electrostatics (part 1-student copy)
TRANSCRIPT
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THEORIGINOFELECTRICITY
The electrical nature of matter is inherent
in atomic structure.
kg10673.1 27pm
kg10675.1 27
nm
kg1011.9 31em
C1060.1 19e
coulombs
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T
HE
O
RIGIN
OF
E
LECTRICITY
-There are two kinds of electric charge:
positive
and
negative.
The SI unit of electric charge is
the Coulomb C). The magnitude of the charge
on an electron or a proton is
e
= 1.60 x 10
-19
C
-The law of conservation of electric charge
states that the
net electric charge of an isolated
system remains constant during any process.
-
Electric repulsion and attraction:
Like charges
repel, unlike charges attract,
and
charged
bodies attract never repel) neutral bodies.
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18.1 THEORIGINOFELECTRICITY
In nature, atoms are normally
found with equal numbers ofprotons and electrons, so they
are electrically neutral.
By adding or removingelectrons
from matter it will acquire a net
electric charge with magnitude
equal to etimes the number of
electrons added or removed, N.
Neq
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18.2 CHARGEDOBJECTSANDTHEELECTRICFORCE
It is possible to transfer electric charge from oneobject to another.
The body that loses electrons has a deficiency of
negative charge (more protons than electrons),
while the body that gains electrons has an excess
of negative charge (more electrons than protons).
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18.2 CHARGEDOBJECTSANDTHEELECTRICFORCE
Like charges repel and
unl ike
charges attract each o ther.
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SAMPLEPROBLEMS 1. An electron has a charge whose magnitude
is e = 1.60 x 10-19C. How many electrons are
there in one Coulomb of negative charge?
Soln:
N = q/e = -1 C/ -1.60 x 10-19C
= 6 x 1018
electrons 2. In a chemical reaction, an aluminum atom
loses three electrons to become an aluminum
ion. An ion is a charged atom. What is the
charge (in Coulomb) on this ion? Soln:
q = N e = 3 (+1.60 x 10-19C)
= + 4.80 x 10-19 C
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SEATWORK 1. A conducting sphere has a net charge of 2.4 1017C.
What is the approximate number of excess electrons onthe sphere?
2. Each of three objects has a net charge. Objects A and B
attract one another. Objects B and C also attract oneanother, but objects A and C repel one another. What is thesign of each of the net charges on these three objects?
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18.3 CONDUCTORSANDINSULATORS
Not only can electric charge exist on an object, but it can also movethrough an object.
Substances that readily conduct electric charge are called electr ical
conduc tors.
Materials that conduct electric charge poorly are called electr icalinsulators .
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METHODSOFCHARGING-Charging by contact (or conduction):
-the process of giving one object a net
electric charge by placing it in contact with
an object that is already charged.
-Charging by induction
-the process of giving an object a net electric
charge without touching it to a charged
object.
-Charging by friction
-the process of giving an object a net electric
charge by rubbingit with another object with
a different ability to lose (or gain) electron(s).
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18.4 CHARGINGBYCONTACTANDBYINDUCTION
Charging by contact.
18 4 CHARGING BY CONTACT AND BY INDUCTION
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18.4 CHARGINGBYCONTACTANDBYINDUCTION
Charging by induction.
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18.4 CHARGINGBYCONTACTANDBYINDUCTION
The negatively charged rod induces a slight positive surface charge
on the plastic.
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COULOMBSLAW
-Apoint charge is a charge that occupies so
little space that it can be regarded as amathematical point.
-Coulombs law gives the magnitude Fof the
electric force that two point charges q1and q2
exert on each other:/
q
1
//
q
2
/
F = k -----------
r
2
Where /q1/ and /q2/ are the magnitudes of thecharges and have no algebraic sign. The term
k is a constant equal to 9 x 109Nm2/C2.
18 5 C L
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18.5 COULOMBSLAW
COULOMBS 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
qqkF
229 CmN1099.841 ok
2212 mNC1085.8
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EXAMPLESOFELECTRICALFORCECALCULATEDI
Example 1 Electric force vs. gravitational force
An particle is the nucleus of a helium atom. It has massm= 6.64 x 10-27kg and charge q= +2e= 3.2 x 10-19C.
Compare the force of the electric repulsion betweentwo particles with the force of gravitational attractionbetween them.
Soln: Fe + Fg + Fe
rFe= kq2/r2 Fg= Gm
2/r2
(9.0 x 109)(3.2 x 10-19C)2
Fe/Fg= kq2/Gm2= -----------------------------------(in SI units)
(6.67 x 10-11)(6.64 x 10-27)2
= 3.1 x 1035
+
+
+
+
+
+
+
+
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18.5 COULOMBSLAW
18 5 COULOMB S LAW
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18.5 COULOMBSLAW
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 5.29x10-11m.
Determine the speed of the electron, assuming the orbit tobe circular.
2
21
r
qqkF
18 5 COULOMB S LAW
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18.5 COULOMBSLAW
N1022.8m1029.5
C1060.1CmN1099.8 8211
219229
2
21
rqqkF
rmvmaF c2
sm1018.2
kg109.11
m1029.5N1022.8 631-
118
mFrv
ECTOR ADDITION OF ELECTRIC FORCES ON A LINE
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ECTORADDITIONOFELECTRICFORCESONALINE
Example 1: Three point charges are located on thepositive x-axis of a coordinate system, as shown in thediagram. Determine the net force on point charge q1.
18 5 COULOMB S LAW
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18.5 COULOMBSLAW
N7.2m20.0
C100.4C100.3CmN1000.92
66229
2
21
12
r
qqkF
N4.8m15.0
C100.7C100.3CmN1000.9
2
66229
2
31
13
r
qqkF
3qtowardsN5.7or5.7NN4.8N7.21312 FFF
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2.Three point charges are arranged on a line. Charge q3=+5.00 nC and is at the origin. Charge q2= -3.00 nC and isat x = +4.00 cm. Charge q1is at x = +2.00 cm. What is q1
(magnitude and sign) if the net force on q3is zero?
Soln: q3= +5.00 nC q1= ? nC q2= -3.00 nC
0 2 cm 4 cm
vector diagram of forces acting on q3:q3
F31 F32= attractive force by q2repulsive force by q1
Fnet= F32+ F31= 0 F32= 9 x 109
(3 x 10-9
)(5 x 10-9
)/(0.04)2
= 8.4375 x 10-5N8.4375 x 10-5N = F31= 9 x 10
9(5 x 10-9) q1/(0.02)2
q1= +0.750 nC
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1. Three point charges are arranged
along the x-axis. Charge q1= +3.00uC is at the origin, and charge q2=-5.00 uC is at x = 0.200 m. Charge
q3= -8.00 uC. Where is q3locatedif the net force on q1is 7.00 N inthe x-direction? Show the free-
body diagram for q1.
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Soln: q3= -8.00 uC q1= +3.00 uC q2= -5.00 uC
r 0.200 m0 m Assume q3is at a distance r
to the left of q1vector diagram of forces acting on q1:
q1F13 F12= attractive force by q2
F13= attractive force by q3Fnet= F12+ F13= 7.00 N to the x axis-7.00 = 9 x 109(3 x 10-6)(5 x 10-6)/(0.2)2
- (9 x 109)(8 x 10-6)(3 x 10-6)/r2
-7.00 = 3.375 0.216 /r2
-10.375 r2= -0.216r = 0.144 m but we take only the negative value as
assumed. Hence, r = -0.144 m
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2. An average human weighs about 650 N. If two suchgeneric humans each carried 1.0 coulomb of excesscharge, one positive and one negative, how far apart would
they have to be for the electric attraction between themto equal their 650-N weight?
Soln:
Fe= 650 N = 9 x 109(1 C)2/r2
r2= 9 x 109(1)/650
r = 3,721 m
18 5 COULOMB S LAW
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18.5 COULOMBSLAW
18.6 THE ELECTRIC FIELD
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18.6 THEELECTRICFIELD
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 chargeshould have a small magnitude so it doesnt affectthe other charge.
18.6 THEELECTRICFIELD
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8 6 C C
Example 6 A Test Charge
The positive test charge has a magnitude of
3.0x10-8C and experiences a force of 6.0x10-8N.
(a) Find the force per coulombthat the test charge
experiences.
(b) Predict the force that a charge of +12x10-8C
would experience if it replaced the test charge.
CN0.2C100.3
N100.6
8
8
oq
F(a)
(b) N1024C100.12CN0.2 88 F
E
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ELECTRICCHARGE
Glass rods, plastic tubes, silk, and fur can be used to demonstratethe movement of electrons and how their presence or absence
make for powerful forces of attraction and repulsion.
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MOVEMENTOFCHARGESCHARGINGBYCONDUCTION
Materials that alloweasy passage ofcharge are calledconductors. Materialsthat resist electronic
flow are calledinsulators. The motionof electrons throughconducts and aboutinsulators allows us to
observe oppositecharges attract andlike charges repel.
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ELECTRONSMOVEFREELYANDCHARGESMAYBEINDUCED
Take a childs toy, a rubber balloon. If you rub the balloon vigorouslyon a fuzzy sweater then bring the balloon slowly toward a painted
concrete or plaster wall, the balloon will stick to the wall and remainfor some time. The electrostatic force between static electrons and theinduced positive charge in the wall attract more strongly than theweight of the balloon.
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STATICELECTRICITYABOUTANINSULATORCANSHIFT
The motion of static charges about a plastic comb and light bits
of paper can cause attractive forces strong enough to overcomethe weight of the paper.
EXAMPLES OF ELECTRICAL FORCE CALCULATED
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EXAMPLESOFELECTRICALFORCECALCULATED
Example 1: An particle (alpha) is the nucleus of a heliumatom. It has mass m = 6.64 x 10-27kg and charge q = +2e =3.2 x 10-19C. Compare the force of the electric repulsion
between two particles with the force of gravitationalattraction between them.
Soln: Fe= 9 x 109(3.2 x 10-19)2/r2
Fg= 6.67 x 10-11(6.64 x 10-27)2/r2
Fe/Fg= 9.2 x 10-28
/2.94 x 10-63
= 3.1 x 1035
E l 2 T i t h l t d th iti
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Example 2: Two point charges are located on the positive x-axis of a coordinate system. Charge q1= 1.0 nC is 2.0 cmfrom the origin, and charge q2= -3.0 nC is 4.0 cm from theorigin. What is the total force exerted by these two charges
on a charge q3= 5.0 nC located at the origin? Gravitationalforces are negligible.
Soln: F31= 9.0 x 109(1.0 x 10-9)(5.0 x 10-9)/(2.0 x 10-2)2
= 1.12 x 10-4N to the left
F32= 9.0 x 109
(3.0 x 10-9
)(5.0 x 10-9
)/(4.0 x 10-2
)2
= 8.4 x 10-5N to the right
= F31+ F32= 28 N, to the left
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Example 1: Two equal positive point chargesq1= q2= 2.0 C are located atx= 0, y =0.30 m andx = 0, y = -0.30 m, respectively.
What are the magnitude and direction ofthe total electric force that these chargesexert on a third point charge Q = 4.0 C at
x = 0.40 m, y = 0 ?
THREE-POINT CHARGES ON A PLANE
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THREE-POINTCHARGESINAPLANE
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Soln:F1= 9 x 10
9(4.0 x 10-6)(2.0 x 10-6)/(0.50)2= 0.29 NF1x= F1cos = (0.29)(0.40/0.50) = 0.23 NF1y= -F1sin = -(0.29)(0.30/0.50) = -0.17 N
The lower charge q2exerts a force with the same magnitudebut at an angle abovethe x-axis. From symmetry, we see thatitsx-component is the same as that due to the upper charge,but its y-component has the opposite sign. So the componentsof the total force Fneton Q are:Fx= 0.23 N + 0.23 N = 0.46 NFy= -0.17 N + 0.17 N = 0The total force on Q is in the +x-direction , with magnitude
0.46 N.
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Example 2: The figure aboveshows three point chargesthat lie in the x,y plane in avacuum. Find the magnitudeand direction of the netelectrostatic force on q1.
THREE-POINT CHARGES ON A PLANE
F12F
q1 F13
F12
F12sin 73o
73o
F12cos 73o
SOLUTION
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SOLUTION
The magnitudes of the forces are:/q1/q2/ (9 x 10
9)(4 x 10-6)(6 x 10-6)
F12= k -------- = ------------------------------------ = 9.6 Nr122 (0.15)2
/q1/q3/ (9 x 109)(4 x 10-6)(5 x 10-6)
F13= k -------- = ------------------------------------ = 18 Nr12
2 (0.10)2
Force x-component y-componentF12 +9.6 (cos 73
o) = +2.8 N +9.6 N (sin 73o) = +9.2 NF13 +18 N 0 NF Fx= +21 N Fy= +9.2 N
The magnitude Fand angle of the net force are:F = Fx
2+ Fy2= (21 N)2+ (9.2 N)2= 23 N
= tan-1(9.2 N/21 N) = 24o
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F32x= cos 60o(16.2 N) = -8.1 N
F32y= sin 60o(16.2 N) = +14 N
Fy= -9.4 N + 14 N = +4.6 N
Fx= -5.4 N 8.1 N = -13.5 N
Fnet1
= (13.5)2+ (4.6)2= 14.3 N, 19oabove the -x-axis