Transcript Chapter 21

Chapter 21
Gauss’s Law
Electric Field Lines
• Electric field lines (convenient for visualizing electric
field patterns) – lines pointing in the direction of the
field vector at any point
• The electric field vector is tangential to the electric
field lines at each point
• The number of lines per unit area through a surface
perpendicular to the lines is proportional to the
strength of the electric field in a given region
Electric Field Line Patterns
• For a positive point charge the lines
will radiate outward equally in all
directions
• A positive test charge would be
repelled away from the positive source
charge
• For a negative point charge the lines
will point inward equally in all
directions
• A positive test charge would be
attracted toward the negative source
charge
Electric Field Line Patterns
• An electric dipole consists of two
equal and opposite point charges
• The number of field lines leaving the
positive charge equals the number of
lines terminating on the negative
charge
• For two equal but like point charges,
at a great distance from the charges,
the field would be approximately that
of a single charge of 2q (bulging out
of the field lines between the charges
– repulsion)
Electric Field Line Patterns
• For these two unequal and unlike
point charges, at a great distance
from the charges, the field would be
approximately that of a single charge
of +q (two lines leave the +2q charge
for each line that terminates on -q)
Rules for Drawing Electric Field Lines
• For a group of charges, the lines must begin on
positive charges and end on negative charges
• In the case of an excess of charge, some lines will
begin or end infinitely far away
• The number of lines drawn leaving a positive charge
or ending on a negative charge is proportional to the
magnitude of the charge
• No two field lines can cross each other
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
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
  EA cos  E  A
• The flux is a maximum (zero) when the surface is
perpendicular (parallel) to the field
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 

surface
E
i
 Ai
E  dA
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 

surface
E
i
 Ai
E  dA
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
Electric Flux, Closed Surface
• 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   En dA
Electric Flux, Closed Surface
• Example: flux through a cube
• The field lines pass perpendicularly through two
surfaces and are parallel to the other four surfaces
• Side 1: Φ = – E l2
• Side 2: Φ = E l2
• For the other sides, Φ = 0
• Therefore, Φtotal = 0
Chapter 21
Problem 23
The electric field on the surface of a 10-cm-diameter sphere is
perpendicular to the sphere and has magnitude 47 kN/C. What’s
the electric flux through the sphere?
Chapter 21
Problem 42
What’s the flux through the hemispherical open surface of radius
R in a uniform field of magnitude E shown in the figure?
Gauss’ Law
Carl Friedrich Gauss
1777 – 1855
• Gauss’ Law: electric flux through any closed surface
is proportional to the net charge Q inside the surface
• εo = 8.85 x 10-12 C2/Nm2 : permittivity of free space
1 /ε 0  4 π k e
• The area in Φ is an imaginary Gaussian surface (does
not have to coincide with the surface of a physical
object)
Gauss’ Law
• A positive point charge q is located at the center of a
sphere of radius r
• The magnitude of the electric field everywhere on the
surface of the sphere is E = keq / r2
• Asphere = 4πr2
 
 E   E  dA  E  dA
q
q
2
 k e 2  4r  4ke q 
r
0
Gauss’ Law
• 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/o and is independent of
the shape of the surface
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
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
Field Due to a Spherically Symmetric
Charge Distribution
• For r > a
 
Q
2
 E   E  dA  E  dA  E  4r 

0
Q
Q
E
2  ke
2
40 r
r
• For r < a
 
2 qin
 E   E  dA  E  dA  E  4r 

 4r / 3

r
E
2
3 0
40 r
3
0
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
Electric Field of a Charged Thin
Spherical Shell
• The calculation of the field outside the shell is
identical to that of a point charge
Q
Q
E
 ke 2
2
4 r  o
r
• The electric field inside the shell is zero
Field Due to a Line of Charge
• Select a cylindrical Gaussian surface
(of radius r and length ℓ)
• Electric field is constant in magnitude
and perpendicular to the surface at
every point on the curved part of the
surface
• 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
Field Due to a Line of Charge
 
Q
 E   E  dA  E  dA 
0
l
E (2rl ) 
0


 2k
E
r
20 r
Field Due to a Plane of Charge
• The uniform field must be perpendicular to the sheet
and directed either toward or away from the sheet
• Use a cylindrical Gaussian surface
• The flux through the ends is EA and there is no field
through the curved part of the surface
• Surface charge density σ = Q / A
Q
EA 
εo
σ
σA 0
E
E2A 0 
2ε o
εo
Field Distance Dependences
for Different Charge Distributions
Chapter 21
Problem 31
The electric field strength outside a charge distribution and 18 cm
from its center has magnitude 55 kN/C. At 23 cm the field strength
is 43 kN/C. Does the distribution have spherical or line symmetry?
Conductors in Electrostatic Equilibrium
• When no net motion of charge occurs within a
conductor, the conductor is said to be in
electrostatic equilibrium
• An isolated conductor has the following properties:
• Property 1: The electric field is zero everywhere
inside the conducting material
• If this were not true there were an electric field inside
the conductor, the free charge there would move and
there would be a flow of charge – the conductor
would not be in equilibrium
Conductors in Electrostatic Equilibrium
• When no net motion of charge occurs within a
conductor, the conductor is said to be in
electrostatic equilibrium
• An isolated conductor has the following properties:
• Property 1: The electric field is zero everywhere
inside the conducting material
Conductors in Electrostatic Equilibrium
• When no net motion of charge occurs within a
conductor, the conductor is said to be in
electrostatic equilibrium
• An isolated conductor has the following properties:
• Property 1: The electric field is zero everywhere
inside the conducting material
Conductors in Electrostatic Equilibrium
• When no net motion of charge occurs within a
conductor, the conductor is said to be in
electrostatic equilibrium
• An isolated conductor has the following properties:
• Property 2: Any excess charge on an isolated
conductor resides entirely on its surface
• The electric field (and thus the flux) inside is zero
whereas the Gaussian surface can be as close to the
actual surface as desired, thus there can be no
charge inside the surface and any net charge must
reside on the surface
Conductors in Electrostatic Equilibrium
• When no net motion of charge occurs within a
conductor, the conductor is said to be in
electrostatic equilibrium
• An isolated conductor has the following properties:
• Property 3: The electric field just outside a charged
conductor is perpendicular to the surface and has a
magnitude of σ/εo
• If this was not true, the component along the surface
would cause the charge to move – no equilibrium
σA
σ
 E  EA 
and E 
εo
εo
Conductors in Electrostatic Equilibrium
• When no net motion of charge occurs within a
conductor, the conductor is said to be in
electrostatic equilibrium
• An isolated conductor has the following properties:
• Property 4: On an irregularly shaped conductor, the
charge accumulates at locations where the radius of
curvature of the surface is smallest
• The forces from the charges at the sharp end
produce a larger resultant force away from the
surface.
Chapter 21
Problem 41
A total charge of is applied to a thin, square metal plate 75 cm on a
side. Find the electric field strength near the plate’s surface.
Answers to Even Numbered Problems
Chapter 21:
Problem 22
± 490 N ⋅ m2/C
Answers to Even Numbered Problems
Chapter 21:
Problem 24
1.81 × 103 N ⋅ m2/C
Answers to Even Numbered Problems
Chapter 21:
Problem 32
−2.0 × 10−4 C/m2
Answers to Even Numbered Problems
Chapter 21:
Problem 38
160 kN/C
Answers to Even Numbered Problems
Chapter 21:
Problem 56
58 nC/m2