Physics chapters 24 - 25 2
Electric charge Able to attract other objects Two kinds
Positive – glass rod rubbed with silk Negative – plastic rod rubbed with fur
Like charges repel Opposite charge attract Charge is not created, it is merely
transferred from one material to another
Physics chapters 24 - 25 3
Elementary particles Proton – positively charged Electron – negatively charged Neutron – no charge Nucleus – in center of atom,
contains protons and neutrons Quarks – fundamental particles –
make up protons and neutrons, have fractional charge
Physics chapters 24 - 25 4
ions
Positive ions – have lost one or more electrons
Negative ions – have gained one or more electrons
Only electrons are lost or gained under normal conditions
Physics chapters 24 - 25 5
Conservation of charge
The algebraic sum of all the electric charges in any closed system is constant.
Physics chapters 24 - 25 6
Electrical interactions
Responsible for many things The forces that hold molecules and
crystals together Surface tension Adhesives Friction
Physics chapters 24 - 25 7
Conductors
Permit the movement of charge through them
Electrons can move freely Most metals are good conductors
Physics chapters 24 - 25 8
Insulators
Do not permit the movement of charge through them
Most nonmetals are good insulators
Electrons cannot move freely
Physics chapters 24 - 25 10
Coulomb’s Law Point charge – has essentially no
volume The electrical force between two
objects gets smaller as they get farther apart.
The electrical force between two objects gets larger as the amount of charge increases
Physics chapters 24 - 25 11
Coulomb’s Law
221
r
qqkF
r is the distance between the charges
q1 and q2 are the magnitudes of the charges
k is a constant 8.99 x 109 N∙m2/C2
Physics chapters 24 - 25 12
Coulombs
SI unit of charge, abbreviated C Defined in terms of current – we
will talk about this later
Physics chapters 24 - 25 13
Coulomb’s law constant
k is defined in terms of the speed of light k = 10-7c
k = 1/4pe0
e0 is another constant that will be more useful later
e0 = 8.85 x 10-12 C2/N∙m2
Physics chapters 24 - 25 14
The coulomb
Very large amount of charge Charge on 6 x 1018 electrons Most charges we encounter are
between 10-9 and 10-6 C 1 mC = 10-6 C
Physics chapters 24 - 25 16
Electric Field
• A field is a region in space where a force can be experienced.
• Or: a region in space where a quantity has a definite value at every point.
Physics chapters 24 - 25 17
Electric Field
• Produced by a charged particle.• The force felt by another charged
particle is caused by the electric field.
• We can check for an electric field with a test charge, qt. If it experiences a force, there is an electric field.
Physics chapters 24 - 25 18
Electric field
• The definite quantity is a ratio of the electric force experienced by a charge to the amount of the charge.
• Vector quantity measured in N/C.
EF tqtq
FE
Physics chapters 24 - 25 19
Electric field
• To determine the field from a point charge, Q, we place a test charge, qt, at some position and determine the force acting on it.
Q qt
F
Physics chapters 24 - 25 20
Direction of E
• If the test charge is positive, E has the same direction as F.
• If the test charge is negative, E has the opposite direction as F.
Physics chapters 24 - 25 22
Electric Field
• The field is there, independent of a test charge or anything else!
• The electric field vector points in the direction a positive charge would be forced.
Physics chapters 24 - 25 23
Example 1
• Two charges, Q1 = +2 x 10-8 C and Q2 = +3 x 10-8 C are 50 mm apart as shown below.
• What is the electric field halfway between them?
Q1Q2
50 mm
E1E2
Physics chapters 24 - 25 24
Example 1
• At the halfway point, r1 = r2 = 25 mm.
• Magnitudes of fields:
E1 kQ1
r12
(9 x 109 N • m2
C2
)(2 x 10 8C)
(2.5 x 10 2 m)2
E2 kQ2
r22
(9 x 109 N • m2
C2
)(3 x 10 8 C)
(2.5 x 10 2 m)2
Physics chapters 24 - 25 25
Example 1
• E1 = 2.9 x 105 N/C
• E2 = 4.3 x 105 N/C
• E1 is to the right and E2 is to the left.
• E1 = 2.9 x 105 N/C
• E2 = - 4.3 x 105 N/C
• E = E1 + E2 = - 1.4 x 105 N/C
Physics chapters 24 - 25 26
Example 2
• For the charges in Example 1, where is the electric field equal to zero?
• Since the fields are in opposite directions between the charges, the point where the field is zero must be between them.
Q1Q2
E1E2
Physics chapters 24 - 25 27
Example 2
E1 E2
kQ1
r12
kQ2
r22
Q1
r12
Q2
r22
r1 + r2 = s, so r2 = s – r1
Physics chapters 24 - 25 28
Example 2
Q1
r12
Q2
r22
Q1
r12
Q2
(s r1)2
(s r1 )2
r12
Q2
Q1
s r1
r1
Q2
Q1
r1 s
1 Q2
Q1
r1 23 mm
Physics chapters 24 - 25 29
Field Diagrams
• To represent an electric field we use lines of force or field lines.
• These represent the sum of the electric field vectors.
Physics chapters 24 - 25 32
Field Diagrams
• At any point on the field lines, the electric field vector is along a line tangent to the field line.
Physics chapters 24 - 25 34
Field Diagrams
• Lines leave positive charges and enter negative charges.
• Lines are drawn in the direction of the force on a positive test charge.
• Lines never cross each other.• The spacing of the lines represents
the strength or magnitude of the electric field.
Physics chapters 24 - 25 35
Point Charges
• Lines leave or enter the charges in a symmetric pattern.
• The number of lines around the charge is proportional to the magnitude of the charge.
Physics chapters 24 - 25 38
Gauss’s Law
• Electric flux through a closed surface is proportional to the total number of field lines crossing the surface in the outward direction minus the number crossing in the inward direction.
0Q
EA
Physics chapters 24 - 25 39
Example 25-9 (see page 563)
Field of a charged sphere is the same as if it were a point charge
204
1
r
qE
Physics chapters 24 - 25 40
Example 25-10 (see page 564)
Field of a infinite line of charge is
rE
02
1
Physics chapters 24 - 25 42
Example 3
• Two parallel metal plates are 2 cm apart.
• An electric field of 500 N/C is placed between them.
• An electron is projected at 107 m/s halfway between the plates and parallel to them.
• How far will the electron travel before it strikes the positive plate?
Physics chapters 24 - 25 43
Example 3
• Two charged parallel plates create a uniform electric field in the space between them.
Physics chapters 24 - 25 44
Example 3
Evo
This is just like a projectile problem except that the acceleration is not a given value.
Physics chapters 24 - 25 45
Example 3
a =F
mF = qE = eE
a =eE
m=
(1.6 x 10–19C)(500 N/C)
9.1 x 10–31kg
= 8.8 x 1013 m/s2
Physics chapters 24 - 25 46
Example 3
• 8.8 x 1013 m/s2 is the vertical acceleration of the electron.
• Horizontally, the acceleration is zero.
• x = vt• v = 1 x 107 m/s & t = ?
Physics chapters 24 - 25 47
Example 3
• Back to vertical direction:• y = yo + vot + 1/2at2
• y = 1/2at2
a
2yt
2(0.01 m)8.8x1013 m / s2
= 1.5 x 10-8 s
Physics chapters 24 - 25 48
Example 3
• Back to horizontal direction:• x = vt• x = (1 x 107 m/s)(1.5 x 10–8 s)
• x = 0.15 m = 15 cm
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