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Chapter 18
Electric Forces and
Electric Fields
18.1 The Origin of Electricity
The electrical nature of matter is inherent
in atomic structure.
kg10673.1 27−
×=pm
kg10675.1 27−
×=nm
kg1011.9 31−
×=em
C1060.1 19−
×=e
coulombs
18.1 The Origin of Electricity
In nature, atoms are normally
found with equal numbers of protons
and electrons, so they are electrically
neutral.
By adding or removing electrons
from matter it will acquire a net
electric charge with magnitude equal
to e times the number of electrons
added or removed, N.
Neq =
18.1 The Origin of Electricity
Example 1 A Lot of Electrons
How many electrons are there in one coulomb of negative charge?
Neq =
18
19-
1025.6
C101.60
C00.1
×=
×
==
e
q
N
18.2 Charged Objects and the Electric Force
The body that loses electrons has an excess of positive charge, while
the body that gains electrons has an excess of negative charge.
It is possible to transfer electric charge
from one object to another.
18.2 Charged Objects and the Electric Force
LAW OF CONSERVATION OF ELECTRIC CHARGE
During any process, the net electric charge of an isolated system remains
constant (is conserved).
18.2 Charged Objects and the Electric Force
Like charges repel and unlike
charges attract each other.
18.2 Charged Objects and the Electric Force
18.3 Conductors and Insulators
Not only can electric charge exist on an object, but it can also move
through and object.
Substances that readily conduct electric charge are called electrical
conductors.
Materials that conduct electric charge poorly are called electrical
insulators.
18.4 Charging by Contact and by Induction
Charging by contact.
18.4 Charging by Contact and by Induction
Charging by induction.
18.4 Charging by Contact and by Induction
The negatively charged rod induces a slight positive surface charge
on the plastic.
18.5 Coulomb’s Law
18.5 Coulomb’s Law
COULOMB’S LAW
The magnitude of the electrostatic force exerted by one point charge
on another point charge is directly proportional to the magnitude of the
charges and inversely proportional to the square of the distance between
them.
2
21
r
qq
kF =
( ) 229
CmN1099.841 ⋅×== ok πε
( )2212
mNC1085.8 ⋅×= −
οε
18.5 Coulomb’s Law
Example 3 A Model of the Hydrogen Atom
In the Bohr model of the hydrogen atom, the electron is in orbit about the
nuclear proton at a radius of 5.29x10-11
m. Determine the speed of the
electron, assuming the orbit to be circular.
2
21
r
qq
kF =
18.5 Coulomb’s Law
( )( )
( )
N1022.8
m1029.5
C1060.1CmN1099.8 8
211
219229
2
21 −
−
−
×=
×
×⋅×
==
r
qq
kF
rmvmaF c
2
==
( )( ) sm1018.2
kg109.11
m1029.5N1022.8 6
31-
118
×=
×
××
==
−−
mFrv
18.5 Coulomb’s Law
Example 4 Three Charges on a Line
Determine the magnitude and direction of the net force on q1.
18.5 Coulomb’s Law
( )( )( )
( )
N7.2
m20.0
C100.4C100.3CmN1099.8
2
66229
2
21
12 =
××⋅×
==
−−
r
qq
kF
( )( )( )
( )
N4.8
m15.0
C100.7C100.3CmN1099.8
2
66229
2
31
13 =
××⋅×
==
−−
r
qq
kF
5.7NN4.8N7.21312 +=+−=+= FFF

18.5 Coulomb’s Law
18.6 The Electric Field
The positive charge experiences a force which is the vector sum of the
forces exerted by the charges on the rod and the two spheres.
This test charge should have a small magnitude so it doesn’t affect
the other charge.
18.6 The Electric Field
Example 6 A Test Charge
The positive test charge has a magnitude of
3.0x10-8
C and experiences a force of 6.0x10-8
N.
(a) Find the force per coulomb that the test charge
experiences.
(b) Predict the force that a charge of +12x10-8
C
would experience if it replaced the test charge.
CN0.2
C100.3
N100.6
8
8
=
×
×
= −
−
oq
F
(a)
(b) ( )( ) N1024C100.12CN0.2 88 −−
×=×=F
18.6 The Electric Field
DEFINITION OF ELECRIC FIELD
The electric field that exists at a point is the electrostatic force experienced
by a small test charge placed at that point divided by the charge itself:
oq
F
E


=
SI Units of Electric Field: newton per coulomb (N/C)
18.6 The Electric Field
It is the surrounding charges that create the electric field at a given point.
18.6 The Electric Field
Example 7 An Electric Field Leads to a Force
The charges on the two metal spheres and the ebonite rod create an electric
field at the spot indicated. The field has a magnitude of 2.0 N/C. Determine
the force on the charges in (a) and (b)
18.6 The Electric Field
( )( ) N1036C100.18CN0.2 88 −−
×=×== EqF o
(a)
(b) ( )( ) N1048C100.24CN0.2 88 −−
×=×== EqF o
18.6 The Electric Field
Electric fields from different sources
add as vectors.
18.6 The Electric Field
Example 10 The Electric Field of a Point Charge
The isolated point charge of q=+15μC is
in a vacuum. The test charge is 0.20m
to the right and has a charge qo=+15μC.
Determine the electric field at point P.
oq
F
E


=
2
21
r
qq
kF =
18.6 The Electric Field
( )( )( )
( )
N7.2
m20.0
C1015C1080.0CmN1099.8
2
66229
2
=
××⋅×
=
=
−−
r
qq
kF o
CN104.3
C100.80
N7.2 6
6-
×=
×
==
oq
F
E
18.6 The Electric Field
2
r
q
kE =
The electric field does not depend on the test charge.
o
o
o qr
qq
k
q
F
E
1
2
==
Point charge q:
18.6 The Electric Field
Example 11 The Electric Fields from Separate Charges May Cancel
Two positive point charges, q1=+16μC and q2=+4.0μC are separated in a
vacuum by a distance of 3.0m. Find the spot on the line between the charges
where the net electric field is zero.
2
r
q
kE =
18.6 The Electric Field
( ) ( )
( )2
6
2
6
m0.3
C100.4C1016
d
k
d
k
−
×
=
× −−
21 EE =
2
r
q
kE =
( ) 22
m0.30.2 dd =−
m0.2+=d
18.6 The Electric Field
Conceptual Example 12 Symmetry and the
Electric Field
Point charges are fixes to the corners of a rectangle in two
different ways. The charges have the same magnitudes
but different signs.
Consider the net electric field at the center of the rectangle
in each case. Which field is stronger?
18.6 The Electric Field
THE PARALLEL PLATE CAPACITOR
Parallel plate
capacitor
oo A
q
E
ε
σ
ε
==
( )2212
mNC1085.8 ⋅×= −
οε
charge density
18.7 Electric Field Lines
Electric field lines or lines of force provide a map of the electric field
in the space surrounding electric charges.
18.7 Electric Field Lines
Electric field lines are always directed away from positive charges and
toward negative charges.
18.7 Electric Field Lines
Electric field lines always begin on a positive charge
and end on a negative charge and do not stop in
midspace.
18.7 Electric Field Lines
The number of lines leaving a positive charge or entering a
negative charge is proportional to the magnitude of the charge.
18.7 Electric Field Lines
18.7 Electric Field Lines
Conceptual Example 13 Drawing Electric
Field Lines
There are three things wrong with part (a) of
the drawing. What are they?
18.8 The Electric Field Inside a Conductor: Shielding
At equilibrium under electrostatic conditions, any
excess charge resides on the surface of a conductor.
At equilibrium under electrostatic conditions, the
electric field is zero at any point within a conducting
material.
The conductor shields any charge within it from
electric fields created outside the condictor.
18.8 The Electric Field Inside a Conductor: Shielding
The electric field just outside the surface of a conductor is perpendicular to
the surface at equilibrium under electrostatic conditions.
18.8 The Electric Field Inside a Conductor: Shielding
Conceptual Example 14 A Conductor in
an Electric Field
A charge is suspended at the center of
a hollow, electrically neutral, spherical
conductor. Show that this charge induces
(a) a charge of –q on the interior surface and
(b) a charge of +q on the exterior surface of
the conductor.
18.9 Gauss’ Law
( )22
4 rqrkqE oπε==
( )oAqE ε=

o
q
EA
ε
=
EAE =Φflux,Electric
18.9 Gauss’ Law
( )∑ ∆=Φ AEE φcos
18.9 Gauss’ Law
GAUSS’ LAW
The electric flux through a Gaussian
surface is equal to the net charge
enclosed in that surface divided by
the permittivity of free space:
( )
o
Q
AE
ε
φ =∆∑ cos
SI Units of Electric Flux: N·m2
/C
18.9 Gauss’ Law
Example 15 The Electric Field of a Charged Thin Spherical Shell
A positive charge is spread uniformly over the shell. Find the magnitude
of the electric field at any point (a) outside the shell and (b) inside the
shell.
( )
o
Q
AE
ε
φ =∆∑ cos
18.9 Gauss’ Law
( ) ( )
( )2
4
0coscos
rEAE
AEAEE
π
φ
=∆=
∆=∆=Φ
∑
∑∑
( )
o
Q
rE
ε
π =2
4
18.9 Gauss’ Law
( )
o
Q
rE
ε
π =2
4
(a) Outside the shell, the Gaussian
surface encloses all of the charge.
or
q
E
επ 2
4
=
(b) Inside the shell, the Gaussian
surface encloses no charge.
0=E
18.9 Gauss’ Law
18.10 Copiers and Computer Printers
18.10 Copiers and Computer Printers
18.10 Copiers and Computer Printers

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Ch18 electric field

  • 1. Chapter 18 Electric Forces and Electric Fields
  • 2. 18.1 The Origin of Electricity The electrical nature of matter is inherent in atomic structure. kg10673.1 27− ×=pm kg10675.1 27− ×=nm kg1011.9 31− ×=em C1060.1 19− ×=e coulombs
  • 3. 18.1 The Origin of Electricity In nature, atoms are normally found with equal numbers of protons and electrons, so they are electrically neutral. By adding or removing electrons from matter it will acquire a net electric charge with magnitude equal to e times the number of electrons added or removed, N. Neq =
  • 4. 18.1 The Origin of Electricity Example 1 A Lot of Electrons How many electrons are there in one coulomb of negative charge? Neq = 18 19- 1025.6 C101.60 C00.1 ×= × == e q N
  • 5. 18.2 Charged Objects and the Electric Force The body that loses electrons has an excess of positive charge, while the body that gains electrons has an excess of negative charge. It is possible to transfer electric charge from one object to another.
  • 6. 18.2 Charged Objects and the Electric Force LAW OF CONSERVATION OF ELECTRIC CHARGE During any process, the net electric charge of an isolated system remains constant (is conserved).
  • 7. 18.2 Charged Objects and the Electric Force Like charges repel and unlike charges attract each other.
  • 8. 18.2 Charged Objects and the Electric Force
  • 9. 18.3 Conductors and Insulators Not only can electric charge exist on an object, but it can also move through and object. Substances that readily conduct electric charge are called electrical conductors. Materials that conduct electric charge poorly are called electrical insulators.
  • 10. 18.4 Charging by Contact and by Induction Charging by contact.
  • 11. 18.4 Charging by Contact and by Induction Charging by induction.
  • 12. 18.4 Charging by Contact and by Induction The negatively charged rod induces a slight positive surface charge on the plastic.
  • 14. 18.5 Coulomb’s Law COULOMB’S LAW The magnitude of the electrostatic force exerted by one point charge on another point charge is directly proportional to the magnitude of the charges and inversely proportional to the square of the distance between them. 2 21 r qq kF = ( ) 229 CmN1099.841 ⋅×== ok πε ( )2212 mNC1085.8 ⋅×= − οε
  • 15. 18.5 Coulomb’s Law Example 3 A Model of the Hydrogen Atom In the Bohr model of the hydrogen atom, the electron is in orbit about the nuclear proton at a radius of 5.29x10-11 m. Determine the speed of the electron, assuming the orbit to be circular. 2 21 r qq kF =
  • 16. 18.5 Coulomb’s Law ( )( ) ( ) N1022.8 m1029.5 C1060.1CmN1099.8 8 211 219229 2 21 − − − ×= × ×⋅× == r qq kF rmvmaF c 2 == ( )( ) sm1018.2 kg109.11 m1029.5N1022.8 6 31- 118 ×= × ×× == −− mFrv
  • 17. 18.5 Coulomb’s Law Example 4 Three Charges on a Line Determine the magnitude and direction of the net force on q1.
  • 18. 18.5 Coulomb’s Law ( )( )( ) ( ) N7.2 m20.0 C100.4C100.3CmN1099.8 2 66229 2 21 12 = ××⋅× == −− r qq kF ( )( )( ) ( ) N4.8 m15.0 C100.7C100.3CmN1099.8 2 66229 2 31 13 = ××⋅× == −− r qq kF 5.7NN4.8N7.21312 +=+−=+= FFF 
  • 20. 18.6 The Electric Field The positive charge experiences a force which is the vector sum of the forces exerted by the charges on the rod and the two spheres. This test charge should have a small magnitude so it doesn’t affect the other charge.
  • 21. 18.6 The Electric Field Example 6 A Test Charge The positive test charge has a magnitude of 3.0x10-8 C and experiences a force of 6.0x10-8 N. (a) Find the force per coulomb that the test charge experiences. (b) Predict the force that a charge of +12x10-8 C would experience if it replaced the test charge. CN0.2 C100.3 N100.6 8 8 = × × = − − oq F (a) (b) ( )( ) N1024C100.12CN0.2 88 −− ×=×=F
  • 22. 18.6 The Electric Field DEFINITION OF ELECRIC FIELD The electric field that exists at a point is the electrostatic force experienced by a small test charge placed at that point divided by the charge itself: oq F E   = SI Units of Electric Field: newton per coulomb (N/C)
  • 23. 18.6 The Electric Field It is the surrounding charges that create the electric field at a given point.
  • 24. 18.6 The Electric Field Example 7 An Electric Field Leads to a Force The charges on the two metal spheres and the ebonite rod create an electric field at the spot indicated. The field has a magnitude of 2.0 N/C. Determine the force on the charges in (a) and (b)
  • 25. 18.6 The Electric Field ( )( ) N1036C100.18CN0.2 88 −− ×=×== EqF o (a) (b) ( )( ) N1048C100.24CN0.2 88 −− ×=×== EqF o
  • 26. 18.6 The Electric Field Electric fields from different sources add as vectors.
  • 27. 18.6 The Electric Field Example 10 The Electric Field of a Point Charge The isolated point charge of q=+15μC is in a vacuum. The test charge is 0.20m to the right and has a charge qo=+15μC. Determine the electric field at point P. oq F E   = 2 21 r qq kF =
  • 28. 18.6 The Electric Field ( )( )( ) ( ) N7.2 m20.0 C1015C1080.0CmN1099.8 2 66229 2 = ××⋅× = = −− r qq kF o CN104.3 C100.80 N7.2 6 6- ×= × == oq F E
  • 29. 18.6 The Electric Field 2 r q kE = The electric field does not depend on the test charge. o o o qr qq k q F E 1 2 == Point charge q:
  • 30. 18.6 The Electric Field Example 11 The Electric Fields from Separate Charges May Cancel Two positive point charges, q1=+16μC and q2=+4.0μC are separated in a vacuum by a distance of 3.0m. Find the spot on the line between the charges where the net electric field is zero. 2 r q kE =
  • 31. 18.6 The Electric Field ( ) ( ) ( )2 6 2 6 m0.3 C100.4C1016 d k d k − × = × −− 21 EE = 2 r q kE = ( ) 22 m0.30.2 dd =− m0.2+=d
  • 32. 18.6 The Electric Field Conceptual Example 12 Symmetry and the Electric Field Point charges are fixes to the corners of a rectangle in two different ways. The charges have the same magnitudes but different signs. Consider the net electric field at the center of the rectangle in each case. Which field is stronger?
  • 33. 18.6 The Electric Field THE PARALLEL PLATE CAPACITOR Parallel plate capacitor oo A q E ε σ ε == ( )2212 mNC1085.8 ⋅×= − οε charge density
  • 34. 18.7 Electric Field Lines Electric field lines or lines of force provide a map of the electric field in the space surrounding electric charges.
  • 35. 18.7 Electric Field Lines Electric field lines are always directed away from positive charges and toward negative charges.
  • 36. 18.7 Electric Field Lines Electric field lines always begin on a positive charge and end on a negative charge and do not stop in midspace.
  • 37. 18.7 Electric Field Lines The number of lines leaving a positive charge or entering a negative charge is proportional to the magnitude of the charge.
  • 39. 18.7 Electric Field Lines Conceptual Example 13 Drawing Electric Field Lines There are three things wrong with part (a) of the drawing. What are they?
  • 40. 18.8 The Electric Field Inside a Conductor: Shielding At equilibrium under electrostatic conditions, any excess charge resides on the surface of a conductor. At equilibrium under electrostatic conditions, the electric field is zero at any point within a conducting material. The conductor shields any charge within it from electric fields created outside the condictor.
  • 41. 18.8 The Electric Field Inside a Conductor: Shielding The electric field just outside the surface of a conductor is perpendicular to the surface at equilibrium under electrostatic conditions.
  • 42. 18.8 The Electric Field Inside a Conductor: Shielding Conceptual Example 14 A Conductor in an Electric Field A charge is suspended at the center of a hollow, electrically neutral, spherical conductor. Show that this charge induces (a) a charge of –q on the interior surface and (b) a charge of +q on the exterior surface of the conductor.
  • 43. 18.9 Gauss’ Law ( )22 4 rqrkqE oπε== ( )oAqE ε=  o q EA ε = EAE =Φflux,Electric
  • 44. 18.9 Gauss’ Law ( )∑ ∆=Φ AEE φcos
  • 45. 18.9 Gauss’ Law GAUSS’ LAW The electric flux through a Gaussian surface is equal to the net charge enclosed in that surface divided by the permittivity of free space: ( ) o Q AE ε φ =∆∑ cos SI Units of Electric Flux: N·m2 /C
  • 46. 18.9 Gauss’ Law Example 15 The Electric Field of a Charged Thin Spherical Shell A positive charge is spread uniformly over the shell. Find the magnitude of the electric field at any point (a) outside the shell and (b) inside the shell. ( ) o Q AE ε φ =∆∑ cos
  • 47. 18.9 Gauss’ Law ( ) ( ) ( )2 4 0coscos rEAE AEAEE π φ =∆= ∆=∆=Φ ∑ ∑∑ ( ) o Q rE ε π =2 4
  • 48. 18.9 Gauss’ Law ( ) o Q rE ε π =2 4 (a) Outside the shell, the Gaussian surface encloses all of the charge. or q E επ 2 4 = (b) Inside the shell, the Gaussian surface encloses no charge. 0=E
  • 50. 18.10 Copiers and Computer Printers
  • 51. 18.10 Copiers and Computer Printers
  • 52. 18.10 Copiers and Computer Printers