General Physics II

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General Physics II
ELECTRICITY AND MAGNETISM
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• General Physics II Electricity Magnetism
• Course Description
• Coulomb's law, the electrostatic field,
  Gausss Law, the electrostatic potential,
  capacitance and dielectrics, electric current,
  resistance and electromotive force, direct
  current circuits, magnetic field and magnetic
  forces, sources of magnetic fields, Ampere's Law,
  Faraday's Law, induction and Maxwell's equations.

II. Course Objectives

• To provide a foundation in physics necessary for
  further study in science, engineering and
  technology.
• To provide an appreciation of the nature of
  physics, its methods and its goals.
• To contribute to the development of the student's
  thinking process through the understanding of the
  theory and application of this knowledge to the
  solution of practical problems.

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Textbook the class notes beside the following
textbooks

1. Physics for Scientists and Engineers, Raymond A.
   Serway, 6th Edition

1. University Physics, Sears, Zemansky and Young

Course Outline

• Charge and Matter Charge and conservation of
  charges, Material and charge, electric forces,
  Coulombs Law
• The Electric Field Electric field, The lines of
  forces, Electric dipole, Continuous charge
  distribution, Effect of electric field on point
  charge, Millikens Experiment.
• Electric Flux and Gausss Law Flux of electric
  field, Gausss Law and its application.

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• Electric Potential Definition of electric
  potential, Potential difference between two
  points, Calculation of electric potential,
  Electric potential energy, Electric field and
  potential.
• Capacitors Capacitor and capacitance, Parallel
  plate capacitor, Cylindrical capacitor, Spherical
  capacitor, Capacitors connections, Energy stored
  in capacitor, Effect of insulator inside a
  capacitor.
• Electric current and Ohms Law Electric
  current, Current density, Resistivity, Ohms Law,
  Electric power, Electromotive force and electric
  circuits, Kirchoffs Law, RC circuit
• The Magnetic Field Definition and introduction,
  Flux of magnetic field, Magnetic force, Hall
  effect, Torque on a current loop, charge in a
  magnetic field, Biot-Savart Law, Helmholtz coils,
  Amperes Law.

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• Faraday's Law Faraday's Law of Induction,
  Motional emf, Lenz's Law, Induced emf and
  Electric Fields, Generators and Motors/ Eddy
  Currents, Maxwell's Equations
• Inductance Self-Inductance, RL Circuits,
  Energy in a Magnetic Field, Mutual Inductance,
  Oscillations in an LC Circuit, The RLC Circuit.
• Alternating Current Circuits AC Sources,
  Resistors in an AC Circuit, Inductors in an AC
  Circuit, Capacitors in an AC Circuit, The RLC
  Series Circuit.
• Power in an AC Circuit, Resonance in a Series RLC
  Circuit, The Transformer and Power Transmission,
  Rectifiers and Filters.

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GRADING POLICY
Your grade will be judged on your performance in
Home work, Quizzes, tow tests and the Lab. Points
will be allocated to each of these in the
following manner
GRADING SCALE
Weight Grade Component
20 HW/Quizzes
20 Midterm Exam
60 Final Exam
100 Total
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Lecture I
Electrostatic
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Introduction

• Knowledge of electricity dates back to Greek
  antiquity (700 BC).
• Began with the realization that amber when rubbed
  with wool, attracts small objects.
• This phenomenon is not restricted to amber/wool
  but may occur whenever two non-conducting
  substances are rubbed together.

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Net Electrical Charge
Matters are made of atoms. An atom is basically
composed of three different components
electrons, protons, and neutrons. An electron can
be removed easily from an atom
Normally, an atom is electrically neutral, which
means that there are equal numbers of protons and
electrons. Positive charge of protons is balanced
by negative charge of electrons. It has no net
electrical charge.
When atoms gain or lose electrons, they are
called "ions." A positive ion is a cation that
misses electrons. A negative ion is an anion
that gains extra electrons.
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What is charge? Objects that exert electric
forces are said to have charge. Charge is the
source of electrical force. There are two kinds
of electrical charges, positive and negative.
Same charges ( and , or - and -) repel and
opposite charges ( and -) attract each other.
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• The Law of Conservation of Charge
• The Law of conservation of charge states that the
  net charge of an isolated system remains
  constant.

Charged Objects
When two objects are rubbed together, some
electrons from one object move to another object.
For example, when a plastic bar is rubbed with
fur, electrons will move from the fur to the
plastic stick. Therefore, plastic bar will be
negatively charged and the fur will be positively
charged.
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Quantization

• Robert Millikan found, in 1909, that
  charged objects may only have an integer multiple
  of a fundamental unit of charge.
• Charge is quantized.
• An object may have a charge e, or 2e, or 3e,
  etc but not say 1.5e.
• Proton has a charge 1e.
• Electron has a charge 1e.
• Some particles such a neutron have no (zero)
  charge.

"charge is quantized" in terms of an equation, we
say q n e
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Unit of Electrical Charge The Coulomb " C "
The symbol for electric charge is written q, - q
or Q. The unit of electric charge is coulomb "C".
The charge of one electron is equal to the charge
of one proton, which is 1.6 10-19 C. This
number is given a symbol "e".
Example How many electrons are there in 1 C of
charge?
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Insulators and Conductors( Material
classification)

• Materials/substances may be classified
• according to their capacity to carry or
  conduct electric charge
• Conductors are material in which electric charges
  move freely.
• Insulator are materials in which electrical
  charge do not move freely.
• Glass, Rubber are good insulators.
• Copper, aluminum, and silver are good conductors.
• Semiconductors are a third class of materials
  with electrical properties somewhere between
  those of insulators and conductors.
• Silicon and germanium are semiconductors used
  widely in the fabrication of electronic devices.

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• Example
• Identify substances or materials that can be
  classified as
• Conductors ?
• Insulators?

Why? is static electricity more apparent in
winter?
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Lecture 2
Coulombs law
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Coulombs Law

• Coulomb discovered in 1785 the fundamental law
• of electrical force between two stationary
  charged particles.
• An electric force has the following properties
• Inversely proportional to the square of the
  separation, r, between the particles, and is
  along a line joining them.
• Proportional to the product of the magnitudes of
  the charges q1 and q2 on the two particles.
• Attractive if the charges are of opposite sign
  and repulsive if the charges have the same sign.

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Coulombs Law (Mathematical Formulation)

• ke known as the Coulomb constant.
• Value of ke depends on the choice of units.
• SI units
• Force the Newton (N)
• Distance the meter (m).
• Charge the coulomb ( C).

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                             where eo  is
known as the Permittivity constant of free
space.  eo  8.85 x 10-12 C2/N.m2  
                                                  
                                              
Experimentally measurement ke 8.9875109
Nm2/C2. Reasonable approximate value ke
8.99109 Nm2/C2.
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Example the Coulomb constant unit
Then the Coulomb constant unit is
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The electrostatic force

• The electrostatic force is often called Coulomb
  force.
• It is a force (thus, a vector)
• a magnitude
• a direction.

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Example

• The electron and proton of a hydrogen atom are
  separated (on the average) by a distance of about
  5.3x10-11 m. Find the magnitude of the electric
  force that each particle exerts on the other.

q1 -1.60x10-19 C q2 1.60x10-19 C r 5.3x10-11 m
Attractive force with a magnitude of 8.2x10-8 N.
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Gravitational vs. Electrical Force
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Superposition of Forces
Q1
Q2
Q0
Q3
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• The net force on q3 is the vector sum of the
• forces F32 and F31.
• The magnitude of the forces F32 and F31 can
• calculated using Coulombs law.

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Example Two fixed charge, 1µC and -3µC are
separated by 10 cm. Where may a third charge
be located so that no force act on it
?
Solve the eq. to find d.
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Examination of the geometry of Figure leads to
If L is much larger than x (which is the case if
? is very small), we may neglect x/2 in the
denominator and write tan ? x/2L. This is
equivalent to approximating tan? by sin?. The
magnitude of the electrical force of one ball on
the other is
by Eq. When these two expressions are used in the
equation mg tan ? Fe, we obtain
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b) We solve x3 for the charge
Thus, the magnitude is
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Lecture 3

• The Electric Field

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Electric Field

• Suggests the notion of electrical field (first
  introduced by Michael Faraday (1791-1867).
• An electric field is said to exist in a region of
  space surrounding a charged object.
• If another charged object enters a region where
  an electrical field is present, it will be
  subject to an electrical force.

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Electric Field Electric Force
Consider a small charge q0 near a larger charge
Q. We define the electric field E at the location
of the small test charge as a ratio of the
electric force F acting on it and the test charge
q0
This is the field produced by the charge Q, not
by the charge q0
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Electric Field Direction

• The direction of E at a point is the
• direction of the electric force that would be
  exerted on a small positive test charge placed at
  that point.

E
E

• -
• - -
• - - -
• - -
• -

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Electric Field from a Point Charge
Suppose we have two charges, q and q0, separated
by a distance r. The electric force between the
two charges is
We can consider q0 to be a test charge, and
determine the electric field from charge q as
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• If q is ve, field at a given point is radially
  outward
• from q.

r
E
qo

q

• If q is -ve, field at a given point is radially
  inward from q.

r
-
qo
E
q
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Electric Field Lines
To visualize electric field patterns, one can
draw lines pointing in the direction of the
electric field vector at any point. These lines
are called electric field lines.

1. The electric field vector is tangent to the
   electric field lines at each point.
2. The number of lines per unit area through a
   surface perpendicular to the lines is
   proportional to the strength of the electric
   field in a given region.
3. No two field lines can cross each other . Why?

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The electric field lines for a point charge. (a)
For a positive point charge, the lines are
directed radially outward. (b) For a negative
point charge, the lines are directed radially
inward.
Note that the figures show only those field lines
that lie in the plane of the page.
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The electric field lines for two positive point
charges.
The electric field lines for two point charges of
equal magnitude and opposite sign (an electric
dipole)
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• Question
• Two charges q1 and q2, fixed along the
  x-axis as shown, produce an electric field E at
  the point (x,y)(0,d), which is the directed
  along the negative y-axis.
• Which of the following is true?
• Both charges are positive
• Both charges are negative
• The charges have opposite signs

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Electric Field from an Electric Dipole
A system of two oppositely charged point
particles is called an electric dipole. The
vector sum of the electric field from the two
charges gives the electric field of the dipole
(superposition principle). We have shown the
electric field lines from a dipole
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Example

• Two charges on the x-axis a distance
• d apart
• Put -q at x -d/2
• Put q at x d/2
• Calculate the electric field at a point P a
  distance x from the origin

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• Principle of superpositionThe electric field
  at any point x is the sum of the electric fields
  from q and -q
• Replacing r and r- we get

• This equation gives the electric field
  everywhere
• on the x-axis (except for x ?d/2)

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Example Electric Field Due to Two Point Charges

• Charge q17.00 mC is at the origin, and charge
  q2-10.00 mC is on the x axis, 0.300 m from the
  origin. Find the electric field at point P, which
  has coordinates (0,0.400) m.

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Example In Figure, determine the point (other
than infinity) at which the total electric field
is zero.
Solution The sum of two vectors can be zero only
if the two vectors have the same magnitude and
opposite directions.
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Motion of charge particles in a uniformelectric
field

• An electron moving horizontally passes
  between two horizontal planes, the upper plane
  charged negatively, and the lower positively. A
  uniform, upward-directed electric field exists in
  this region. This field exerts a force on the
  electron. Describe the motion of the electron in
  this region.

- - - - - - - - - - - -
- - - - - - - - - -
ve
-


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• Horizontally
• No electric field
• No force
• No acceleration
• Constant horizontal velocity

• Vertically
• Constant electric field
• Constant force
• Constant acceleration
• Vertical velocity increase linearly with time.

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- - - - - - - - - - - -
- - - - - - - - - -
-

• Conclusions
• The charge will follow a parabolic path downward.
• Motion similar to motion under gravitational
  field only except the downward acceleration is
  now larger.

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Phosphor Screen
This device is known as a cathode ray tube (CRT)
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Continuous Charge Distributions
Single charge
Single piece of a charge distribution
Discrete charges
Continuous charge distribution
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Cartesian
Polar
Line charge
Surface charge
Volume charge
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Example Electric Field Due to a Charged Rod
A rod of length l has a uniform positive charge
per unit length ? and a total charge Q. Calculate
the electric field at a point P that is located
along the long axis of the rod and a distance a
from one end.
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Example Infinitely Long Line of Charge


y-components cancel by symmetry






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Example Charged Ring
perpendicular-components cancel by symmetry







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When
The charged ring must look like a point source.
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Example Uniformly Charged Disk
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Two Important Limiting Cases
Large Charged Plate
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Lecture 4 Discussion
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• 1 In figure, two equal positive charges
  q2x10-6C interact with a third charge
  Q4x10-6C.  Find the magnitude and direction of
  the resultant force on Q.

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2 A charge Q is fixed at each of two opposite
corners of a square as shown in figure.  A charge
q is placed at each of the other two corners. If
the resultant electrical force on Q is Zero, how
are Q and q related.
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• 3 Two fixed charges, 1µC and -3µC are separated
  by 10cm as shown in figure (a) where may a third
  charge be located so that no force acts on it? 
  (b) is the equilibrium stable or unstable for the
  third charge?

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• 4 Find the electric field at point p in figure
  due to the charges shown.

Solution
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• 5 A charged cord ball of mass 1g is suspended
  on a light string in the presence of a uniform
  electric field as in figure.  When E(3i5j)
  105N/C, the ball is in equilibrium at T37o.
  Find (a) the charge on the ball and (b) the
  tension in the string.

Substitute T from equation (1) into equation (2)
Substitute T from equation (1) into equation (2)
Substitute by q into equation (1) to find
T5.4410-3N
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• 6 A 1.3µC charge is located on the x-axis at
  x-0.5m, 3.2µC charge is located on the x-axis at
  x1.5m, and 2.5µC charge is located at the
  origin.  Find the net force on the 2.5µC charge

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• 7 Two free point charges q and 4q are a
  distance 1cm apart.  A third charge is so placed
  that the entire system is in equilibrium.  Find
  the location, magnitude and sign of the third
  charge.  Is the equilibrium stable?

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• 8 Two protons in a molecule are separated by a
  distance of 3.810-10m. Find the electrostatic
  force exerted by one proton on the other.
• 9 The electric force on a point charge of 4.0mC
  at some point is 6.910-4N in the positive x
  direction.  What is the value of the electric
  field at that point?

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• 10 Two point charges are a distance d apart . 
  Find  E points to the left P.  Assume
  q11.010-6C, q23.010-6C, and d10cm

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• 11 Calculate E (direction and magnitude) at
  point P in Figure.

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• 12 A uniform electric field exists in a region
  between two oppositely charged plates.  An
  electron is released from rest at the surface of
  the negatively charged plate and strikes the
  surface of the opposite plate, 2.0cm away, in a
  time 1.510-8s. (a) What is the speed of the
  electron as it strikes the second plate? (b) What
  is the magnitude of the electric field .

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• 13 Three charges are placed on corners of an
  equilateral triangle as shown in Figure 1. An
  electron is placed at the center of the triangle.
  What is the magnitude of the net force on the
  electron?

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• 14 A uniform electric field exists in the
  region between two oppositely charged plane
  parallel plates. An electron is released from
  rest at the surface of the negatively charged
  plate and strikes the surface of the opposite
  plate 2x10-8 s later. If the magnitude of the
  electric field is 4x103 N/C, what is the
  separation between the plates?

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Lecture 5 Electric Flux and Gausss Law
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Electric Flux

• Electric flux quantifies the notion number of
  field lines crossing a surface.
• The electric flux ? through a flat surface in a
  uniform electric field depends on the field
  strength E, the surface area A, and the angle ?
  between the field and the normal to the surface.
• Mathematically, the flux is given by
• Here A is a vector whose magnitude is the
  surface area A and whose orientation is normal to
  the surface.

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• When q lt 90, the flux is positive (out of the
  surface), and when q gt 90, the flux is negative.
• Units Nm2/C in SI units, the electric flux is
  a SCALAR quantity
• Find the electric flux through the area A 2 m2,
  which is perpendicular to an electric field E22
  N/C
• Answer F 44 Nm2/C.

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Example

• Calculate the flux of a constant E field (along
  x) through a cube of side L.

y
2
1
Solution
E
x
z
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• Question
• The flux through side B of the cube in the
  figure is the
• same as the flux through side C. What is
  a correct expression for the flux through each of
  these sides?

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When we have a complicated surface, we can divide
it up into tiny elemental areas
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Example

• Whats the total flux on a closed surface with a
  charge inside?
• The shape and size dont matter!
• Just use a sphere

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What is Gausss Law?

• Gausss Law does not tell us anything new, it is
  NOT a new law of physics, but another way of
  expressing Coulombs Law

• Gausss law makes it possible to find the
  electric field easily in highly symmetric
  situations.

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Gauss Law

• The precise relation between flux and the
  enclosed charge is given by Gauss Law
• e0 is the permittivity of free space in the
  Coulombs law
• The symbol has a little circle to indicate
  that the integral is over a closed surface.

The flux through a closed surface is equal to the
total charge contained divided by permittivity of
free space
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• A few important points on Gauss Law
• The integral is over the value of E on a closed
  surface of our choice in any given situation
• The charge Qencl is the net charge enclosed by
  the arbitrary close surface of our choice.
• It does NOT matter where or how much charge is
  distributed inside the surface
• The charge outside the surface does not
  contribute. Why?

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Question

• Whats the total flux with the charge outside?
  Why?
• Solution
• Zero.
• Because the surface surrounds no charge

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Gauss? Coulomb

• Calculate E of point like () charge Q
• Consider sphere radius r centered at the charge
• Spherical symmetry E is the same everywhere on
  the sphere, perpendicular to the sphere

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Application of Gausss lawGausss law can be
used to calculate the electric field if the
symmetry of the charge distribution is high. Here
we concentrate in three different ways of charge
distribution

• A linear charge distribution
• A surface charge distribution
• A volume charge distribution

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A linear charge distribution

• Lets calculate the electric field from a
  conducting wire with charge per unit length ?
  using Gauss Law
• We start by assuming a Gaussian surface in the
  form of a right cylinder with radius r and length
  L placed around the wire such that the wire is
  along the axis of the cylinder

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• From symmetry we can see that the electric field
  will extend radially from the wire.
• How?
• If we rotate the wire along its axis, the
  electric field must look the same
• Cylindrical symmetry
• If we imagine a very long wire, the electric
  field cannot be different anywhere along the
  length of the wire
• Translational symmetry
• Thus our assumption of a right cylinder as a
  Gaussian surface is perfectly suited for the
  calculation of the electric field using Gauss
  Law.

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• The electric flux through the ends of the
  cylinder
• is zero because the electric field is always
  
• parallel to the ends.
• The electric field is always perpendicular to the
  wall of the cylinder so
• and now solve for the electric field

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A surface charge distribution

• Assume that we have a thin, infinite
• non-conducting sheet of positive charge

The charge density in this case is the charge per
unit area, ? From symmetry, we can see that the
electric field will be perpendicular to the
surface of the sheet
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• To calculate the electric field using Gauss Law,
  
• we assume a Gaussian surface in the form of
  a right
• cylinder with cross sectional area A and
  height 2r, chosen
• to cut through the plane perpendicularly.
• Because the electric field is perpendicular to
  the planeeverywhere, the electric field will be
  parallel to the walls of the cylinder and
  perpendicular to the ends of the cylinder.
• Using Gauss Law we get
• so the electric field from an
  infinitenon-conducting sheet with charge density
  ?

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• Assume that we have a thin, infinite conductor
• (metal plate) with positive charge

• The charge density in this case is also the
  charge per unit area, ?, on either surface there
  is equal surface charge on both sides.
• From symmetry, we can see that the electric field
  will be perpendicular to the surface of the sheet

• To calculate the electric field using Gauss
  Law, we assume a Gaussian surface in the form of
  a right cylinder with cross sectional area A and
  height r, chosen to cut through one side of the
  plane perpendicularly.

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• The field inside the conductor is zero so the end
  inside the
• conductor does not contribute to the
  integral.
• Because the electric field is perpendicular to
  the plane
• everywhere, the electric field will be
  parallel to the walls of the cylinder and
  perpendicular to the end of the cylinder outside
  the conductor.
• Using Gauss Law we get
• so the electric field from an
  infiniteconducting sheet with surface charge
  density ? is

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A volume charge distribution

• lets calculate the electric field from
• charge distributed uniformly throughout
  charged sphere.
• Assume that we have insolating a solid sphere of
  charge Q with radius r with constant charge
  density per unit volume ?.
• We will assume two different spherical
• Gaussian surfaces
• r2 gt r (outside)
• r1 lt r (inside)

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• Lets start with a Gaussian surface with
• r1 lt r.
• From spherical symmetry we know that the electric
  field will be radial and perpendicular to the
  Gaussian surface.
• Gauss Law gives us
• Solving for E we find

inside
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In terms of the total charge Q
inside
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• Now consider a Gaussian surface with
• radius r2 gt r.
• Again by spherical symmetry we know that the
  electric field will be radial and perpendicular
  to the Gaussian surface.
• Gauss Law gives us
• Solving for E we find

outside
same as a point charge!
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Electric field vs. radius for a conducting sphere
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Properties of Conductors

• E is zero within conductor
• If there is a field in the conductor, then the
  free electrons would feel a force and be
  accelerated. They would then move and since
  there are charges moving the conductor would not
  be in electrostatic equilibrium. Thus E0
• net charge within the surface is zero
• How ?

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Lecture 6 Application (Gausss Law)
101
1 A solid conducting sphere of radius a has a
net charge 2Q.  A conducting spherical shell of
inner radius b and outer radius c is concentric
with the solid sphere and has a net charge Q as
shown in figure.  Using Gausss law find the
electric field in the regions labeled 1, 2, 3, 4
and find the charge distribution on the spherical
shell.
Region (1) r lt a To find the E inside the solid
sphere of radius a we construct a Gaussian
surface of radius r lt a E 0 since no charge
inside the Gaussian surface
Region (3) b gt r lt c E0 How?
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Region (2) a lt r lt b we construct a spherical
Gaussian surface of radius r                    
         
Region (4) r gt c we construct a spherical
Gaussian surface of radius r gt c, the total net
charge inside the Gaussian surface is q 2Q
(-Q) Q Therefore Gausss law gives
                                                 
103
2 A long straight wire is surrounded by a
hollow cylinder whose axis coincides with that
wire as shown in figure.  The solid wire has a
charge per unit length of ? , and the hollow
cylinder has a net charge per unit length of 2?
.  Use Gauss law to find (a) the charge per unit
length on the inner and outer surfaces of the
hollow cylinder and (b) the electric field
outside the hollow cylinder, a distance r from
the axis.
(a) Use a cylindrical Gaussian surface S1 within
the conducting cylinder where E0
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(b) For a Gaussian surface S2 outside the
conducting cylinder                             
                                 
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3 Consider a long cylindrical charge
distribution of radius R with a uniform charge
density ?.  Find the electric field at distance r
from the axis where r lt R.
If we choose a cylindrical Gaussian surface of
length L and radius r, Its volume is pr²L , and
it encloses a charge ?pr²L . By applying Gausss
law we get,
radially outward from the cylinder axis
Thus
Notice that the electric field will increase as r
increases, and also the electric field is
proportional to r for rltR.  For the region
outside the cylinder (rgtR), the electric field
will decrease as r increases.
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Two Parallel Conducting Plates

• When we have the situation shown in the left two
  panels (a positively charged plate and another
  negatively charged plate with the same magnitude
  of charge), both in isolation, they each have
  equal amounts of charge (surface charge density
  s) on both faces.
• But when we bring them close together, the
  charges on the far sides move to the near sides,
  so on that inner surface the charge density is
  now 2s.
• A Gaussian surface shows that the net charge is
  zero (no flux through sides dA perpendicular to
  E, or ends E 0). E 0 outside, too, due to
  shielding, in just the same way we saw for the
  sphere.

107
Two Parallel Nonconducting Sheets

• The situation is different if you bring two
  nonconducting sheets of charge close to each
  other.
• In this case, the charges cannot move, so there
  is no shielding, but now we can use the principle
  of superposition.
• In this case, the electric field on the left due
  to the positively charged sheet is canceled by
  the electric field on the left of the negatively
  charged sheet, so the field there is zero.
• Likewise, the electric field on the right due to
  the negatively charged sheet is canceled by the
  electric field on the right of the positively
  charged sheet.

• The result is much the same as before, with the
  electric field in between being twice what it was
  previously.

108
4 Two large non-conducting sheets of ve charge
face each other as shown in figure.  What is E at
points (i) to the left of the sheets (ii) between
them and (iii) to the right of the sheets?
We know previously that for each sheet, the
magnitude of the field at any point is
a) At point to the left of the two parallel sheets
109
b) At point between the two sheets
 (c) At point to the right of the two parallel
sheets                               
110
5 A square plate of copper of sides 50cm is
placed in an extended electric field of 8104N/C
directed perpendicular to the plate.  Find (a)
the charge density of each face of the plate
111

• 6 An electric field of intensity 3.5103N/C is
  applied the x axis.  Calculate the electric flux
  through a rectangular plane 0.35m wide and 0.70m
  long if (a) the plane is parallel to the yz
  plane, (b) the plane is parallel to the xy plane,
  and (c) the plane contains the y axis and its
  normal makes an angle of 40o with the x axis.

(a) the plane is parallel to the yz plane
(b) the plane is parallel to the xy plane The
angel 90
c) the plane is parallel to the xy plane The
angel 40
112
7 A long, straight metal rod has a radius of
5cm and a charge per unit length of 30nC/m.  Find
the electric field at the following distances
from the axis of the rod (a) 3cm, (b) 10cm, (c)
100cm.
113

• 8 The electric field everywhere on the surface
  of a conducting hollow sphere of radius 0.75m is
  measured to be equal to 8.90102N/C and points
  radially toward the center of the sphere. What is
  the net charge within the surface?

114
9 A point charge of 5mC is located at the
center of a sphere with a radius of 12cm.  What
is the electric flux through the surface of this
sphere?
10 (a) Two charges of 8mC and -5mC are inside a
cube of sides 0.45m.  What is the total electric
flux through the cube? (b) Repeat (a) if the same
two charges are inside a spherical shell of
radius 0. 45 m.
115

• 12 A solid copper sphere 15cm in radius has a
  total charge of 40nC.  Find the electric field at
  the following distances measured from the center
  of the sphere (a) 12cm, (b) 17cm, (c) 75cm.

(a) At 12 cm the charge in side the Gaussian
surface is zero so the electric field E0
(b)
116
13 Two long, straight wires are separated by a
distance d 16 cm, as shown below. The top wire
carries linear charge density 3 nC/m while the
bottom wire carries -5 nC/m.
1- What is the electric field (including
direction) due to the top wire at a point exactly
half-way between the two wires?
2- Find the electric field due to the bottom wire
at the same point, exactly half-way between the
two wires (including direction).
3- Work out the total electric field at that
point.
117
using Gauss Law and cylindrical Gaussian surface
as Lec.5 page 15
118
Lecture 7 The Electric Potential
119
Electric Potential Energy

• The electric force, like the gravitational force,
  is a conservative force. (Conservative force The
  work is path-independent.)
• As in mechanics, work is
• Work done on the positive charge by moving it
  from A to B

B
A
d
120

• The work done by a conservative force equals
• the negative of the change in potential
  energy,
• DPE
• This equation is valid only for the case of a
  uniform electric field

If a charged particle moves perpendicular to
electric field lines, no work is done.
121

• The potential difference between points A and B,
  VB-VA, is defined as the change in potential
  energy (final minus initial value) of a charge,
  q, moved from A to B, divided by the charge
• Electric potential is a scalar quantity
• Electric potential difference is a measure of
  electric energy per unit charge
• Potential is often referred to as voltage

If the work done by the electric field is zero,
then the electric potential must be constant
122

• Electric potential difference is the work done to
• move a charge from a point A to a point B
• divided by the magnitude of the charge. Thus
  the
• SI units of electric potential difference
• In other words, 1 J of work is required to move a
  1 C of charge between two points that are at
  potential difference of 1 V
• Question How can a bird stand on a high voltage
  line without getting zapped?

123

• Units of electric field (N/C) can be expressed in
• terms of the units of potential (as volts per
  meter)
• Because the positive tends to move in the
  direction of the electric field, work must be
  done on the charge to move it in the direction,
  opposite the field. Thus,
• A positive charge gains electric potential energy
  when it is moved in a direction opposite the
  electric field
• A negative charge looses electrical potential
  energy when it moves in the direction opposite
  the electric field

124

• Example A uniform electric field of magnitude
  250 V/m is directed in the positive x direction.
  A 12µC charge moves from the origin to the point
  (x,y) (20cm, 50cm). (a) What was the change in
  the potential energy of this charge? (b) Through
  what potential difference did the charge move?
• Begin by drawing a picture of the situation,
  including the direction of the electric field,
  and the start and end point of the motion.
• (a) The change potential energy is given by
  the charge times the field times the distance
  moved parallel to the field. Although the charge
  moves 50cm in the y direction, the y direction is
  perpendicular to the field. Only the 20cm moved
  parallel to the field in the x direction matters
  for determining the change of potential energy.
  ? PE -qEd -(12µC)(250 V/m)(0.20 m)
  -6.010-4 .
• (b) The potential difference is the
  difference of electric potential,.
• ? V ? PE / q -6.010-4 J / 12µC -50 V.

125
Analogy between electric and gravitational fields

• The same kinetic-potential energy theorem works
  here
• If a positive charge is released from A, it
  accelerates in the direction of electric field,
  i.e. gains kinetic energy
• If a negative charge is released from A, it
  accelerates in the direction opposite the
  electric field

A
A
d
d
q
m
B
B
126
Example motion of an electron
What is the speed of an electron accelerated
from rest across a potential difference of 100V?
Vab
Given DV100 V me 9.1110-31 kg mp
1.6710-27 kg e 1.6010-19 C Find ve? vp?
127

• Problem
• A proton is placed between two parallel
  conducting plates in a vacuum as shown.
• The potential difference between the two plates
  is 450 V. The proton is released from rest close
  to the positive plate.
• What is the kinetic energy of the proton when it
  reaches the negative plate?

Example Through what potential difference would
an electron need to accelerate to achieve a speed
of 60 of the speed of light, starting from rest?
(The speed of light is 3.00108 m/s.) The final
speed of the electron is vf 0.6(3.00108 m/s)
1.80108 m/s. At this speed, the energy
(non-relativistic) is the kinetic energy,KE KEf
½mvf² 0.5(9.1110-31 kg)(1.80108 m/s)²
1.4810-14 J.This energy must equal the change
in potential energy from moving through a
potential difference, KEf ?PE -q ?V.
Therefore?V KE/q (1.4810-14 J)/(-1.610-19
C) -9.25104 V.
128
Electric potential and potential energy due to
point charges

• Electric circuits point of zero potential is
  defined by grounding some point in the circuit
• Electric potential due to a point charge at a
  point in space point of zero potential is taken
  at an infinite distance from the charge
• With this choice, a potential can be found as
• Note the potential depends only on charge of an
  object, q, and a distance from this object to a
  point in space, r.

129
Superposition principle for potentials

• If more than one point charge is present, their
• electric potential can be found by applying
  superposition principle
• The total electric potential at some point P due
  to several point charges is the algebraic sum of
  the electric potentials due to the individual
  charges.
• Remember that potentials are scalar quantities!

130
Potential energy of a system of point charges

• Consider a system of two particles
• If V1 is the electric potential due to charge q1
  at a point P, then work required to bring the
  charge q2 from infinity to P without acceleration
  is q2V1. If a distance between P and q1 is r,
  then by definition
• Potential energy is positive if charges are of
  the same sign.

q2
q1
r
P
A
131
Example potential energy of an ion
Three ions, Na, Na, and Cl-, located such,
that they form corners of an equilateral
triangle of side 2 nm in water. What is the
electric potential energy of one of the Na ions?
Cl-
?
Na
Na
132
Potentials and charged conductors

• Recall that work is opposite of the change in
• potential energy,
• No work is required to move a charge between two
  points that are at the same potential. That is,
  W0 if VBVA
• Recall
• all charge of the charged conductor is located on
  its surface
• electric field, E, is always perpendicular to its
  surface, i.e. no work is done if charges are
  moved along the surface
• Thus potential is constant everywhere on the
  surface of a charged conductor in equilibrium

but thats not all!
133

• Because the electric field is zero inside the
  conductor, no work is required to move charges
  between any two points, i.e.
• If work is zero, any two points inside the
  conductor have the same potential, i.e. potential
  is constant everywhere inside a conductor
• Finally, since one of the points can be
  arbitrarily close to the surface of the
  conductor, the electric potential is constant
  everywhere inside a conductor and equal to its
  value at the surface!
• Note that the potential inside a conductor is not
  necessarily zero, even though the interior
  electric field is always zero!

134
The electron volt

• A unit of energy commonly used in atomic,
• nuclear and particle physics is electron volt
  (eV)
• The electron volt is defined as the energy that
  electron (or proton) gains when accelerating
  through a potential difference of 1 V
• Relation to SI
• 1 eV 1.6010-19 CV 1.6010-19 J

Vab1 V
135
Example ionization energy of the electron in a
hydrogen atom
In the Bohr model of a hydrogen atom, the
electron, if it is in the ground state, orbits
the proton at a distance of r 5.2910-11 m.
Find the ionization energy of the atom, i.e. the
energy required to remove the electron from the
atom.
Note that the Bohr model, the idea of electrons
as tiny balls orbiting the nucleus, is not a very
good model of the atom. A better picture is one
in which the electron is spread out around the
nucleus in a cloud of varying density however,
the Bohr model does give the right answer for the
ionization energy
136
In the Bohr model of a hydrogen atom, the
electron, if it is in the ground state, orbits
the proton at a distance of r 5.29 x 10-11 m.
Find the ionization energy, i.e. the energy
required to remove the electron from the atom.
The ionization energy equals to the total energy
of the electron-proton system,
Given r 5.292 x 10-11 m me 9.1110-31 kg
mp 1.6710-27 kg e 1.6010-19
C Find E?
with
The velocity of e can be found by analyzing the
force on the electron. This force is the Coulomb
force because the electron travels in a circular
orbit, the acceleration will be the centripetal
acceleration
or
or
Thus, total energy is
137
Calculating the Potential from the Electric Field

• To calculate the electric potential from the
• electric field we start with the definition
  of the work dW done on a particle with charge q
  by a force F over a displacement ds
• In this case the force is provided by the
  electric fieldF qE
• Integrating the work done by the electric force
  on the particle as it moves in the electric field
  from some initial point i to some final point f
  we obtain

138

• Remembering the relation between the change
• in electric potential and the work done
• we find
• Taking the convention that the electric potential
  is zero at infinity we can express the electric
  potential in terms of the electric field as

139
Example - Charge moves in E field

• Given the uniform electric field E, find the
  potential difference Vf-Vi by moving a test
  charge q0 along the path icf.
• Idea Integrate E ? ds along the path connecting
  ic then cf. (Imagine that we move a test charge
  q0 from i to c and then from c to f.)

140
Example - Charge moves in E field
141
Question

• We just derived Vf-Vi for the path i ? c ? f.
  What is Vf-Vi when going directly from i to f ?
• A 0
• B -Ed
• C Ed
• D -1/2 Ed

142
Lecture 8 Application (The Electric Potential)
143
1 What potential difference is needed to stop
an electron with an initial speed of 4.2105m/s?
2 An ion accelerated through a potential
difference of 115V experiences an increase in
potential energy of 7.3710-17J. Calculate the
charge on the ion.
144
3 An infinite charged sheet has a surface
charge density s of 1.010-7 C/m2.  How far apart
are the equipotential surfaces whose potentials
differ by 5.0 V?
4 At what distance from a point charge of 8µC
would the potential equal 3.6104V?
145
5 At a distance r away from a point charge q,
the electrical potential is V400v and the
magnitude of the electric field is E150N/C. 
Determine the value of q and r.
146
6 Calculate the value of the electric potential
at point P due to the charge configuration shown
in Figure. Use the values q15mC, q2-10mC,
a0.4m, and b0.5m.                               
                                   
By substitute find V
147
7 Two large parallel conducting plates are 10cm
apart and carry equal and opposite charges on
their facing surfaces.  An electron placed midway
between the two plates experiences a force of
1.61015N.  What is the potential difference
between the plates?
148
8 Two point charges are located as shown in,
where ql4mC, q2-2mC, a0.30m, and b0.90m.
Calculate the value of the electrical potential
at points P1, and P2.  Which point is at the
higher potential?                                 
                                   
149
9 In figure prove that the work required to put
four charges together on the corner of a square
of radius a is given by
150

• 10 Assume we have a system of three point
  chargesq1 1.50 ?Cq2 2.50 ?Cq3 -3.50
  ?C.
• q1 is located at (0,a)q2 is located at (0,0)q3
  is located at (b,0)a 8.00 m and b 6.00 m.
• What is the electric potential at point P located
  at (b,a)?

151

• The electric potential at point P is given by the
  sum of the electric potential from the three
  charges

r1
r2
r3
152
11 An electron initially has velocity 5 x 105
m/s. It is accelerated through a potential of 2
V. What is its final velocity?
153

• 12 A charge of -1x10-8C weighs 1 g. It is
  released at rest from point P and moves to point
  Q. It's velocity at point Q is 1 cm/s. What is
  the potential difference, VP - VQ?

154
13 A proton with a speed of vi 2.0 x 105 m/s
enters a region of space where source charges
have created an electric potential. What is the
protons speed after it has moved through a
potential difference of ?V 100 V?
155
14 Potential Due to a Charged Rod

• A rod of length L located along the x axis has a
  uniform linear charge density ?. Find the
  electric potential at a point P located on the y
  axis a distance d from the origin.
• Start with
• then,
• So

156
15 The Electric Potential of a Charged Ring
Find the potential of a thin uniformly charged
ring of radius R and charge Q at point P on the z
axis?
157
16 The Electric Potential of a Charged Disk
Find the potential V of a thin uniformly charged
disk of radius R and charge density s at point P
on the z axis?
158

• Lecture 9
• The Electric Field Electric Potential Due to
  Continuous Charge Distributions

159
Field Potential Due to a Continuous Charge
Distribution
The electric field and electric potential due to
a continuous charge distribution is found by
treating charge elements as point charges and
then summing via integrating, the electric field
vectors and the electric potential produced by
all the charge elements.
160
1 Electric Field Due to a Charged Rod
A rod of length L has a uniform positive charge
per unit length ? and a total charge Q. Calculate
the electric field at a point P that is located
along the long axis of the rod and a distance a
from one end.
161
2 Infinitely Long Line of Charge


y-components cancel by symmetry






162
3 Electric Field on the Z-Axis of a Charged Ring
determine the field at point P on the axis of the
ring.
perpendicular-components cancel by symmetry







163
When
The charged ring must look like a point source.
Note that for z gtgt R (the radius of the ring),
this reduces to a simple Coulomb field.
164
4 Electric Field on the Axis of an Uniformly
Charged Disk
Using the charged ring result,
165
(No Transcript)
166
Two Important Limiting Cases
Large Charged Plate
167
5 Potential Due to a Charged Rod

• A rod of length L located along the x axis has a
  uniform
• linear charge density ?. Find the electric
  potential at a point P located on the y axis a
  distance d from the origin.
• Start with
• then,
• So

168
6 The Electric Potential of a Charged Ring
Find an expression for the electric potential at
a point P located on the perpendicular central
axis of a uniformly charged ring of radius a and
total charge Q
169
7 The Electric Potential of a Charged Disk
Find the potential V of a thin uniformly charged
disk of radius R and charge density s at point P
on the z axis?
170
Lecture 10 Capacitance and capacitors
171
Capacitors

• Capacitors are devices that store energy in an
  electric field.
• Capacitors are used in many every-day
  applications
• Heart defibrillators
• Camera flash units
• Capacitors are an essential part of electronics.
• Capacitors can be micro-sized on computer chips
  or super-sized for high power circuits such as FM
  radio transmitters.

172
Definition of Capacitance

• The definition of capacitance is
• The units of capacitance are coulombs per volt.
• The unit of capacitance has been given the name
  farad (abbreviated F) named after British
  physicist Michael Faraday (1791 - 1867)
• A farad is a very large capacitance
• Typically we deal with ?F (10-6 F), nF (10-9
  F),or pF (10-12 F)

173
The parallel-plate capacitor

• The capacitance of a device depends on the area
  of the plates and the distance between the plates
• where A is the area of one of the plates, d is
  the separation, e0 is a constant (permittivity of
  free space),
• e0 8.8510-12 C2/Nm2

A
Q
d
A
-Q
174

• Example A parallel plate capacitor has plates
  2.00 m2 in area, separated by a distance of 5.00
  mm. A potential difference of 10,000 V is
  applied across the capacitor. Determine
• the capacitance
• the charge on each plate

Solution
Since we are dealing with the parallel-plate
capacitor, the capacitance can be found as
Given DV10,000 V A 2.00 m2 d 5.00 mm
Find C? Q?
Once the capacitance is known, the charge can be
found from the definition of a capacitance via
charge and potential difference
175
Cylindrical Capacitor

• Consider a capacitor constructed of two collinear
  conducting cylinders of length L.
• The inner cylinder has radius r1 andthe outer
  cylinder has radius r2.
• Both cylinders have charge perunit length ? with
  the inner cylinderhaving positive charge and the
  outercylinder having negative charge.
• We will assume an ideal cylindrical capacitor
• The electric field points radially from the inner
  cylinder to the outer cylinder.
• The electric field is zero outside the collinear
  cylinders.

176

• We apply Gauss Law to get the electric field
  between
• the two cylinder using a Gaussian surface
  with radius r
• and length L as illustrated by the red lines
• which we can rewrite to get anexpression for
  the electric fieldbetween the two cylinders

177

• As we did for the parallel plate capacitor, we
  define the
• voltage difference across the two cylinders
  to be VV1 V2.
• The capacitance of a cylindrical capacitor is

178
Spherical Capacitor

• Consider a spherical capacitor formed by two
• concentric conducting spheres with radii r1
  and r2

• Lets assume that the inner sphere has charge q
  and the outer sphere has charge q.
• The electric field is perpendicular to the
  surface of both spheres and points radially
  outward

179

• To calculate the electric field, we use a
  Gaussian surfaceconsisting of a concentric
  sphere of radius r such that r1 lt r lt r2
• The electric field is always perpendicular to the
  Gaussian surface so
• which reduces to

180

• To get the electric potential we follow a method
  similar to the one we used for the cylindrical
  capacitor and integrate from the negatively
  charged sphere to the positively charged sphere
• Using the definition of capacitance we find
• The capacitance of a spherical capacitor is then

181
Combinations of capacitors

• It is very often that more than one capacitor is
• used in an electric circuit
• We would have to learn how to compute the
  equivalent capacitance of certain combinations of
  capacitors

C2
C1
C3
182
a. Parallel combination
Connecting a battery to the parallel combination
of capacitors is equivalent to introducing the
same potential difference for both capacitors,
A total charge transferred to the system from the
battery is the sum of charges of the two
capacitors,
183
Parallel combination

• Analogous formula is true for any number of
• capacitors,
• It follows that the equivalent capacitance of a
  parallel combination of capacitors is greater
  than any of the individual capacitors

(parallel combination)
184
Example A 3 mF capacitor and a 6 mF capacitor
are connected in parallel across an 18 V battery.
Determine the equivalent capacitance and total
charge deposited.
Given V 18 V C1 3 mF C2 6
mF Find Ceq? Q?
First determine equivalent capacitance of C1 and
C2
Next, determine the charge
185
Series combination
Connecting a battery to the serial combination of
capacitors is equivalent to introducing the same
charge for both capacitors,
A voltage induced in the system from the battery
is the sum of potential differences across the
individual capacitors,
186
Series combination

• Analogous formula is true for any number of
• capacitors,
• It follows that the equivalent capacitance of a
  series combination of capacitors is always less
  than any of the individual capacitance in the
  combination

(series combination)
187
Example A 3 mF capacitor and a 6 mF capacitor
are connected in series across an 18 V battery.
Determine the equivalent capacitance and total
charge deposited.
Given V 18 V C1 3 mF C2 6
mF Find Ceq? Q?
First determine equivalent capacitance of C1 and
C2
Next, determine the charge
188
Example Capacitors in Series and Parallel
Three capacitors are connected as shown. (a)
Find the equivalent capacitance of the
3-capacitor combination. (b) The capacitors,
initially uncharged, are connected across a 6.0 V
battery. Find the charge and voltage drop for
each capacitor.
189
(