Download Chapter 21 - Chabot College

Chapter 21
Electric Charge and
Electric Field
PowerPoint® Lectures for
University Physics, Thirteenth Edition
– Hugh D. Young and Roger A. Freedman
Lectures by Wayne Anderson
Copyright © 2012 Pearson Education Inc.
Introduction
• Water makes life possible
as a solvent for biological
molecules. What electrical
properties allow it to do
this?
• We now begin our study
of electromagnetism, one
of the four fundamental
forces in Nature.
• We start with electric
charge and electric fields.
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Goals for Chapter 21
• Study electric charge & charge conservation
• Learn how objects become charged
• Calculate electric force between objects using
Coulomb’s law
F = k|q1q2|/r2 = (1/4π0)|q1q2|/r2
• Learn distinction between electric force and
electric field
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Goals for Chapter 21
• Calculate the electric
field due to many
charges
• Visualize and interpret
electric fields
• Calculate the
properties of electric
dipoles
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Goals for Chapter 21
• Be able to solve this kind of problem: (page 715)
Charge Q is distributed
uniformly around a semicircle
of radius a.
y
Charge Q
a
x
P
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What is the magnitude and
direction of the resulting E field
at point P, at the center of
curvature of the semicircle?
Physics from 4A you will need to know!
• Forces as vectors
• Establish coordinate frame
• Break into components Fx, Fy, Fz
• Add like components!
• Resolve net vector
• Answers must have three things!
1. Magnitude
2. Direction
3. UNITS
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Physics from 4A you will need to know!
• Chapter 4: Forces
• “Links in a Chain” page 127
• Exercises/Problems 4.2, 37, 54, 62* (calculus)
• Chapter 5: Applications
• Section 5.2 (Dynamics) ex 5.10, 5.11, 5.12
• Section 5.4 (Circular) ex 5.20, 5.21
• “In a Rotating Cone” page 162
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Math from 4A you will need to know!
• Integration of continuous variables
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Electric charge
• Two positive or two
negative charges repel
each other.
A positive charge and a
negative charge attract
each other.
• Check out:
http://www.youtube.com/wat
ch?v=45AAIl9_lsc
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Electric charge
• Two positive or two
negative charges repel
each other.
A positive charge and a
negative charge attract
each other.
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Electric charge
• Two positive or two
negative charges repel
each other.
A positive charge and
a negative charge
attract each other.
• Check out Balloons in
PhET simulations
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Electric charge and the structure of matter
• The particles of the
atom are the negative
electron, the positive
proton, and the
uncharged neutron.
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You should know this already: Atoms and ions
• A neutral atom has the same number of protons as electrons.
• A positive ion is an atom with one or more electrons removed.
A negative ion has gained one or more electrons.
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You should know this already: Atoms and ions
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Conservation of charge
• Proton & electron have same
magnitude charge.
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Conservation of charge
• Proton & electron have same
magnitude charge.
• All observable charge is
quantized in this unit.
“½ e”
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Conservation of charge
• Proton & electron have same
magnitude charge.
• Universal principle of charge
conservation states algebraic
sum of all electric charges in
any closed system is constant.
+3e – 5e +12e – 43e = -33 e
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Conductors and insulators
• A conductor permits the
easy movement of charge
through it. An insulator
does not.
• Most metals are good
conductors, while most
nonmetals are insulators.
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Conductors and insulators
• A conductor permits the easy movement of charge
through it.
• An insulator does not.
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Conductors and insulators
• Semiconductors are
intermediate in their
properties between good
conductors and good
insulators.
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Charging by induction
• Start with UNCHARGED conducting ball…
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Charging by induction
• Bring a negatively charged rod near – but not touching.
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Charging by induction
• Bring a negatively charged rod near – but not touching.
• The negative rod is able to charge the metal ball without losing
any of its own charge.
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Charging by induction
• Now connect the conductor to the ground (or neutral “sink”)
• What happens?
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Charging by induction
• Now connect the conductor to the ground (or neutral “sink”)
• Conductor allows electrons to flow from ball to ground…
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Charging by induction
• Connect the conductor to the ground (or neutral “sink”)
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Charging by induction
• The negative rod is able to charge the metal ball without losing
any of its own charge.
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Electric forces on uncharged objects
• The charge within an insulator can shift slightly. As a result, an
electric force *can* be exerted upon a neutral object.
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Electrostatic painting
• Induced positive charge on the metal object attracts the
negatively charged paint droplets. Check out
http://www.youtube.com/watch?feature=endscreen&v=zTwkJBtCcBA&NR=1
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Q21.1
When you rub a plastic rod with fur, the plastic rod becomes
negatively charged and the fur becomes positively charged.
As a consequence of rubbing the rod with the fur,
A. the rod and fur both gain mass.
B. the rod and fur both lose mass.
C. the rod gains mass and the fur loses mass.
D. the rod loses mass and the fur gains mass.
E. none of the above
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A21.1
When you rub a plastic rod with fur, the plastic rod becomes
negatively charged and the fur becomes positively charged.
As a consequence of rubbing the rod with the fur,
A. the rod and fur both gain mass.
B. the rod and fur both lose mass.
C. the rod gains mass and the fur loses mass.
D. the rod loses mass and the fur gains mass.
E. none of the above
Copyright © 2012 Pearson Education Inc.
Q21.2
A positively charged piece of plastic exerts an attractive force
on an electrically neutral piece of paper. This is because
A. electrons are less massive than atomic nuclei.
B. the electric force between charged particles
decreases with increasing distance.
C. an atomic nucleus occupies only a small part of
the volume of an atom.
D. a typical atom has many electrons but only one
nucleus.
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A21.2
A positively charged piece of plastic exerts an attractive force
on an electrically neutral piece of paper. This is because
A. electrons are less massive than atomic
nuclei.
B. the electric force between charged particles
decreases with increasing distance.
C. an atomic nucleus occupies only a small part
of the volume of an atom.
D. a typical atom has many electrons but only
one nucleus.
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Coulomb’s law – Electric FORCE
• The magnitude of electric
force between two point
charges is directly
proportional to the
product of their charges
and
inversely proportional
to the square of the
distance between them.
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Coulomb’s law
• Mathematically:
F = k|q1q2|/r2
= (1/4π0)|q1q2|/r2
• A VECTOR
•
Magnitude
•
Direction
•
Units
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Coulomb’s law
• Mathematically:
|F| = k|q1q2|/r2
• “k” = 9 x 109 Newton
meter2/Coulomb2
• “k” = 9 x 109 Nm2/C2
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Coulomb’s law
• Mathematically:
|F| = k|q1q2|/r2
= (1/4π0)|q1q2|/r2
• 0 = 8.85 x 10 – 12 C2/Nm2
0 = “Electric Permittivity of Free Space”
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Measuring the electric force between point charges
Example 21.1 compares the
electric and gravitational
forces.
An alpha particle has mass m = 6.64 x 10-27 kg
and charge q = +2e = 3.2 x 10-19 C.
Compare the magnitude of the electric repulsion
between two alpha particles and their
gravitational attraction
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Measuring the electric force between point charges
DRAW the VECTORS!!
An alpha particle has mass m = 6.64 x 10-27 kg
and charge q = +2e = 3.2 x 10-19 C.
Compare the magnitude of the electric repulsion
between two alpha particles and their
gravitational attraction
Copyright © 2012 Pearson Education Inc.
Measuring the electric force between point charges
An alpha particle has mass
m = 6.64 x 10-27 kg and
charge q = +2e =
3.2 x 10-19 C.
Fe/Fg = ?
Remember Fg = Gm1m2/r2 & G= 6.67 x 10-11 Nm2/kg2
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Measuring the electric force between point charges
An alpha particle has mass m
= 6.64 x 10-27 kg and charge
q = +2e = 3.2 x 10-19 C.
Fe/Fg = 3.1 x 1035!!!!
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Force between charges along a line
• Example 21.2 for two charges:
Two point charges, q1 = +25nC, and q2 = -75 nC,
separated by r = 3.0 cm.
What is the Force of q1 on q2?
What is the Force of q2 on q1?
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Force between charges along a line
• Example 21.2 for two charges:
Two point charges, q1 = +25nC, and q2 = -75 nC,
separated by r = 3.0 cm.
What is the Force of q1 on q2?
Step 1: Force is a vector – create a coordinate
system FIRST!
x
Copyright © 2012 Pearson Education Inc.
Force between charges along a line
• Example 21.2 for two charges:
Two point charges, q1 = +25nC, and q2 = -75 nC,
separated by r = 3.0 cm. What is the Force of q1 on
q2? What is the force of q2 on q1?
Copyright © 2012 Pearson Education Inc.
Force between charges along a line
• Example 21.2 for two charges:
Two point charges, q1 = +25nC, and q2 = -75 nC,
separated by r = 3.0 cm. What is the Force of q1 on
q2? What is the force of q2 on q1?
Copyright © 2012 Pearson Education Inc.
Force between charges along a line
• Example 21.2 for two charges:
Two point charges, q1 = +25nC, and q2 = -75 nC,
separated by r = 3.0 cm.
What is the Force of q1 on q2?
F of q1 on q2 = F12 = 0.019N
F12
x
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<-x>
Force between charges along a line
• Example 21.3 for three charges:
Two point charges, q1 = +1.0nC at x = +2.0 cm, and
q2 = -3.0 nC at x = +4.0 cm. What is the Force of q1
and q2 on q3 = + 5.0 nC at x = 0?
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Force between charges along a line
• Example 21.3 for three charges:
Two point charges, q1 = +1.0nC at x = +2.0 cm, and
q2 = -3.0 nC at x = +4.0 cm. What is the Force of q1
and q2 on q3 = + 5.0 nC at x = 0?
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Vector addition of electric forces
• Example 21.4 shows that we must use vector addition when
adding electric forces.
Two equal positive charges, q1 = q2 = +2.0mC are
located at x=0, y = 0.30 m and x=0, y = -.30 m
respectively.
What is the Force of q1 and q2 on Q = + 5.0 mC
at x = 0.40 m, y = 0?
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Vector addition of electric forces
• Example 21.4 shows that we must use vector addition when
adding electric forces.
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Vector addition of electric forces
• Example 21.4 shows that we must use vector addition when
adding electric forces.
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Vector addition of electric forces
• Example 21.4 shows that we must use vector addition when
adding electric forces.
Copyright © 2012 Pearson Education Inc.
Q21.3
Three point charges lie at the
vertices of an equilateral triangle as
shown. All three charges have the
same magnitude, but charges #1
and #2 are positive (+q) and charge
#3 is negative (–q).
The net electric force that charges
#2 and #3 exert on charge #1 is in
Charge #2
+q
Charge #1
+q
y
–q
x
A. the +x-direction.
B. the –x-direction.
C. the +y-direction.
D. the –y-direction.
E. none of the above
© 2012 Pearson Education, Inc.
Charge #3
A21.3
Three point charges lie at the
vertices of an equilateral triangle as
shown. All three charges have the
same magnitude, but charges #1
and #2 are positive (+q) and charge
#3 is negative (–q).
The net electric force that charges
#2 and #3 exert on charge #1 is in
Charge #2
+q
Charge #1
+q
y
–q
x
A. the +x-direction.
B. the –x-direction.
C. the +y-direction.
D. the –y-direction.
E. none of the above
© 2012 Pearson Education, Inc.
Charge #3
Q21.4
Three point charges lie at the
vertices of an equilateral triangle as
shown. All three charges have the
same magnitude, but charge #1 is
positive (+q) and charges #2 and #3
are negative (–q).
The net electric force that charges
#2 and #3 exert on charge #1 is in
Charge #2
–q
Charge #1
+q
y
–q
x
A. the +x-direction.
B. the –x-direction.
C. the +y-direction.
D. the –y-direction.
E. none of the above
© 2012 Pearson Education, Inc.
Charge #3
A21.4
Three point charges lie at the
vertices of an equilateral triangle as
shown. All three charges have the
same magnitude, but charge #1 is
positive (+q) and charges #2 and #3
are negative (–q).
The net electric force that charges
#2 and #3 exert on charge #1 is in
Charge #2
–q
Charge #1
+q
y
–q
x
A. the +x-direction.
B. the –x-direction.
C. the +y-direction.
D. the –y-direction.
E. none of the above
© 2012 Pearson Education, Inc.
Charge #3
Electric field
• A charged body produces an electric field in the space around it
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Electric field
• We use a small test charge q0 to find out if an electric field is
present.
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Electric field
• We use a small test charge q0 to find out if an electric field is
present.
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Definition of the electric field
• E fields are VECTOR fields – and solutions to
problems require magnitude, direction, and units.
Copyright © 2012 Pearson Education Inc.
Definition of the electric field
• E fields are VECTOR fields – and solutions to
problems require magnitude, direction, and units.
Copyright © 2012 Pearson Education Inc.
Definition of the electric field
• E fields are VECTOR fields – and solutions to
problems require magnitude, direction, and units.
– Need Coordinate System for direction!
– Need Units Force/Charge = Newtons/Coulomb = N/C
– E = (kq0q/r2)/q0 = (kq/r2) (-r direction!) N/C
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Electric field of a point charge
• E fields from positive charges point AWAY from the charge
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Electric field of a point charge
• E fields point TOWARDS a negative charge:
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Electric-field vector of a point charge
• Example 21.6 - the vector
nature of the electric field.
• Charge of -8.0 nC at
origin.
What is E field at P =
(Px,Py) = (1.2, -1.6)?
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Electric-field vector of a point charge
• What is E field at P =
(Px,Py) = (1.2, -1.6)?
• E = -11N/C x + 14 N/C y
(2 sig figs)
• |E| = 18 N/C
in direction
q = arctan Ey/Ex
= 127 degrees from +x
or 53˚ from –x axis
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Electric-field vector of a point charge
• What is E field at P =
(Px,Py) = (1.2, -1.6)?
• E = -11N/C x + 14 N/C y
• |E| = 18 N/C
in direction
q = arctan Ey/Ex
= -53 degrees
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Q21.5
A positive point charge +Q is released from rest in an electric
field. At any later time, the velocity of the point charge
A. is in the direction of the electric field at the position
of the point charge.
B. is directly opposite the direction of the electric field at
the position of the point charge.
C. is perpendicular to the direction of the electric field at
the position of the point charge.
D. is zero.
E. not enough information given to decide
© 2012 Pearson Education, Inc.
A21.5
A positive point charge +Q is released from rest in an electric
field. At any later time, the velocity of the point charge
A. is in the direction of the electric field at the position
of the point charge.
B. is directly opposite the direction of the electric field at
the position of the point charge.
C. is perpendicular to the direction of the electric field at
the position of the point charge.
D. is zero.
E. none of the above are true.
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Electron in a uniform field
• Example 21.7: Determine force on a charge in a known
uniform electric field.
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Electron in a uniform field
• Plates 1.0 cm apart, connected to 100 V battery creating
a uniform E field of 100V/0.01 m = 10,000 N/C
• What’s a VOLT?
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Electron in a uniform field
• Plates 1.0 cm apart, connected to
100 V battery creating a uniform
E field of 100V/0.01 m = 10,000
N/C
• What’s a VOLT?
– A unit of potential
energy/charge (Joules/Coulomb)
– Like “electrical water
pressure”
– 1 Volt/meter =
1 Newton/Coulomb (E field)
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Electron in a uniform field
• Plates 1.0 cm apart, connected to 100 V battery creating
a uniform E field of 100V/0.01 m = 10,000 N/C
• Electron released from rest; what is acceleration?
Final velocity? Total KE? Time to travel 1.0 cm?
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Superposition of electric fields
• The total electric field at a point is the vector sum of the fields due
to all the charges present.
• Example 21.8 for an electric dipole.
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Q21.6
Two point charges and a point P lie
at the vertices of an equilateral
triangle as shown. Both point
charges have the same magnitude q
but opposite signs. There is nothing
at point P.
The net electric field that charges
#1 and #2 produce at point P is in
Charge #1
–q
P
y
+q
x
A. the +x-direction.
B. the –x-direction.
C. the +y-direction.
D. the –y-direction.
E. none of the above
© 2012 Pearson Education, Inc.
Charge #2
A21.6
Two point charges and a point P lie
at the vertices of an equilateral
triangle as shown. Both point
charges have the same magnitude q
but opposite signs. There is nothing
at point P.
The net electric field that charges
#1 and #2 produce at point P is in
Charge #1
–q
P
y
+q
x
A. the +x-direction.
B. the –x-direction.
C. the +y-direction.
D. the –y-direction.
E. none of the above
© 2012 Pearson Education, Inc.
Charge #2
Q21.7
Two point charges and a point P lie
at the vertices of an equilateral
triangle as shown. Both point
charges have the same negative
charge (–q). There is nothing at
point P.
The net electric field that charges
#1 and #2 produce at point P is in
Charge #1
–q
P
y
–q
x
A. the +x-direction.
B. the –x-direction.
C. the +y-direction.
D. the –y-direction.
E. none of the above
© 2012 Pearson Education, Inc.
Charge #2
A21.7
Two point charges and a point P lie
at the vertices of an equilateral
triangle as shown. Both point
charges have the same negative
charge (–q). There is nothing at
point P.
The net electric field that charges
#1 and #2 produce at point P is in
Charge #1
–q
P
y
–q
x
A. the +x-direction.
B. the –x-direction.
C. the +y-direction.
D. the –y-direction.
E. none of the above
© 2012 Pearson Education, Inc.
Charge #2
Field of a charged line segment Example 21.10
• Line of charge => use l as Charge/Meter = Q/2a
• Set up dQ = l dy; find field dEx and dEy at P (in x and y
separately) from kdQ/r2
• Integrate from y = -a to +a (Check out Youtube)
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Field of a charged line segment Example 21.10
• Find E = Ex only = kQ/x(x2+a2)½ (+x direction)
• For a >>x, E = k(Q/a)/x[(x/a)2+1)]½ ~ k(Q/a)/x ~ 2k l/ x
• So for LONG wire, field nearby goes as 1/r…
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Q21.9
Positive charge is uniformly
distributed around a semicircle.
The electric field that this
charge produces at the center
of curvature P is in
A. the +x-direction.
B. the –x-direction.
C. the +y-direction.
D. the –y-direction.
E. none of the above
© 2012 Pearson Education, Inc.
A21.9
Positive charge is uniformly
distributed around a semicircle.
The electric field that this
charge produces at the center
of curvature P is in
A. the +x-direction.
B. the –x-direction.
C. the +y-direction.
D. the –y-direction.
E. none of the above
© 2012 Pearson Education, Inc.
Field of a ring of charge
• Example 21.9 - a uniform ring of charge.
• Any continuous charge distribution – INTEGRALS!
• Always start with a small “dQ”, and calculate dF or dE
created from that dQ.
• Remember dF & dE are still VECTORS!
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Field of a ring of charge
• Example 21.9 - a uniform ring of charge.
• Use l as Charge/Meter for “charge density” [C/m]
• dQ = l (charge density) x ds (length of segment)
• dQ = l (Coulombs/meter) x ds (meters) = Coulombs
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Field of a uniformly charged disk
• Example 21.11 – Superposition of multiple rings!
• Surface of charge – use s = Charge/Area = Q/pR2
• Find dQ = s dA where dA = (2pr)dr
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Field of a uniformly charged disk
• Find dQ = s dA where dA = (2pr)dr
• dEx = [kdQ/(x2 + r2)]cos(q)
q
q
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Field of a uniformly charged disk
• Ex = k2sp[ 1 -
1
[(R2/x2) +1]
• At R >>x, Ex = k2sp = s/20
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] (+x direction)
Field of two oppositely charged infinite sheets
• Example 21.12- Superposition of two sheets
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Electric field lines
•
An electric field line is an imaginary line or curve
whose tangent at any point is the direction of the electric
field vector at that point.
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Electric field lines of point charges
•
Figure 21.28 below shows the electric field lines of a single point
charge and for two charges of opposite sign and of equal sign.
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Q21.8
The illustration shows the electric field lines due to three
point charges. The electric field is strongest
A. where the field lines are closest
together.
B. where the field lines are
farthest apart.
C. where adjacent field lines are
parallel.
D. none of the above
Copyright © 2012 Pearson Education Inc.
A21.8
The illustration shows the electric field lines due to three
point charges. The electric field is strongest
A. where the field lines are closest
together.
B. where the field lines are farthest
apart.
C. where adjacent field lines are
parallel.
D. none of the above
Copyright © 2012 Pearson Education Inc.
Electric dipoles
• An electric dipole is a pair
of point charges having
equal but opposite sign and
separated by a distance.
• Figure 21.30 at the right
illustrates the water
molecule, which forms an
electric dipole.
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Force and torque on a dipole
•
Figure below left shows the force on a dipole in an electric field.
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Electric field of a dipole
• Follow Example 21.14 using Figure 21.33.
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Laser printer
• A laser printer makes use of forces between
charged bodies.
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