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Fall 2008Lecture 1-1Physics 231
Electric Charges,Forces, and
Fields
Fall 2008Lecture 1-2Physics 231
Electric ChargesElectric charge is a basic property of matterTwo basic charges
Positive and NegativeEach having an absolute value of
1.6 x 10-19 CoulombsExperiments have shown that
Like signed charges repel each otherUnlike signed charges attract each other
For an isolated system, the net charge of the system remains constant
Charge Conservation
Fall 2008Lecture 1-3Physics 231
Two basics type of materialsConductors
Materials, such as metals, that allow the free movement of charges
InsulatorsMaterials, such as rubber and glass, that don’t
allow the free movement of charges
Fall 2008Lecture 1-4Physics 231
Coulomb’s LawCoulomb found that the electric force between
two charged objects isProportional to the product of the charges on the objects, andInversely proportional to the separation of the objects squared
221
rqq
kF =
k being a proportionality constant, having a value of 8.988 x 109 Nm2/c2
Fall 2008Lecture 1-5Physics 231
Electric ForceAs with all forces, the electric force is a Vector
12221
12 r̂rqqkF =
r
This gives the force on charged object 2 due to charged object 1
12r̂ is a unit vector pointing from object 1 to object 2
So we rewrite Coulomb’s Law as
q2q1
The direction of the force is either parallel or antiparallel to this unit vector depending upon the relative signs of the charges
Fall 2008Lecture 1-6Physics 231
Electric ForceThe force acting on each charged object has the same magnitude - but acting in opposite directions
2112 FFrr
= (Newton’s Third Law)
Fall 2008Lecture 1-7Physics 231
Example 1A charged ball Q1 is fixed to a horizontal surface
as shown. When another massive chargedball Q2 is brought near, it achieves an equilibrium position at a distance d12 directly above Q1.
When Q1 is replaced by a different charged ball Q3, Q2 achieves an equilibrium position at a distance d23 (< d12) directly above Q3.
For 1a and 1b which is the correct answer
1a: A) The charge of Q3 has the same sign of the charge of Q1
B) The charge of Q3 has the opposite sign as the charge of Q1
C) Cannot determine the relative signs of the charges of Q3 & Q1
1b: A) The magnitude of charge Q3 < the magnitude of charge Q1
B) The magnitude of charge Q3 > the magnitude of charge Q1
C) Cannot determine relative magnitudes of charges of Q3 & Q1
Q2
Q1
gd12
Q2
d23
Q3
Fall 2008Lecture 1-8Physics 231
Example 1Q2
Q1
gd12
Q2
d23
Q3
A charged ball Q1 is fixed to a horizontal surface as shown. When another massive chargedball Q2 is brought near, it achieves an equilibrium position at a distance d12 directly above Q1.
When Q1 is replaced by a different charged ball Q3, Q2 achieves an equilibrium position at a distance d23 (< d12) directly above Q3.
1a: A) The charge of Q3 has the same sign of the charge of Q1
B) The charge of Q3 has the opposite sign as the charge of Q1
C) Cannot determine the relative signs of the charges of Q3 & Q1
• To be in equilibrium, the total force on Q2 must be zero.• The only other known force acting on Q2 is its weight.• Therefore, in both cases, the electrical force on Q2 must be directed upward
to cancel its weight.• Therefore, the sign of Q3 must be the SAME as the sign of Q1
Fall 2008Lecture 1-9Physics 231
Example 1Q2
Q1
gd12
Q2
d23
Q3
A charged ball Q1 is fixed to a horizontal surface as shown. When another massive charged ball Q2 is brought near, it achieves an equilibrium position at a distance d12 directly above Q1.
When Q1 is replaced by a different charged ball Q3, Q2 achieves an equilibrium position at a distance d23 (< d12) directly above Q3.
1b: A) The magnitude of charge Q3 < the magnitude of charge Q1
B) The magnitude of charge Q3 > the magnitude of charge Q1
C) Cannot determine relative magnitudes of charges of Q3 & Q1
• The electrical force on Q2 must be the same in both cases … it just cancels the weight of Q2
• Since d23 < d12 , the charge of Q3 must be SMALLER than the charge of Q1so that the total electrical force can be the same!!
Fall 2008Lecture 1-10Physics 231
More Than Two ChargesGiven charges q, q1, and q2
q
q1
q2
qqF1
r
qqF2
rnetFr
If q1 were the only other charge, we would know the force on qdue to q1 - qqF
1
r
If q2 were the only other charge, we would know the force on qdue to q2 - qqF
2
r
What is the net force if both charges are present?
The net force is given by the Superposition Principle
21 FFFnetrrr
+=
Fall 2008Lecture 1-11Physics 231
Superposition of ForcesIf there are more than two charged objects
interacting with each otherThe net force on any one of the charged
objects is The vector sum of the individual Coulomb
forces on that charged object
∑≠
=ji
rrqkqF ijij
ijj ˆ2
r
Fall 2008Lecture 1-12Physics 231
Example Two
x (cm)
y (cm)
1 2 3 4 5
4321
qo
q2q1 θ
qo, q1, and q2 are all point charges where qo = -1µC, q1 = 3µC, and q2 = 4µC
What is the force acting on qo?
We have that 20100 FFFrrr
+=
What are F0x and F0y ?
210
1010
rqqkF = yFF ˆ1010 −=
r
220
2020
rqqkF =
202020 rFF ˆ−=r
2010 FFrr
and calculate to NeedDecompose into its x and ycomponents
20Fr
( ) ( ) yFxFF ˆsinˆcos θθ 202020 −=r
20
02r
xx −=θcos
20
20r
yy −=θsin
Fall 2008Lecture 1-13Physics 231
Example Two - Continued
10FrNow add the components of and to find and xF0 yF020F
r
X-direction:
θcos200 FF x =
xxx FFF 20100 +=
010 =xF
yyy FFF 20100 x (cm)
y (cm)
1 2 3 4 5
4321
qo
q2q1
20Fr
10Fr
0Fr
Y-direction: +=
θsin20100 FFF y −−=
Fall 2008Lecture 1-14Physics 231
Example Two - ContinuedPutting in the numbers . . .
x (cm)
y (cm)
1 2 3 4 5
4321
qo
q2q1
20Fr
10Fr
0Fr
cm310 =r cm520 =r
8.0cos =θ
N4.1420 =FN3010 =F
We then get for the componentsN52.110 =xF N64.380 −=yF
The magnitude of is0Fr
N32.4020
200 =+= yx FFF
At an angle given by
( ) o40.73)52.11/64.38(tantan 100
1 −=== −−xy FFθ
Fall 2008Lecture 1-15Physics 231
Note on constantsk is in reality defined in terms of a more
fundamental constant, known as the permittivity of free space.
2
212
0
0
C 10854.8
41
Nmxwith
k
−=
=
ε
πε
Fall 2008Lecture 1-16Physics 231
Electric Field
The Electric Force is like the Gravitational Force
Action at a Distance
The electric force can be thought of as being mediated by an electric field.
Fall 2008Lecture 1-17Physics 231
What is a Field?A Field is something that can be defined anywhere in space
A field represents some physical quantity (e.g., temperature, wind speed, force)
It can be a scalar field (e.g., Temperature field)
It can be a vector field (e.g., Electric field)
It can be a “tensor” field (e.g., Space-time curvature)
Fall 2008Lecture 1-18Physics 231
A Scalar Field
7782
8368
5566
8375 80
90 91
757180
72
84
73
82
8892
7788
887364
A scalar field is a map of a quantity that has only a magnitude, such as temperature
Fall 2008Lecture 1-19Physics 231
A Vector Field77
82
8368
5566
8375 80
90 91
757180
72
84
73
57
8892
7756
887364
A vector field is a map of a quantity that is a vector, a quantity having both magnitude and direction, such as wind
Fall 2008Lecture 1-20Physics 231
Electric FieldWe say that when a charged object is put at
a point in space,The charged object sets up an Electric Field throughout the space surrounding the charged object
It is this field that then exerts a force on another charged object
Fall 2008Lecture 1-21Physics 231
Electric FieldLike the electric force,
the electric field is also a vector
If there is an electric force acting on an object having a charge qo, then the electric field at that point is given by
0qFEr
r= (with the sign of q0 included)
Fall 2008Lecture 1-22Physics 231
Electric FieldThe force on a positively charged object is in the same direction as the electric field at that point,
While the force on a negative test charge is in the opposite direction as the electric field at the point
Fall 2008Lecture 1-23Physics 231
Electric Field
A positive charge sets up an electric field pointing away from the charge
A negative charge sets up an electric field pointing towards the charge
Fall 2008Lecture 1-24Physics 231
Electric Field
⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜
⎝
⎛
= ∑≠ ji
rrqk ijij
ijj qF ˆ2
rEarlier we saw that the force on a charged object is given byThe term in parentheses remains the same if we change the charge on the object at the point in question
The quantity in the parentheses can be thought of as the electric field at the point where the test object is placed
rrqkE ˆ2=
r
The electric field of a point charge can then be shown to be given by
Fall 2008Lecture 1-25Physics 231
Electric Field
As with the electric force, if there are several charged objects, the net electric field at a given point is given by the vector sum of the individual electric fields
∑=i
iEErr
Fall 2008Lecture 1-26Physics 231
Electric FieldIf we have a continuous charge distribution the summation becomes an integral
∫= rrdqkE ˆ2
r
Fall 2008Lecture 1-27Physics 231
Hints
1) Look for and exploit symmetries in the problem.
2) Choose variables for integrationcarefully.
3) Check limiting conditions for appropriate result
Fall 2008Lecture 1-28Physics 231
Electric FieldRing of Charge
Fall 2008Lecture 1-29Physics 231
Electric FieldLine of Charge
Fall 2008Lecture 1-30Physics 231
Example 3
Two equal, but opposite charges are placed on the x axis. The positive charge is placed at x = -5 m and the negative charge is placed at x = +5m as shown in the figure above.
1) What is the direction of the electric field at point A?a) up b) down c) left d) right e) zero
2) What is the direction of the electric field at point B?a) up b) down c) left d) right e) zero
Fall 2008Lecture 1-31Physics 231
Example 4
Q2Q1 x
y
Ed
Two charges, Q1 and Q2, fixed along the x-axis asshown produce an electric field, E, at a point(x,y) = (0,d) which is directed along the negativey-axis.
Which of the following is true?
(a) Both charges Q1 and Q2 are positive
(b) Both charges Q1 and Q2 are negative
(c) The charges Q1 and Q2 have opposite signs
E
Q2Q1
(a)
Q2Q1
(b)E
Q2Q1
(c) E
Fall 2008Lecture 1-32Physics 231
Electric Field Lines
Possible to map out the electric field in a region of spaceAn imaginary line that at any given point has its tangent being in the direction of the electric field at that point
The spacing, density, of lines is related to the magnitude of the electric field at that point
Fall 2008Lecture 1-33Physics 231
Electric Field Lines
At any given point, there can be only one field line
The electric field has a unique direction at any given point
Electric Field LinesBegin on Positive ChargesEnd on Negative Charges
Fall 2008Lecture 1-34Physics 231
Electric Field Lines
Fall 2008Lecture 1-35Physics 231
Electric Dipole
An electric dipole is a pair of point charges having equal magnitude but opposite sign that are separated by a distance d.
Two questions concerning dipoles:1) What are the forces and torques acting on a dipole when placed in an external electric field?2) What does the electric field of a dipole look like?
Fall 2008Lecture 1-36Physics 231
Force on a DipoleGiven a uniform external field
Then since the charges are of equal magnitude, the force on each charge has the same value
However the forces are in opposite directions!
Therefore the net force on the dipole is
Fnet = 0
Fall 2008Lecture 1-37Physics 231
Torque on a Dipole
The individual forces acting on the dipole may not necessarily be acting along the same line.
If this is the case, then there will be a torque acting on the dipole, causing the dipole to rotate.
Fall 2008Lecture 1-38Physics 231
Torque on a Dipole
( )Edqrrr ×=τ
The torque is then given by τ = qE dsinφ
d is a vector pointing from the negative charge to the positive charge
Fall 2008Lecture 1-39Physics 231
Potential Energy of a DipoleGiven a dipole in an external field:
Dipole will rotate due to torqueElectric field will do workThe work done is the negative of the change in potential energy of the dipole
The potential energy can be shown to be
( )EdqUrr
⋅−=
Fall 2008Lecture 1-40Physics 231
Electric Field of a Dipole