1 lecture 1 electric charge structure of matter 08/30/2010 conductors and insulators charging...
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1
Lecture 1
Electric ChargeStructure of Matter
08/30/2010
Conductors and InsulatorsCharging MechanismsCoulomb’s lawSuperposition principle
2Electric charge - experiments
Repulsion Attraction
Plexiglass
Plexiglass
Plexiglass
Plastic
3
4
Plexiglass Plastic
Plexiglass Repels Attracts
Plastic Attracts Repels
New physical property of matter: electric chargeelectric charge
Electric charge
Charged bodies interact through: electrical forceselectrical forces
Need two flavors to explain the existence of repulsive as well as
attractive forcespositive (+) and negative (-)positive (+) and negative (-)
5Microscopic view of matter
Elementary particles:
Number of electrons = Number of protons
The atom is neutral
6Electric charges
Like charges repel; opposite charges attract.
Examples of charges: electrons (negative); protons (positive)
Two types of charges: positive (+) and negative (-)
Type of materials: insulators, conductors (Ex metals), semiconductors
7Electric charge properties
Like charges repel; opposite charges attract.
Charge is discrete (quantized)- The smallest charge possible is = - Any charge is a integer multiple of the elementary unit of
charge
1.602 x 10-19 C (Coulombs)
Charge is conserved - charge can be exchanged between different parts of a closed system, but the total charge of the system cannot change
8Three pithballs are suspended from thin threads. It is found thatpithballs 1 and 2 repel each other and that pithballs 2 and 3 repel each other. From this we can conclude that:
1. 1 and 3 carry charges of opposite sign. 2. 1 and 3 carry charges of equal size.3. all three carry charges of the same sign.4. one of the objects carries no charge.5. we need to do more experiments
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21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40
9
Types of materials
Conductors - charges move freely (Ex. metals). One consequence
of this property is that the charge that is transferred to a conductor will spread out uniformly on its surface. Insulators - charges do not move freely (Ex. glass, plastic). Semiconductors - intermediate case - charges can move freely under in some special cases (higher temperature, applied voltage) (silicon, germanium). Superconductors – extremely good conductors = zero
resistance. Restricted to very low temperatures. (Ex. niobium, BISCO)
10Electroscope and Van de Graaff generator
Electroscope: used to measured charge by measuring the deflection of charged metal foils
Van de Graaff generator.
11Charging mechanisms
Objects can become charged when elementary charged particles (most probably electrons) are transferred from one object to the other.
- When a glass rod and fur are rubbed together some electrons are transferred to the fur (triboelectricity)
12Charging by contact
- Electrons are transferred from the plastic rod to the metal ball
13When a neutral metal sphere is charged by contact with a positively charged glass rod, the sphere:
1. loses electrons 2. gains electrons3. loses protons4. gains protons
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14
Charging by induction
- The copper rod is attracted to the charged glass rod even if initially is uncharged and does not contact the glass rod at any time.
15The diagram below shows a neutral metal sphere on a isolating pedestal. Which of the other diagrams shown, describes best the charge distribution on the sphere when a negatively charged rod is brought in its vicinity?
1. A 2. B3. C4. D
A B
C D
++++
----
++++
+++
++-++--
-++
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16The diagram below shows three identical neutral metal spheres on a isolating pedestals, which are in contact. Which of the other diagrams shown, describes best the charge distribution on the spheres when a negatively charged rod is brought in its vicinity?
1. A 2. B3. C4. D
A B
C D
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17
A negatively charged plastic rod is brought in the vicinity of a neutral metal ball placed on a isolating pedestal. If the opposite side of the sphere is briefly connected to the ground and then the plastic rod is removed, what will the final charge on the ball be?1. The metal ball will be
neutral as initially. 2. The metal ball will be
positively charged3. The metal ball will be
negatively charged.
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18Solution
- Due to the interaction with the plastic rod, positive and negative charges will be separated on the ball. Through the grounding loop the electrons are transferred to the Earth which acts as an infinite sink or source of electrons, leaving the sphere with a deficit of electrons (positively charged)
19Three metal balls are suspended from thin threads. It is found that balls 1 and 2 attract each other and that balls 2 and 3 repel each other. From this we can definitely conclude that:
1. 1 and 3 carry charges of opposite sign. 2. 1 and 3 carry charges of the same sign.3. all three carry charges of the same sign.4. one of the objects carries no charge.5. we need to do more experiments
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21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40
20
Coulomb's Law
Explored by Charles Augustin de Coulomb
rrqqkF 2
21
constanty permitivit - mN
C1085.8
CmN1099.8
41k
0
2
212
0
2
29
0
21
Coulomb's law properties
Inverse square law
2112 FF
Direction: along the line joining the charges
Attractive for unlike charges and repulsive for like charges
rrqqkF 2
21
Newton’s 3rd Law action reaction pair:
22Two uniformly charged spheres are firmly fastened to and electrically insulated from frictionless pucks on an air table. The charge on sphere 2 is three times the charge on sphere 1. Which force diagram correctly shows the magnitude and direction of the electrostatic forces?
20
1. 1 2. 23. 34. 45. 56. 6
1 2 3 4 5 6
0% 0% 0%0%0%0%
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21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40
41 42 43 44 45 46 47 48 49 50
23Which graph best represents the magnitude of the interaction force between two positively charged ions as a function of the distance separating them?
20
A B
C D
1 2 3 4
0% 0%0%0%
1. A 2. B3. C4. D
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24
Coulomb's law vs. Newton law of gravitation
2
210-
221
2
29
221
kgmN106.6742 G ,
rmmGF :ngravitatio of law sNewton'
CmN108.99 k ,
rqq
kF :law sCoulomb'
Inverse square laws - force acting along the line joining the particles.
There are two kinds of charges, but only one type of mass.
Gravity is always attractive.
The electrostatic force, (e.g., between an electron and proton), is enormously stronger (~ 1035 times stronger)
25A hydrogen atom is composed of a nucleus containing a single proton, about which a single electron orbits. The electric force between the two particles is 2.3 x 1039 greater than the gravitational force! If we can adjust the distance between the two particles, can we find a separation at which the electric and gravitational forces are equal?
20
1. Yes, we must move the particles further apart. 2. Yes, we must move the particles closer, together.3. No, at any distance.
1 2 3
0% 0%0%
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26
Superposition principle
When two or more charges each exert a force on a charge, the total force on that charge is the vector sum of the forces exerted by the individual charges.
Example: - Linear distribution
n21 F...FFF
What is the net force acting on q3?
27Solution:
cmrwhererqq
kF
cmrwhererqq
kF
on
on
0.4,
0.2,
23223
3232
13213
3131
Step 1 : - Calculate the force produce by each charge
opposite. are they since forces individual thebetween difference by thegiven is force total theof magnitude the-
213
312
23
3231323 r
qqk
rqq
kFFF ononontotal
Step 2 : - Add all the forces using vector summation rules
28
Example 2: - Planar distribution (vector nature of Coulomb's law)Find the magnitude and direction of the force on Q.
29
0 N 0.17N 0.17
N 0.46 N 0.23N 0.23
21
21
yonQyonQytotal
xonQxonQxtotal
FFF
FFF
the total force on charge Q acts only along x
Step 1 : - find the force exerted by each charge on Q
N 0.29
(0.5m)C)10C)(4.010(2
CNm109.0
rQq
kF 2
66
2
29
21
1onQQ)1(q
The components on the x and y axis of are:1onQF
1onQ x 1onQ
1onQ y 1onQ
F F cos 360 - α
F F sin 360 - α
with
-1
0.3sin α =0.60.5
α=sin (0.6) 36.86o
o1onQ x 1onQ
o1onQ y 1onQ
F F cos 323.13 0.232 N
F F sin 323.13 0.17 N
N 17.0αsinFF
N 232.0αcosFF
2onQy2onQ
2onQx2onQ
Step 2 : - Add the two forces (addition using components)
Obtain the F2onQ components in a similar way
30
+q +4q
L
Extra credit (5 points) – due Wednesday, September 7
Problem
Two free point charges +q and +4q are located a distanceL, apart. A third charge is placed so that the whole system is in equilibrium. Find the location, magnitude, and sign of the third charge.
31Review – dealing with vector summation
y
x
1F
2F
F1x F2x
F2y
F1y
1 2
theangles are always measured clockwise towards the positive x axis
x
1F
2F
Fx
Fy
y
F
111
111
sincos
FFFF
y
x
222
222
sincos
FFFF
y
x
For the sum vector
221121
221121
sinsincoscos
FFFFFFFFFF
yyy
xxx
angle tantan
magnitude
1
22
x
y
x
y
yx
FF
FF
FFF
Question: How do we express the sum vector parameters (F, ), if we know them for the individual vectors, i.e. (F1, 1) and (F2,
32Extracredit Problem – Assigned on 09/03/2010Two free point charges +q and +4q are located a distance L, apart. A third charge is placed so that the whole system is in equilibrium. Find the location, magnitude, and sign of the third charge.Solution:
Step 1: - determine the condition for the charge qo to be at equilibrium.
Call the new charge, q0, and let it be distance x from +q. The free-body diagram shows relationship of the two forces on the new charge:
xLxxLxxL
xLx
xLqq
kxqq
k
324)(
)(41
)(4
22
22
20
20
33
Step 2: - determine the condition for the charge q or 4q to be at equilibrium.
qqqqqLq
Lq
xLwhereLqq
kxqq
k
94
9449
)(4
9
3)(4
00220
220
qq94
0
34Problem (Chapter 21, Problem 4 – Page 574)
Identical isolated conducting spheres 1 and 2 have equal charges and are separated by a distance that is large compared with their diameters (Figure a). The electrostatic force acting on sphere 2 due to sphere 1 is F. Suppose now that a third identical sphere 3, having an insulating handle and initially neutral, is touched first to sphere 1 (Figure b), then to sphere 2 (Figure c), and finally removed (Figure d). The electrostatic force that now acts on sphere 2 has magnitude F’. What is the ratio F’/F?
1 2 3 F12
q q 0 Fq/2 q q/2 F/2q/2 3q/4 3q/4 3F/8
35In the following diagram what is the direction of the electrostatic force on the negative charge -q?
1. 2. 3. 4. 5. none of the above
Q
Q
- q
201 2 3 4 5
0% 0% 0%0%0%
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36In the figure a charged particle of charge (- q) is surrounded by two circular rings of charged particles. What is the net force on the central particle due to the other particles?
1.
2.
3.
4.
5.
j
j
j
j
j
2o
2
2o
2
2o
2
2o
2
2o
2
2o
2
r4q
R4q2
R4q2
R4q2
r4q2
r4q2
201 2 3 4 5
0% 0% 0%0%0%
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
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41 42 43 44 45 46 47 48 49 50
37Two small charged objects attract each other with a force F when separated by a distance d. If the charge on each object is reduced to one-fourth of its original value and the distance between them is reduced to d/2 the force becomes:
1. F/16 2. F/83. F/4 4. F/25. F
201 2 3 4 5
0% 0% 0%0%0%
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38
Electric Field
Why do we need this concept?
- Coulomb's law describes action at distance (like gravitation).- Any charge changes the properties of the space around it.- We need a field to describe this change.
The notion of field is not a new one: - A field is a quantity that
is assigned a value at each point in a region of space
39
Vector field
Wind speed - the absolute value and direction is known.
© http://maps.wunderground.com/
40
Back to the electric field
0qFE
- Units: N/C - E does not depend on the test charge q0.
00q
00q
F
F
Note: The rest of this chapter and chapter 23 are mostly dedicated tohow to determine the magnitude and orientation of different electric.
To calculate the electric field of a charge distribution use superposition
...EEEE 321total
41
Force on a charge in an electric field
EqF
The direction of the force acting on a point charge in an electric field depends on both the direction of the electric field and the sign of the charge.
Positive charge q0 placed in an electric field
Negative charge q0 placed in an electric field
-
42
rrq
41E 2
0
Electric field due to a point charge
rrq qkF 2
0
0qFE
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