Download four slides per page

Electric Flux

Chapter 24
Gauss’s Law

Defining Electric Flux

EFM06AN1
Electric flux is the
product of the
magnitude of the
electric field and the
surface area, A,
perpendicular to the
field
ΦE = EA
Electric Flux, General Area



The electric flux is
proportional to the
number of electric field
lines penetrating some
surface
The field lines may
make some angle θ
with the perpendicular
to the surface
Then ΦE = EA cos θ
1
Electric Flux, Interpreting the
Equation



The flux is a maximum when the
surface is perpendicular to the field
The flux is zero when the surface is
parallel to the field
If the field varies over the surface, then
Φ = EA cos θ is valid for only a small
element of the area
Electric Flux, final



The surface integral means the integral
must be evaluated over the surface in
question
In general, the value of the flux will
depend both on the field pattern and on
the surface
The units of electric flux will be N.m2/C

i.e. [N/C] x [m2]
Electric Flux, General
In the more general
case, look at a small
area element
 E  E iA i cos  i  E i  A i
 In general, this
becomes


 E  lim
 Ai  0
E
i

  Ai 
E  dA
surface
Electric Flux, Closed Surface


Assume a closed
surface
The vectors ΔAi
point in different
directions


At each point, they
are perpendicular to
the surface
By convention, they
point outward
2
Quick Quiz 24.1
Flux Through Closed Surface

The net flux through the surface is
proportional to the net number of lines leaving
the surface


This is the number of lines leaving the surface
minus the number entering the surface
If En is the component of E perpendicular to
the surface, then
E 
 E  dA   E
n
dA
Suppose the radius of a sphere, with a charge at its centre, is
halved. What happens to the flux through the sphere and the
magnitude of the electric field at its surface?
(a) The flux and field both increase.
(b) The flux and field both decrease.
(c) The flux increases and the field decreases.
(d) The flux decreases and the field increases.
(e) The flux remains the same and the field increases.
(f) The flux decreases and the field remains the same.

Quick Quiz 24.1
Answer: (e). The same number of field lines pass through a
sphere of any size. Because points on the surface of the
sphere are closer to the charge, the field is stronger.
Quick Quiz 24.2
In a charge-free region of space, a closed container is placed
in an electric field. A requirement for the total electric flux
through the surface of the container to be zero is that
(a) the field must be uniform
(b) the container must be symmetric
(c) the container must be oriented in a certain way
(d) The requirement does not exist – the total electric flux is
zero no matter what.
3
Quick Quiz 24.2
Gauss’s Law, Introduction
Answer: (d). All field lines that enter the container also leave
the container so that the total flux is zero, regardless of the
nature of the field or the container.

Gauss’s law is an expression of the
general relationship between the net
electric flux through a closed surface
and the charge enclosed by the surface


Gauss’s Law – General


Gauss’s law is of fundamental
importance in the study of electric fields
Gauss’s Law – General, cont.
A positive point
charge, q, is located
at the centre of a
sphere of radius r
The magnitude of
the electric field
everywhere on the
surface of the
sphere is
E = keq / r2
The closed surface is often called a
Gaussian surface

The field lines are directed radially outward
and are perpendicular to the surface at
every point
E 



 E  dA  E  dA
This will be the net flux through the
gaussian surface, the sphere of radius r
We know E = keq/r2 and Asphere = 4πr2,
q
 E  E  A sphere  4  k e q 
0

4
Gauss’s Law – General, notes



The net flux through any closed surface
surrounding a point charge, q, is given by q/εo
and is independent of the shape of that surface.
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
Gauss’s Law – Final



 E  dA   E
1
 E2 
  dA
E 
 E  dA 
q in
0
qin is the net charge inside the surface
E represents the electric field at any point on
the surface


E 
Gauss’s law states

E is the total electric field and may have contributions
from charges both inside and outside of the surface
Although Gauss’s law can, in theory, be solved
to find E for any charge configuration, in
practice it is limited to symmetric situations

Finding an electric field using
Gauss’s law

EFA06AN1
Quick Quiz 24.3
If the net flux through a gaussian surface is zero, the
following four statements could be true. Which of the
statements must be true?
(a) There are no charges inside the surface.
(b) The net charge inside the surface is zero.
(c) The electric field is zero everywhere on the surface.
(d) The number of electric field lines entering the surface
equals the number leaving the surface.
5
Quick Quiz 24.3
Answer: (b) and (d).
Statement (a) is not necessarily true because an equal
number of positive and negative charges could be present
inside the surface.
Statement (c) is not necessarily true - the number of field
lines entering the surface will be equal to the number
leaving the surface, but this does not mean the field is zero
on the surface.
Quick Quiz 24.4
Consider the charge distribution shown in the figure. The
charges contributing to the total electric flux through surface
S’ are
(a) q1 only
(b) q4 only
(c) q2 and q3
(d) all four charges
(e) none of the charges
Quick Quiz 24.4
Answer: (c). The charges q1 and q4 are outside the surface
and contribute zero net flux through S’.
Quick Quiz 24.5
Again consider the charge distribution shown in this figure.
The charges contributing to the total electric field at a chosen
point on the surface S’ are
(a) q1 only
(b) q4 only
(c) q2 and q3
(d) all four charges
(e) none of the charges
6
Conditions for a Gaussian
Surface
Quick Quiz 24.5
Answer: (d). We don't need the surfaces to realize that any
given point in space will experience an electric field due to
all local source charges.
Choose a surface over which the integral can
be simplified and the electric field determined.
Try to choose a surface that satisfies one or
more of these conditions:






Field Due to a Spherically
Symmetric Charge Distribution
Field Due to a Point Charge

Choose a sphere as the
Gaussian surface
E 
 E  dA
E  E
E 


E is parallel to dA at each
point on the surface



E dA 
 dA  E (4  r
q
4  0 r
2
 ke
2
q

E 
0
)
Select a sphere as the
Gaussian surface
For r >a
 E  dA  
2
 E  4 r 
q
r
The value of the electric field can be argued from
symmetry to be constant over the surface
The dot product of E.dA can be expressed as a
simple algebraic product EdA because E and dA
are parallel
The dot product is 0 because E and dA are
perpendicular
The field can be argued to be zero over the surface
Q
0
E dA  Q
so E 
0
Q
4 0r
2
 ke
Q
r
2
Q is the total charge inside the surface
2

7
Spherically Symmetric
Distribution, final
Spherically Symmetric, cont.



Select a sphere as the Gaussian
surface, r < a
qenclosed =  (4/3πr3)
Charge density
 = Q / [4/3πa3]
E 
E 
 E  dA  
q enc
4  0 r
2
 ke
E dA 
q enc
r
2
 ke
q enc
2

4 k e r
3
Inside the sphere, E
varies linearly with r


0
4

   r 3 
 3

r


k e Qr
a
E → 0 as r → 0
The field outside the
sphere is equivalent
to that of a point
charge located at
the centre of the
sphere
3

Field Due to a Thin Spherical
Shell



Use spheres as the Gaussian surfaces
When r > a, the charge inside the surface is Q and
E = keQ / r2
When r < a, the charge inside the surface is 0 and E = 0
Field at a Distance from a Line
of Charge

Select a cylindrical
charge distribution


The cylinder has a
radius of r and a length
of ℓ
E is constant in
magnitude and
perpendicular to the
surface at every point
on the curved part of
the surface
8
Field Due to a Line of Charge


Field Due to a Line of Charge
The end view
confirms the field is
perpendicular to the
curved surface
The field through the
ends of the cylinder
is 0 since the field is
parallel to these
surfaces

Use Gauss’s law to find the field
E 
 E  dA  E  dA  EA
A  2  rl, so E (2  rl )   l
E 

2  0 r
 2ke

q encl
0
 l
0
0

r

Field Due to a Plane of Charge




E must be perpendicular to the plane
and must have the same magnitude at
all points equidistant from the plane
Choose a small cylinder whose axis is
perpendicular to the plane for the
Gaussian surface
E is parallel to the curved surface, with
no contribution to the surface area from
this curved part of the cylinder
The flux through each end of the
cylinder is EA and so the total flux is
2EA
Field Due to a Plane of Charge


The total charge in the surface is σA
Applying Gauss’s law
 E  2 EA 



A
0
, so E 

2 0
Note, this does not depend on r
Therefore, the field is uniform everywhere
9
Properties of a Conductor in
Electrostatic Equilibrium
Electrostatic Equilibrium

1. The electric field is zero everywhere inside
the conductor.
2. If an isolated conductor carries a charge, the
charge resides on its surface.
3. The electric field just outside a charged
conductor is perpendicular to the surface
and has a magnitude of σ/εo.
4. On an irregularly shaped conductor, the
surface charge density is greatest at
locations where the radius of curvature is
the smallest.
When there is no net motion of charge
within a conductor, the conductor is said
to be in electrostatic equilibrium
Property 2: Charge Resides
on the Surface
Property 1: Einside = 0





Consider a conducting slab in
an external field E
If the field inside the conductor
were not zero, free electrons in
the conductor would
experience an electrical force
These electrons would
accelerate
These electrons would not be
in equilibrium
Therefore, there cannot be a
field inside the conductor





Choose a Gaussian surface
inside but close to the actual
surface
The electric field inside is
zero (property 1)
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
Therefore, any charge must
be on the surface.
10
Demo EA10: Gauss’s law; field
inside insulator, not conductor

Property 3: Field’s Magnitude
and Direction
Excess charges resides on the outer surface of a
conductor. A metal can is placed on top of a van
der Graaff generator and filled with styrofoam
packing. When the generator is turned on, the
packing remains within the cup. One can show that
the outside of the can has charge by drawing off a
spark with the earth rod. The metal can and the
generator are all at one equipotential and the
charge is on the outside of the conductor. There is
no field inside the can. The can is replaced by a
polystyrene cup. In this case, the packing is ejected
out of the cup. Because the cup is an insulator, a
field can exist within it – the cup makes little
disturbance to the equipotential surfaces.


Choose a cylinder as the
Gaussian surface
The field must be
perpendicular to the
surface, for, 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.
 E  EA 
A
0
, and E 

0

Property 4: Charge density
highest at sharp points

The charge density is high
where the radius of curvature is
small


End of Chapter
And low where the radius of
curvature is large
The electric field is large near
the convex points having small
radii of curvature and reaches
very high values at sharp points
11
Quick Quiz 24.6
Quick Quiz 24.6
Your little brother likes to rub his feet on the carpet and then
touch you to give you a shock. While you are trying to
escape the shock treatment, you discover a hollow metal
cylinder in your basement, large enough to climb inside. In
which of the following cases will you not be shocked?
Answer: (a). Charges added to the metal cylinder by your
brother will reside on the outer surface of the conducting
cylinder. If you are on the inside, these charges cannot
transfer to you from the inner surface. For this same reason,
you are safe in a metal automobile during a lightning storm.
(a) You climb inside the cylinder, making contact with the
inner surface, and your charged brother touches the outer metal
surface.
(b) Your charged brother is inside touching the inner metal
surface and you are outside, touching the outer metal surface.
(c) Both of you are outside the cylinder, touching its outer
metal surface but not touching each other directly.
12