetry of the problem, the field in
this region is radial and everywhere
perpendicular to the spherical Gaussian surface.
Furthermore, the field has the same value at
every point on the Gaussian surface so the
solution then proceeds exactly as in Ex. 2, but
Qin=2Q-Q.
2
E
d
A

E
d
A

E
(
4

r )


Gauss’s law now gives:
E  4 r
Norah Ali Al moneef
2
 
Qin
0

2Q  Q
0

Q
0
55
1
Q
Q
or E 
 ke 2
2
4 0 r
r
Example
A spherical shell of radius 10 cm has a charge 2×10–6 C
distributed uniformly over its surface. Find the electric field
(a) Inside the shell
(b) Just outside the shell
(c) At a point 15 cm away from the centre
q = 2 ×10–6 C,
R = 0.1 m,
r = 0.15 m
(a) Inside the shell, electric field is zero
1
q
2×10–6
9
(b) E =
= 9×10 ×
2
4ε0 R
0.12
= 1.8×106 NC–1
1
q
2×10–6
9
(c) E' =
= 9×10 ×
4ε0 r2
0.152
=8×105 NC–1
Norah Ali Al moneef
56
Example
An infinite line charge produces a field of 9×104 N C–1 at
a distance of 2 cm. Calculate the linear charge density.
E = 9×104 N C–1,
E=
r = 2×10–2 m
1 λ
2 ε 0 r
 λ = 4 ε 0 ×
Er
2
1
9×104 ×2×10–2
–7
–1
=
×
=10
Cm
9×109
2
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57
Conductor
+ +
+- +
++++++
Norah Ali Al moneef
- -
+
+
+
+
Conductors.. In these
materials, the charges ARE
FREE TO MOVE.
58
24-4Conductors in Electrostatic Equilibrium
By electrostatic equilibrium we mean a situation where
there is no net motion of charge within the conductor
The electric field is zero everywhere inside the
conductor
Any net charge resides on the conductor’s surface
The electric field just outside a charged conductor is
perpendicular to the conductor’s surface
•On an irregularly shaped conductor, the surface charge
density is greatest at locations where the radius of
curvature of the surface is smallest.
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59
24-4Conductors in Electrostatic Equilibrium
The electric field is zero everywhere inside the conductor
Why is this so?
 Place a conducting slab in an external field, E.
If there was a field in the conductor the charges would
accelerate under the action of the field.
++++++++++++
---------------------
The charges in the conductor move
creating an internal electric field that
cancels the applied field on the inside of
the conductor
Ein
E
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E
60
When electric charges are at rest, the electric field within a
conductor is zero.
On an irregularly shaped conductor, the surface charge
density is greatest at locations where the radius of
curvature of the surface is smallest.
The electric field is stronger where the
surface is more sharply curved.
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61
•Charge Resides on the Surface
 Choose a Gaussian surface inside but close to the
actual surface
 The electric field inside is zero
 There is no net flux through the gaussian surface
 Because the gaussian surface can be as close to
the actual surface as desired, there can be no
charge inside the surface
Since no net charge can be inside the surface, any net charge must reside
on the surface
Gauss’s law does not indicate the distribution of these charges, only that it
must be on the surface of the conductor
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62
Charges on Conductors
Field within conductor
E=0
Charge on solid conductor resides
on surface.
Charge in cavity makes a equal but
opposite charge reside on inner
surface of conductor.
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63
E-Field’s Magnitude and Direction
 Choose a cylinder as the Gaussian surface
 The field must be perpendicular to the
surface
 If there were a parallel component to
E, charges would experience a force
and accelerate along the surface and it
would not be in equilibrium
 The net flux through the surface is through only
the flat face outside the conductor
 The field here is perpendicular to the surface
 Applying Gauss’s law
A
F E  EA 
0
Norah Ali Al moneef

FE 
0
64
Under electrostatic conditions the
electric field inside a solid
conducting sphere is zero. Outside
the sphere the electric field drops
off as 1 / r2, as though all the
excess charge on the sphere were
concentrated at its center.
Electric field = zero (electrostatic)
inside a solid conducting sphere
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65
Example
• A spherical conducting shell has an excess charge
of +10 C.
• A point charge of 15 C is located at center of the
sphere.
• Use Gauss’ Law to calculate the charge on inner
and outer surface of sphere
R1
R2
-15 C
• Since E = 0 inside the metal, flux through this surface = 0
• Gauss’ Law says total charge enclosed = 0
-5 C
• Charge on inner surface = +15 C
Since TOTAL charge on shell is +10 C,
Charge on outer surface = +10 C  15 C = 5 C!
-15C
a) Inner: +15 C; outer: 0
(b) Inner: 0; outer: +10 C
(c) Inner: +15 C; outer: -5 C
Norah Ali Al moneef
+15C
66
 A line charge  (C/m) is placed along
the axis of an uncharged conducting
cylinder of inner radius ri = a, and
outer radius ro = b as shown.
 What is the value of the charge
density o (C/m2) on the outer
surface of the cylinder?
(a)
Norah Ali Al moneef
b
a

(c)
(b)
67
Example
The uniform surface charge density on a spherical copper
shell is . What is the electric field strength on the surface
of the shell?
The electric field on the surface of a uniformly charged
spherical conductor is given by,
E=
1 q 

2
4ε0 R
o
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68
Example
A spherical charged conductor has a uniform surface charge
density s . The electric field on its surface is E. If the radius
of the sphere is doubled keeping the surface density of
charge unchanged, what will be the electric field on the
surface of the new sphere ?
The electric field on the surface of a uniformly charged
spherical conductor is given by,
1 q 4R2 

E=


4ε0 R2 40R2 o
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69
Example
Consider a thin spherical shell of radius 14.0 cm with a total charge of
32.0 μC distributed uniformly on its surface. Find the electric field
(a) 10.0 cm and
(b) 20.0 cm from the center of the charge distribution.
E
Norah Ali Al moneef
keQ
2
r
8.99  10  32.0  10 


 7.19 M N
6
9
 0.200
70
2
C
Example
A uniformly charged, straight filament 7.00 m in length has a total
positive charge of 2.00 μC. An uncharged cardboard cylinder 2.00 cm in
length and 10.0 cm in radius surrounds the filament at its center, with the
filament as the axis of the cylinder. Using reasonable approximations,
find (a) the electric field at the surface of the cylinder and
(b) the total electric flux through the cylinder.

9
2ke 2 8.99  10 N  m
E

r
2


C 2  2.00  106 C 7.00 m 


0.100 m
E  51.4 kN C ,radially outw ard
FE  EA cos  E  2 r  cos0


FE  5.14  104 N C 2  0.100 m   0.020 0 m   1.00  646 N  m 2 C
Norah Ali Al moneef
71
Example
A square plate of copper with 50.0-cm sides has no net charge and is placed in a
region of uniform electric field of 80.0 kN/C directed perpendicularly to the
plate. Find
(a) the charge density of each face of the plate and
(b) the total charge on each face.



  8.00  104 8.85  1012  7.08  107 C m 2


Q   A  7.08  107  0.500 C
2
Q  1.77  107 C  177 nC
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72
Two infinite parallel sheets of charge
surface
E ^ of the sheet
Due to uniform surface charge density,
Region II
E = E1 – E2
Region I

E = – E1 + E2
=–
=

(σ A + σ B )
2ε0
A
Norah Ali Al moneef
(σ A – σB )
2ε0
73
B
Two infinite parallel sheets of
charge
A
B
Region III
E = E1 + E2
=
Norah Ali Al moneef
74
(σ A + σ B )
2ε0
Two Parallel Nonconducting Sheets
 The situation is different if you bring two nonconducting sheets of charge
close to each other.
 In this case, the charges cannot move, so there is no shielding, but now we
can use the principle of superposition.
 In this case, the electric field on the left due to the positively charged sheet is
canceled by the electric field on the left of the negatively charged sheet, so
the field there is zero.
 Likewise, the electric field on the right due to the negatively charged sheet is
canceled by the electric field on the right of the positively charged sheet.

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75
The result is much the same as
before, with the electric field in
between being twice what it
was previously.
Special Case
The sheets have equal and opposite charge density
If
A  
and
B  – 
I
In regions I and III, E = 0
In region II,
Norah Ali Al moneef
E
E0
then
E
II
as


0
A  B  (  –)  0
 – (– ) 

20
0
76
III
–
E0
Example
Two large thin metal plates with surface charge densities of opposite
signs but equal magnitude of 44.27 ×10–20 Cm–2 are placed
parallel and close to each other. What is the field
(i) To the left of the plates?
(ii) To the right of the plates?
(iii) Between the plates?
Electric field exists only in the region between the
plates. Therefore,
(i), (ii) E = 0
(iii) σ = 44.27×10–20 Cm–2 ,
σ 44.27×10–20
E= =
=5×108 NC–1
–12
ε0 8.854×10
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77
Example:
A uniform electric field of magnitude 1 N/C is
pointing in the positive y direction. If the
cube has sides of 1 meter, what is the flux
through sides A, B, C?
A) 1, 0, 1
B) 0, 0, 1
B
1, 0, 0
D) 0, 0, 0
E) 1, 1, 1
C)
Norah Ali Al moneef
A
78
C
Example:
Consider three Gaussian surfaces. Surface 1
encloses a charge of +q, surface 2 encloses a
charge of –q and surface 3 encloses both
charges. Rank the 3 surfaces according to the
flux, greatest first.
A) 1, 2, 3
B) 1, 3, 2
+q
C) 2, 1, 3
-q
1
2
D) 2, 3, 1
E) 3, 2, 1
Norah Ali Al moneef
3
79
Example:
Rank the following Gaussian surfaces by the amount of
flux that passes through them, greatest first (q is at
the center of each).
A) 1, 2, 3
B) 1, 3, 2
C) 2, 1, 3
D) 3, 2, 1
E) All tie
q
q
q
1
2
3
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80
Rank the following Gaussian surfaces by the
strength of the field at the surface at the point
direction below q (where the numbers are).
A) 1, 2, 3
B) 1, 3, 2
C) 2, 1, 3
D) 3, 2, 1
q
q
1
2
E) All tie
Norah Ali Al moneef
q
3
81
Problem
What is the electric flux through the right
the face, the left face,and the top face?

E  3.0iˆ  4.0 ˆj
Right face:

dA  dAiˆ

 
F r   E  dA   3.0 iˆ  4.0 ˆj

 3.0 xdA  3.0 3.0dA  9.0 dA
 36 N  m 2 / C
Left face:
Top face:
Norah Ali Al moneef
F l  12 N  m 2 / C
Ft 
 3.0iˆ  4.0 ˆj  dAˆj 
 16 N  m 2 / C
82
Example:
A uniform electric field of magnitude 6000 N/C points upward in
a charge-free region. What is the electric flux through the shaded
side of the square pyramid shown if its base is 80 m on a side and
it is 50 m high?
.
All flux lines entering the base must leave the top,
so the flux through the four sides is the flux
through the base:
ФE = EA= (6000 N/C)(80 m)2= 38.4 MNm2/C.
By symmetry, the flux through each
side is the same: ФE= 9.6 MNm2/C
Norah Ali Al moneef
83
Example:
Consider a closed triangular box resting within a horizontal
electric field of magnitude E = 7.80 x104 N/C as shown in
Figure. Calculate the electric flux through
(a) the vertical rectangular surface,
(b) (b) the slanted surface,
(c) (c) the entire surface of the box
Norah Ali Al moneef
84
Norah Ali Al moneef
85
Example:
A uniform electric field aiˆ + bjˆ intersects a surface of
area
A. What is the flux through this area if the surface lies
(a) in the yz plane? (b) in the xz plane? (c) in the xy plane?
Norah Ali Al moneef
86
Example:
Consider a thin spherical shell of radius 14.0 cm with a
total charge of 32.0 μC distributed uniformly on its
surface. Find the electric field
(a) 10.0 cm
(b) 20.0 cm from the center of the charge distribution
Norah Ali Al moneef
87
Example - Charge in a Cube
 Q=3.76 nC is at the
Q
center of a cube. What is
the electric flux through
one of the sides?
 Gauss’ Law:
F  Q / 0
 Since a cube has 6 identical sides and the point charge is
at the center
F one face
F 1Q
Nm 2
 
 (substitute the numerical values)  70.8
6 6 0
C
Norah Ali Al moneef
88
Example:
 Shown is an arrangement of five charged pieces of
plastic (q1=q4=3nC, q2=q5=-5.9nC and q3=-3.1nC). A
Gaussian surface S is indicated. What is the net electric
flux through the surface?
A: F=-6 x 10-9C/0= -678 Nm2/C
B: F= x 10-9C/0= -1356 Nm2/C
C: F=0
D: F= .9 x 10-9C/0= 328 Nm2/C
enclosed charge
Norah Ali Al moneef
89
Example
Two large thin metal plates with surface charge densities of opposite
signs but equal magnitude of 44.27 ×10–20 Cm–2 are placed parallel
and close to each other. What is the field
(i) To the left of the plates?
(ii) To the right of the plates?
(iii) Between the plates?
Electric field exists only in the region between the plates.
(i), (ii) E = 0
(iii) σ = 44.27×10–20 Cm–2 ,
σ
44.27×10–20
8
–1
E=
=
=5×10
NC
ε0
8.854×10–12
Norah Ali Al moneef
90
Example
A spherical shell of radius 10 cm has a charge 2×10–6 C distributed
uniformly over its surface. Find the electric field
(a) Inside the shell
(b) Just outside the shell
(c) At a point 15 cm away from the centre
q = 2 ×10–6 C, R = 0.1 m,
r = 0.15 m
(a) Inside the shell, electric field is zero
1
q
2×10–6
9
(b) E =
= 9×10 ×
2
4ε 0 R
0.12
= 1.8×106 NC–1
1
q
2×10–6
9
(c) E' =
= 9×10 ×
4ε0 r2
0.152
=8×105 NC–1
Norah Ali Al moneef
91
Example
Ex) What is the electric flux through this disk of radius r = 0.10 m if
the uniform electric field has a magnitude
E = 2.0x103 N/C?
F E  EAcos  (2.0 10 )  ( r )  cos30
3
2
2
0
 (2.0 10 )  (3.14 10 )  0.866  54 N  m / C
3
Norah Ali Al moneef
2
92
Example
A sphere of radius 8.0 cm carries a uniform volume charge
density  = 500 nC/m3. What is the electric field
magnitude at r = 8.1 cm?
A. 0.12 kN/C
B. 1.5 kN/C
C. 0.74 kN/C
D.2.3 kN/C
E. 12 kN/C
Norah Ali Al moneef
93
Example
A spherical shell of radius 9.0 cm carries a
uniform surface charge density  = 9.0 nC/m2.
The electric field magnitude at r = 4.0 cm is
approximately
A. 0.13 kN/C
B. 1.0 kN/C
C. 0.32 kN/C
D.0.75 kN/C
E. zero
Norah Ali Al moneef
94
Example
A spherical shell of radius 9.0 cm carries a uniform surface
charge density  = 9.0 nC/m2. The electric field magnitude
at r = 9.1 cm is approximately
A. zero
B. 1.0 kN/C
C. 0.65 kN/C
D.0.32 kN/C
E. 0.13 kN/C
Norah Ali Al moneef
95
Example
An infinite plane of surface charge density  = +8.00 nC/m2 lies
in the yz plane at the origin, and a second infinite plane of
surface charge density  = –8.00 nC/m2 lies in a plane parallel
to the yz plane at x = 4.00 m. The electric field magnitude at x =
3.50 m is approximately
A. 226 N/C
B. 339 N/C
C. 904 N/C
D.452 N/C
E. zero
Norah Ali Al moneef
96
Example
An infinite plane of surface charge density  = +8.00 nC/m2
lies in the yz plane at the origin, and a second infinite plane of
surface charge density  = –8.00 nC/m2 lies in a plane parallel
to the yz plane at x =4.00 m. The electric field magnitude at x =
5.00 m is approximately
A. 226 N/C
B. 339 N/C
C. 904 N/C
D.452 N/C
E. zero
Norah Ali Al moneef
97
Example
A solid conducting sphere of radius ra is placed concentrically inside
a conducting spherical shell of inner radius rb1 and outer radius rb2.
The inner sphere carries a charge Q while the outer sphere does not
carry any net charge. The electric field for ra < r < rb1 is
A.
B.
C.
D.
E.
kQ
rˆ
2
r
kQ
rˆ
r2
2kQ

rˆ
r
2kQ
rˆ
r
zero

Norah Ali Al moneef
ra
Q
rb1
rb2
98
Example
A solid conducting sphere of radius ra is placed concentrically
inside a conducting spherical shell of inner radius rb1 and outer
radius rb2. The inner sphere carries a charge Q while the outer
sphere does not carry any net charge. The electric field for rb1 < r <
rb2 is
kQ
A.
B.
C.
D.
E.
Norah Ali Al moneef

2
rˆ
r
kQ
rˆ
2
r
2kQ

rˆ
r
2kQ
rˆ
r
zero
ra
Q
rb1
rb2
99
Example
A solid conducting sphere of radius ra is placed concentrically
inside a conducting spherical shell of inner radius rb1 and outer
radius rb2. The inner sphere carries a charge Q while the outer
sphere does not carry any net charge. The electric field for r > rb1 is
A.
B.
C.
D.
E.
Norah Ali Al moneef
kQ
 2 rˆ
r
kQ
rˆ
2
r
2kQ

rˆ
r
2kQ
rˆ
r
zero
ra
Q
rb1
rb2
100
Example
Consider a thin spherical shell of radius 14.0 cm with a
total charge of 32.0 μC distributed uniformly on its
surface. Find the electric field
(a) 10.0 cm and
(b) 20.0 cm from the center of the charge distribution.
E
keQ
2
r
8.99  10  32.0  10 


 7.19 M N
6
9
 0.200
2
C
Summary:
Gauss’ Law
· Gauss’ Law depends on the enclosed charge only
  qenc
F   E  dA 
o
1.
2.
3.

If there is a positive net flux there is a net positive charge
enclosed
If there is a negative net flux there is a net negative charge
enclosed
If there is a zero net flux there is no net charge enclosed
Gauss’ Law works in cases of symmetry
Norah Ali Al moneef
102
GAUSS LAW – SPECIAL SYMMETRIES
SPHERICAL
CYLINDRICAL
PLANAR
(point or sphere)
(line or cylinder)
(plane or sheet)
Depends only on
radial distance
from central point
Depends only on
perpendicular
distance from line
Depends only on
perpendicular
distance from plane
GAUSSIAN
SURFACE
Sphere centered
at point of
symmetry
Cylinder centered
at axis of
symmetry
Pillbox or cylinder
with axis
perpendicular to plane
ELECTRIC
FIELD E
E constant at
surface
E ║A - cos  = 1
E constant at
curved surface and
E║A
E ┴ A at end
surface
cos  = 0
E constant at end
surfaces and E ║ A
E ┴ A at curved
surface
cos  = 0
CHARGE
DENSITY
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103
Geometry
Charge
Density
Gaussian surface
Linear
 = q/L
Cylindrical, with
axis along line of
charge
Electric field
E
Sheet or
Plane
 = q/A
Cylindrical, with
axis along E.
E
Spherical
 = q/V
Spherical, with
center on center
of sphere
E
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104

0

20 r
Conducting
Line of Charge
E

Nonconducting
2 0


q
q

r
E

r

R
3 
2

4

R
40 r
0


r<R
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105