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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



2
Name
Charge per length
Charge per area
Charge per volume
Unit
C/m
C/m 2
C/m 3
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Electric Flux
• Electric flux is the product of the magnitude of
the electric field and the surface area, A,
perpendicular to the field
ΦE = EA
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24.1 Electric Flux
• 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
• The flux is a maximum
when the surface is
perpendicular to the field
• The flux is a minimum
(zero) when the surface is (parallel) to the field
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Electric Flux
• If the field varies over the surface, Φ = EA cos θ is
valid for only a small element of the area
• In the more general case, look at a small area
element
E  Ei Ai cos θi  Ei  Ai
• In general, this becomes
E  lim
Ai 0
E 

 E  A
i
i
E  dA
surface
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Electric Flux
• The surface integral means the integral must
be evaluated over the surface in question
• The value of the flux depends both on the
field pattern and on the surface
• SI units: N.m2/C
E  lim
Ai 0
E 

 E  A
i
i
E  dA
surface
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Electric Flux, Closed Surface
• For a closed surface, by convention, the A vectors are
perpendicular to the surface at each point and point outward
• (1) θ < 90o, Φ > 0
• (2) θ = 90o, Φ = 0
• (3) 180o > θ > 90o, Φ < 0
• The net flux through the surface is
proportional to the number of lines leaving
the surface minus the number entering the
surface
 
 E   E  dA   E dA
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Example 24-1: 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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24-2 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 =    E  dA   E cosdA 
For a Point charge
 
  EdA 
 EdA
E=kq/r2
2
kq/r
  dA
  kq/r 2  dA  kq/r 2 (4r 2 )
  4kq
4k  1/0 where 0  8.85x10
12
net 
qenc
0
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.
9
9
• Gaussian surfaces of various shapes
can surround the charge (only S1 is
spherical)
• The electric flux is proportional to the
number of electric field lines
penetrating these surfaces, and this
number is the same
• Thus the net flux through any closed
surface surrounding a point charge q is
given by q/εo and is independent of
the shape of the surface
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Gauss’ Law
• If the charge is outside the
closed surface of an arbitrary
shape, then any field line
entering the surface leaves at
another point
• Thus the electric flux through a
closed surface that surrounds no
charge is zero
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Gauss’ Law
• 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



 
 
 E   E  dA   E1  E2  ...  dA
• Although Gauss’s law can, in theory, be solved to find for
any charge configuration, in practice it is limited to symmetric
situations
• One should choose a Gaussian surface over which the surface
integral can be simplified and the electric field determined
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Problem 24-9
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
E 



6.
89

10
N

m
C
0
8.85  1012 C 2 N  m 2
(b) Since the net electric flux is negative, more
lines enter than leave the surface.
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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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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
 E   E  dA   E n dA   EdA  E  dA  E  4r 2
E 


 E .dA 
Qin

E has the has same value
at all points on the surface
0
E is can be moved out
A spherical Gaussian surface
centered on a point charge q
Integral is the sum of
surface area
Coulomb’s Law
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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
(B) Find the magnitude of the E-field at a point inside the sphere
• For
E
r>a


  E  dA  E  dA  E  4r 2 Q
0
E 
Q
Q
2  ke
4 0 r
r2
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b--Find the magnitude of the E-field at a point inside the
sphere
r< a
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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Field Due to a Spherically Symmetric
Charge Distribution
• Inside the sphere, E varies
linearly with r (E → 0 as r →
0)
• The field outside the sphere
is equivalent to that of a
point charge located at the
center of the sphere
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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
1
4 0
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q
r2
20
 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   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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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
• Select a cylinder as Gaussian
surface. The cylinder has a radius of
r and a length of ℓ
• E is constant in magnitude and
parallel to the surface (the
direction of a surface is its normal!)
at every point on the curved part of
the surface (the body of the
cylinder.
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q
 
L


Qin
 E .dA  
E
Gauss s law
0
d A  E  d A  0  E ( 2 r L ) 

2k
E

r
2 0 r
L

0
 Line of Charge
E
20r

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2k 
E
r̂
r
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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 nonconducting 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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 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   E   E  dA  q /  0
(Gauss)
EA  EA  A /  0
2 EA  A /  0
E   / 2 0
 the electric field from an infinite
 non-conducting sheet with charge density 
E   / 2 0
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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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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
E
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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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•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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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
 E  EA 
0

E 
0
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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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homework
problem1 - Charge in a Cube
 Q=3.76 nC is at the center of a cube. What
Q
is the electric flux through one of the
sides?
Problem 2 :
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?
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problem3 :
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
problem4 :
A 5 nC charge is placed inside of a cubic surface. The cube is
measured to be 0.5 m per side. What is the flux through the
cube?
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problem5 :
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?
problem6 :
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.
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problem7 :
A sphere of radius 8.0 cm carries a uniform volume charge
density r = 500 nC/m3. What is the electric field magnitude at
r = 8.1 cm?
Problem 8:
A long, straight wire is surrounded by a hollow metal cylinder
whose axis coincides with that of the wire. The wire has a
charge per unit length of λ, and the cylinder has a net charge
per unit length of 2λ. From this information, use Gauss’s law
to find (a) the charge per unit length on the inner and outer
surfaces of the cylinder and (b) the electric field outside the
cylinder, a distance r from the axis.
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Problem 9:
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.
Problem 10:
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.
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King Saud university
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Problem 11:
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.
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Summary:
Gauss’ Law
· Gauss’ Law depends on the enclosed charge only
  qenc
   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
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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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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

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


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r<R
40
1- The electric field has a magnitude of 3.0 N/m at a distance of 60 cm from
a point charge. What is the charge?
(a) 1.4 nC
(b) 120 pC
(c) 36 mC
(d) 12 C
(e) 3.0 nC
2- An electron traveling horizontally from North to South enters a region
where a uniform electric field is directed downward. What is the direction
of the electric force exerted on the electron once it has entered the field?
(a) downward
(b) upward
(c) to the east
(d) to the west
(e) to the north
5/25/2017
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A conducting sphere has a net charge of −4.8 × 10−17 C. What is
the approximate number of excess electrons on the sphere?
(a) 100
(b) 200
(c) 300
(d) 400
(e) 500
Two point charges, 8x10-9 C and -2x10-9 C are separated by 4
m. The electric field magnitude (in units of V/m) midway
between them is:
A)9x109
B) 13,500
C) 135,000
D) 36x10-9
5/25/2017
E) 22.5
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