Unformatted text preview: Chapter 16
Electric Charge and
Electric Field Units of Chapter 16
• Static Electricity; Electric Charge and Its Conservation
• Electric Charge in the Atom
• Insulators and Conductors
• Induced Charge; the Electroscope
Charge; the Electroscope
• Coulomb’s Law
• Solving Problems Involving Coulomb’s Law and Vectors
• The Electric Field
• Field Lines
•Electric Fields and Conductors 16.1 Static Electricity; Electric Charge
and Its Conservation
and Its Conservation
Objects can be charged by rubbing
can be charged by rubbing 16.1 Static Electricity;
Electric Charge and Its
Electric Charge and Its
Conservation
Charge comes in two
types positive and
types, positive and
negative; like charges
repel and opposite
repel and opposite
charges attract 16.1 Static Electricity; Electric Charge
and Its Conservation
and Its Conservation Electric charge is conserved – the
arithmetic sum of the total charge cannot
change in any interaction. 16.2 Electric Charge in the Atom
Atom:
Nucleus (small,
massive, positive
charge)
charge)
Electron cloud (large,
very low density,
negative charge) 16.2 Electric Charge in the Atom
Atom is electrically neutral.
Rubbing charges objects by moving electrons
from one to the other. 16.2 Electric Charge in the Atom
Polar molecule: neutral overall, but charge not
evenly distributed
evenly distributed 16.3 Insulators and Conductors
Conductor: Insulator: Charge flows freely Almost no charge flows Metals Most other materials Some materials are semiconductors. 16.4 Induced Charge; the Electroscope
Metal objects can be charged by conduction: 16.4 Induced Charge; the Electroscope
They can also be charged by induction: 16.4 Induced Charge; the Electroscope
Nonconductors won’t become charged by
conduction or induction but will experience
conduction or induction, but will experience
charge separation: 16.4 Induced Charge; the Electroscope
The electroscope
can be used for
can be used for
detecting charge: 16.4 Induced Charge; the Electroscope
The electroscope can be charged either by
conduction or by induction. 16.4 Induced Charge; the Electroscope
The charged electroscope can then be used to
determine the sign of an unknown charge.
determine the sign of an unknown charge. ConcepTest
ConcepTest 16.1a Electric Charge I
Two charged balls are
repelling each other as
they hang from the ceiling.
What can you say about
their charges? 1) one is positive, the other
is positive the other
is negative
2) both are positive
3) both are negative
4) both are positive or both
are negative ConcepTest
ConcepTest 16.1a Electric Charge I
Two charged balls are
repelling each other as
they hang from the ceiling.
What can you say about
their charges? 1) one is positive, the other
is positive the other
is negative
2) both are positive
3) both are negative
4) both are positive or both
are negative The fact that the balls repel each
ba
eac
other only can tell you that they
have the same charge but you do
charge,
not know the sign So they can
not know the sign. So they can
be either both positive or both
negative.
FollowFollowup: What does the picture look like if the two balls are oppositely
charged? ConcepTest 16.1b Electric Charge II
From the picture,
what can you
conclude about
the charges? 1)
1) have opposite charges 2) have the same charge 3) all have the same charge 4) one ball must be neutral (no charge) ConcepTest 16.1b Electric Charge II
From the picture,
what can you
conclude about
the charges? 1)
1) have opposite charges 2) have the same charge 3) all have the same charge 4) one ball must be neutral (no charge) The GREEN and PINK balls must
have the same charge, since they
repel each other The YELLOW
repel each other. The YELLOW
ball also repels the GREEN, so it
must also have the same charge
as the GREEN (and the PINK). ConcepTest
ConcepTest 16.2a Conductors I
A metal ball hangs from the ceiling 1) positive by an insulating thread. The ball is 2) negative attracted
attracted to a positivecharged rod
positive
rod 3) neutral
neutral held near the ball. The charge of 4) positive or neutral the ball must be:
the ball must be: 5) negative or neutral
negative or neutral ConcepTest 16.2a Conductors I
A metal ball hangs from the ceiling 1)
1) positive by an insulating thread. The ball is 2) negative attracted
attracted to a positivecharged rod
positive
rod 3) neutral
neutral held near the ball. The charge of 4) positive or neutral the ball must be:
the ball must be: 5) negative or neutral
negative or neutral Clearly, the ball will be attracted if its
the ball will be attracted if its
charge is negative However, even if
negative.
the ball is neutral the charges in the
neutral,
ball can be separated by induction
(polarization), leading to a net
attraction.
tt remember
the ball is
the ball is a
conductor! FollowFollowup: What happens if the metal ball is replaced by a plastic ball?
ball? ConcepTest 16.2b Conductors II
Two neutral conductors are connected
Two neutral conductors are connected 1) 0 0 by a wire and a charged rod is brought 2) + – 3) – + 4) + + 5) – – near, but does not touch. The wire is
taken away, and then the charged rod
is removed. What are the charges on
the conductors?
the conductors? 0 0 ? ? ConcepTest 16.2b Conductors II
Two neutral conductors are connected
Two neutral conductors are connected 1) 0 0 by a wire and a charged rod is brought 2) + – 3) – + 4) + + 5) – – near, but does not touch. The wire is
taken away, and then the charged rod
is removed. What are the charges on
the conductors?
conductors? While the conductors are connected, positive 0 0 ? ? charge will flow from the blue to the green
ball due to polarization Once disconnected,
polarization.
th
the charges will remain on the separate
th
conductors even when the rod is removed.
FollowFollowup: What will happen when the
conductors are reconnected with a wire? 16.5 Coulomb’s Law
Experiment shows that the electric force
between two charges is proportional to the
between two charges is proportional to the
product of the charges and inversely
proportional to the distance between them. 16.5 Coulomb’s Law
Coulomb’s law:
(161) This equation gives the magnitude of
the force.
the force. 16.5 Coulomb’s Law
The force is along the line connecting the
li
charges, and is attractive if the charges are
opposite and repulsive if they are the same
opposite, and repulsive if they are the same. 16.5 Coulomb’s Law
Unit of charge: coulomb, C
The proportionality constant in Coulomb’s
law is then: Charges produced by rubbing are
typically around a microcoulomb: 16.5 Coulomb’s Law
Charge on the electron: Electric charge is quantized in units
charge is quantized in units
of the electron charge. 16.5 Coulomb’s Law
The proportionality constant k can also be
written in terms of
written in terms of
, the permittivity of free
the permittivity of free
space: (162) ConcepTest 16.3a Coulomb’s Law I
If F1 = 3N, what is the
what
magnitude of the force F2? 1)
1) 1.0 N
2) 1.5 N
3) 2.0 N F1 = 3N Q Q F2 = ? 4) 3.0 N
5) 6.0 N ConcepTest 16.3a Coulomb’s Law I
If F1 = 3N, what is the
what
magnitude of the force F2? 1)
1) 1.0 N
2) 1.5 N
3) 2.0 N F1 = 3N Q Q F2 = ? 4) 3.0 N
5) 6.0 N The force F2 must have the same magnitude as F1. This is
same
due to the fact that the form of Coulomb’s Law is totally
symmetric with respect to the two charges involved. The
force of one on the other of a pair is the same as the reverse
reverse.
Note that this sounds suspiciously like Newton’s 3rd Law!!
Note that this sounds suspiciously like Newton’s 3rd Law!! ConcepTest
ConcepTest 16.3b Coulomb’s Law II
F1 = 3N
3N Q Q F2 = ? 2) 3.0 N If we increase one charge to 4Q
If we increase one charge to 4Q,
what
what is the magnitude of F1?
F1 = ? 4Q Q 1) 3/4 N F2 = ? 3) 12 N
12
4) 16 N
5) 48 N
48 ConcepTest
ConcepTest 16.3b Coulomb’s Law II
F1 = 3N
3N Q Q F2 = ? 2) 3.0 N If we increase one charge to 4Q
If we increase one charge to 4Q,
what
what is the magnitude of F1?
F1 = ? 4Q Q 1) 3/4 N F2 = ? 3) 12 N
12
4) 16 N
5) 48 N
48 Originally we had:
F1 = k(Q)(Q)/r2 = 3 N
Now we have:
F1 = k(4Q)(Q)/r2
which is 4 times bigger than before.
is times bigger
before
FollowFollowup: Now what is the magnitude of F2? MCAT
MCAT Test 161
16The force between two charges
force between two charges 1) 9 F separated by a distance d is F. If 2) 3 F the charges are pulled apart to a 3)
3) F distance 3d, what is the force on 4) 1/3 F each charge? 5) 1/9 F
1/9 F F Q Q
d ? ? Q Q
3d ConcepTest 16.3c Coulomb’s Law III
The force between two charges
force between two charges 1)
1) 9 F separated by a distance d is F. If 2) 3 F the charges are pulled apart to a 3)
3) F distance 3d, what is the force on 4) 1/3 F each charge? 5) 1/9 F
1/9
F Originally we had: F Q Q Fbefore = k(Q)(Q)/d2 = F
Now we have:
Fafter = k(Q)(Q)/(3d)2 = 1/9 F d
? ? Q Q
3d FollowFollowup: What is the force if the original distance is halved? 16.5 Coulomb’s Law
Coulomb’s law strictly applies only to point charges.
Superposition: for multiple point charges, the forces
on each charge from every other charge can be
on each charge from every other charge can be
calculated and then added as vectors. 16.6 Solving Problems Involving
Coulomb Law and Vectors
Coulomb’s Law and Vectors The net force on a charge is the vector
sum of all the forces acting on it. 16.6 Solving Problems Involving
Coulomb’s Law and Vectors
Vector addition review: Particles of charge +75, +48, and 85 μC are
placed in a line (Fig. 16–49). The center one is
0.35 m from each of the others. Calculate the
from each of the others Calculate the
net force on each charge due to the other two. Let the right be the positive direction on the line
of charges. Use the fact that like charges repel
and unlike charges attract to determine the
di
direction of the forces. In the following
th
th
expressions, k = 8.988 ×10 N ⋅ m C .
9 2 2 ( 75μ C ) ( 48μ C ) ( 75μ C ) ( 85μ C )
F+75 = −k
+k
= −147.2 N ≈ −1.5 ×102 N
2
2
0.35 m )
0.70 m )
(
(
( 75μ C ) ( 48μ C ) ( 48μ C ) ( 85μ C )
+k
= 563.5 N ≈ 5.6 ×102 N
F+48 = k
2
2
0.35 m )
0.35 m )
(
(
( 85μ C ) ( 75μ C ) ( 85μ C ) ( 48μ C )
−k
= −416.3 N ≈ −4.2 ×102 N
F−85 = −k
( 0.70 m )2
( 0.35 m )2 1 2 ConcepTest 16.6 Forces in 2D 3 Which of the arrows best 4 represents the direction
of the net force on charge
of the net force on charge d
+2Q +Q +Q due to the other two
charges? d
+4Q 5 1 2 ConcepTest 16.6 Forces in 2D 3 Which of the arrows best 4 represents the direction
of the net force on charge
of the net force on charge d
+2Q +Q 5 +Q due to the other two
d charges? +4Q The charge +2Q repels +Q towards
charge +2
the right. The charge +4Q repels +Q
upwards, but with a stronger force.
but with stronger force.
Therefore, the net force is up and to +2Q the right, but mostly up
up.
FollowFollowup: What happens if the
yellow charge would be +3Q? +4Q 16.7 The Electric Field
The electric field is the
force on small charge,
force on a small charge,
divided by the charge: (163) 16.7 The Electric Field
For a point charge:
(164a) (164b) 16.7 The Electric Field
Force on a point charge in an electric field:
(165) Superposition principle for electric fields: 16.7 The Electric Field
Problem solving in electrostatics: electric
forces and electric fields
forces and electric fields
1. Draw a diagram; show all charges, with
signs and electric fields and forces with
signs, and electric fields and forces with
directions
2. Calculate forces using Coulomb’s law
forces using Coulomb’s law
3. Add forces vectorially to get result ConcepTest 16.7 Electric Field
You are sitting certain distance from
You are sitting a certain distance from
a point charge, and you measure an
point
electric field of E0. If the charge is
doubled and your distance from the
di
th
charge is also doubled, what is the
doubled,
electric field strength now? (1)
(1) 4 E0
(2) 2 E0
(3) E0
(4) 1/2 E0
(5) 1/4 E0
4 ConcepTest 16.7 Electric Field
You are sitting certain distance from
You are sitting a certain distance from
a point charge, and you measure an
point
electric field of E0. If the charge is
doubled and your distance from the
di
th
charge is also doubled, what is the
doubled,
electric field strength now? (1)
(1) 4 E0
(2) 2 E0
(3) E0
(4) 1/2 E0
(5) 1/4 E0
4 Remember that the electric field is: E = kQ/r2.
that the electric field is:
kQ/r
Doubling the charge puts a factor of 2 in the
numerator, but doubling the distance puts a factor
of
of 4 in the denominator, because it is distance
squared!! Overall, that gives us a factor of 1/2
1/2. FollowFollowup: If your distance is doubled, what must you do to
the charge to maintain the same E field at your new position? ConcepTest 16.8a Field and Force I
Between the red and the
the red
the
blue charge, which of
them experiences the
greater electric field due
to the green charge? +2 +1
d 1)
1) +1 2) +2 3) the same for both +1 +1
d ConcepTest 16.8a Field and Force I
Between the red and the
the red
the
blue charge, which of
them experiences the
greater electric field due
to the green charge? 1)
1) +1 2) +2 3) the same for both +2 +1
d Both charges feel the same electric
field due to the green charge because
they are at the same point in space
space! +1 +1
d Q
E=k 2
r ConcepTest 16.8b Field and Force II
Between the red and the
the red
the
blue charge, which of
them experiences the
greater electric force due
to the green charge? +2 +1
d 1)
1) +1 2) +2 3) the same for both +1 +1
d ConcepTest 16.8b Field and Force II
Field
Between the red and the
the red
the
blue charge, which of
them experiences the
greater electric force due
to the green charge? +2 +1 1)
1) +1 2) +2 3) the same for both +1 d The electric field is the same for both charges,
but the force on a given charge also depends
on the magnitude of that specific charge
charge. +1
d F = qE ConcepTest 16.9a Superposition I
2 C
2 What is the electric field at 1 the center of the square?
3
5) E = 0 2 C 4 ConcepTest 16.9a Superposition I
2 C
2 White dot = 1 C 1 What is the electric field at the
3 center of the square?
5) E = 0 2 C For the upper charge, the E field vector at the center
the upper charge the
vector at the center
of the square points towards that charge. For the
lower charge, the same thing is true. Then the vector
sum of these two E field vectors points to the left
left. FollowFollowup: What if the lower charge was +2 C?
What if both charges were +2 C? 4 ConcepTest 16.9b Superposition II
2 C 2 C
2 What is the electric field at 1 the center of the square?
3
5) E = 0 2 C 4
2 C ConcepTest 16.9b Superposition II
2 C 2 C
2 What is the electric field at 1 the center of the square?
3
5) E = 0 2 C The four E field vectors all point outwards
from the center of the square toward their
respective charges. Because they are all
respective charges. Because they are all
equal, the net E field is zero at the center
center!!
FollowFollowup: What if the upper two charges were +2 C?
What if the righthand charges were +2 C?
right 4
2 C ConcepTest 16.11 Uniform Electric Field
In a uniform electric field in empty 1) 12 N space, a 4 C charge is placed and it 2) 8 N feels an electrical force of 12 N. If 3) 24
3) 24 N this charge is removed and a 6 C
charge is placed at that point
instead, what force will it feel?
it Q 4) no force
5) 18 N
18 ConcepTest 16.11 Uniform Electric Field
In a uniform electric field in empty 1)
1) 12 N space, a 4 C charge is placed and it 2) 8 N feels an electrical force of 12 N. If 3) 24 N
24 this charge is removed and a 6 C
charge is placed at that point
instead, what force will it feel?
it 4) no force
5) 18 N
18 Since the 4 C charge feels a force, there must
the
charge feels force, there must
be an electric field present, with magnitude:
E = F / q = 12 N / 4 C = 3 N/C
12
Once the 4 C charge is replaced with a 6 C Q charge, this new charge will feel a force of:
F = q E = (6 C)(3 N/C) = 18 N
FollowFollowup: What if the charge is placed at a different position in the field?
different 16.8 Field Lines
The electric field can be represented by field
lines. These lines start on a positive charge
and end on a negative charge. 16.8 Field Lines The number of field lines starting (ending)
on a positive (negative) charge is
proportional to the magnitude of the charge
proportional to the magnitude of the charge. The electric field is stronger where the field
lines are closer together. 16.8 Field Lines
Electric dipole: two equal charges, opposite in
sign:
sign: 16.8 Field Lines
The electric field between
two closely spaced,
oppositely charged parallel
plates is constant
plates is constant. MCAT Test 161
16An electron is located between
An electron is located between
a pair of oppositely charged
pair
parallel plates. As the electron
parallel plates. As the electron
approaches
approaches the positively
charged plate, the kinetic
energy
energy of the electron?  A: Increases
B: Decreases
C: Remains the same +
+
+
+
+
+ 16.8 Field Lines
Summary of field lines:
1. Field lines indicate the direction of the
lines indicate the direction of the
field; the field is tangent to the line.
2. The magnitude of the field is proportional
magnitude of the field is proportional
to the density of the lines.
3. Field lines start on positive charges and
li
end on negative charges; the number is
proportional to the magnitude of the
proportional to the magnitude of the
charge. 16.9 Electric Fields and Conductors
The static electric field inside a conductor is
zero – if it were not, the charges would move. The net charge on a conductor is on its
surface
surface. 16.9 Electric Fields and Conductors
The electric field is
perpendicular to the
perpendicular to the
surface of a conductor –
again, if it were not,
charges would move. What are the magnitude and direction of the
electric field at a point midway between a 8.0
μC and a +7.0 μC charge 8.0 cm apart? Assume
cm apart? Assume
no other charges are nearby. +7.0 8.0
8cm What are the magnitude and direction of the
electric field at a point midway between a 8.0
μC and a +7.0 μC charge 8.0 cm apart? Assume
cm apart? Assume
no other charges are nearby. +7.0 8.0
8cm The electric field due to the negative charge will
point toward the negative charge, and the
electric field due to the positive charge will
point away from the positive charge. Thus both
fields point in the same direction, towards the
negative charge and so can be added
negative charge, and so can be added.
The direction is towards the negative charge .
E = E1 + E2 = k
= ( Q1
2
1 r +k Q2
r 4 8.988 ×109 N ⋅ m2 C2 (8.0 ×10 m)
−2 2 =k 2
2 ) ( Q1 ( d / 2) 2 +k Q2 ( d / 2) 2 = ) 4k
d 2 (Q 1 + Q2 ) 8.0 ×10−6 C + 7.0 ×10−6 C = 8.4 ×107 N C Summary of Chapter 16
• Two kinds of electric charge – positive and
negative
ti
• Charge is conserved
• Charge on electron: • Conductors: electrons free to move
• Insulators: nonconductors Summary of Chapter 16
• Charge is quantized in units of e
• Objects can be charged by conduction or
induction
• Coulomb’s law: • Electric field is force per unit charge: Summary of Chapter 16
• Electric field of a point charge: • Electric field can be represented by electric
field lines
field lines
• Static electric field inside conductor is zero;
surface field is perpendicular to surface
surface field is perpendicular to surface Demonstration 16.1
One person should touch the ball on top of the
electroscope with their finger and hold their finger
th
there, while another person brings a charged rubber
rod near, but not touching the leaves of the
electroscope. While the rubber rod is near the
electroscope, the first person should remove their
finger from the top of the electroscope. Then remove
the rubber rod from near the electroscope
the rubber rod from near the electroscope. Demonstration 16.1
One person should touch the ball on top of the
electroscope with their finger and hold their finger
there, while another person brings charged rubber
there, while another person brings a charged rubber
rod near, but not touching the leaves of the
electroscope. Demonstration 16.1
One person should touch the ball on top of the
electroscope with the...
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 Fall '09
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 Charge, Electric charge

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