Electrostatics

1
Electrostatics

• October 5, 2013

2
iClicker

• An object is placed on the axis of a converging
  thin lens of focal length 2 cm, at a distance of
  8 cm from the lens. The distance between the
  image and the lens is most nearly
• a. 0.4 cm
• b. 0.8 cm
• c. 1.6 cm
• d. 2.0 cm
• e. 2.7 cm

3
Atomic Charges

• The atom has positive charge in the nucleus,
  located in the protons.
• The positive charge cannot move from the atom
  unless there is a nuclear reaction.
• The atom has negative charge in the electron
  cloud on the outside of the atom.
• Electrons can move from atom to atom without all
  that much difficulty.

4
Charge

• Charge comes in two forms, which Ben Franklin
  designated as positive () and negative().
• Charge is quantized
• It only comes in packets
• The smallest possible stable charge, which we
  designate as e, is the magnitude of the charge on
  1 electron or 1 proton.
• e is referred to as the elementary charge.
• e 1.602 10-19 Coulombs.
• We say a proton has charge of e, and an electron
  has a charge of e.

5
Problem

• A certain static discharge delivers -0.5 Coulombs
  of electrical charge. How many electrons are in
  this discharge?

6
Problem

• The total charge of a system composed of 1800
  particles, all of which are protons or electrons,
  is 31x10-18 C. How many protons are in the
  system? How many electrons are in the system?

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Coulombs Law andElectrical Force
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Electric Force

• Charges exert forces on each other.
• Like charges (two positives, or two negatives)
  repel each other, resulting in a repulsive force.
• Opposite charges (a positive and a negative)
  attract each other, resulting in an attractive
  force.

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Coulombs Law

• Coulombs law tells us how the magnitude of the
  force between two particles varies with their
  charge and with the distance between them.
• k 9.0 109 N m2 / C2
• q1, q2 are charges (C)
• r is distance between the charges (m)
• F is force (N)
• Coulombs law applies directly only to
  spherically symmetric charges.

10
Problem

• A point charge of positive 12.0 µC experiences an
  attractive force of 51 mN when it is placed 15 cm
  from another point charge. What is the other
  charge?

11
iClicker

• Two isolated charges, q and - 2q, are 2
  centimeters apart. If F is the magnitude of the
  force acting on charge -2q, what are the
  magnitude and direction of the force acting on
  charge q ?
• Magnitude Direction
• (A) (1/2) F Toward charge - 2q
• (B) 2 F Away from charge -2q
• (C) F Toward charge - 2q
• (D) F Away from charge - 2q
• (E) 2F Toward charge - 2q

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Superposition

• Electrical force, like all forces, is a vector
  quantity.
• If a charge is subjected to forces from more than
  one other charge, vector addition must be
  performed.
• Vector addition to find the resultant vector is
  sometimes called superposition.

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Problem
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Electric Fields
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The Electric Field

• The presence of or charge modifies empty
  space. This enables the electrical force to act
  on charged particles without actually touching
  them.
• We say that an electric field is created in the
  space around a charged particle or a
  configuration of charges.
• If a charged particle is placed in an electric
  field created by other charges, it will
  experience a force as a result of the field.
• We can easily calculate the electric force from
  the electric field.

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Electric Fields

• Forces exist only when two or more particles are
  present.
• Fields exist even if no force is present.

17
Field Lines Around Charges

• The arrows in a field are not vectors, they are
  lines of force.
• The lines of force indicate the direction of the
  force on a positive charge placed in the field.
• Negative charges experience a force in the
  opposite direction.

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Field Line Rules
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Single Charges
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Double Charges
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Plate Charges
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Field Vectors from Field Lines

• The electric field at a given point is not the
  field line itself, but can be determined from the
  field line.
• The electric field vectors, and thus force
  vector, is always tangent to the line of force at
  that point.

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Force due to Electric Field

• The force on a charged particle placed in an
  electric field is easily calculated.
• F Eq
• F Force (N)
• E Electric Field (N/C)
• q Charge (C)

24
Problem

• The electric field in a given region is 4000 N/C
  pointed toward the north. What is the force
  exerted on a 400 µg Styrofoam bead bearing 600
  excess electrons when placed in the field? Ignore
  gravitational effects.

25
iClicker

• An electron e and a proton p are simultaneously
  released from rest in a uniform electric field E.
  Assume that the particles are sufficiently far
  apart so that the only force acting on each
  particle after it is released is that due to the
  electric field. At a later time when the
  particles are still in the field, the electron
  and the proton will have the same
• a. direction of motion
• b. speed
• c. displacement
• d. magnitude of acceleration
• e. magnitude of force acting on them

26
Problem

• A 400 mg Styrofoam bead has 600 excess electrons
  on its surface. What is the magnitude and
  direction of the electric field that will suspend
  the bead in midair?

27
Problem

• A proton traveling at 440 m/s in the x direction
  enters an electric field of magnitude 5400 N/C
  directed in the y direction. Find the
  acceleration.

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Electric Fields

• The Electric Field surrounding a point charge or
  a spherical charge can be calculated by
• E kq / r2
• E Electric Field (N/C)
• k 9 x 109 N m2/C2
• q Charge (C)
• r distance from center of charge q (m)

29
iClicker

• A point P is 0.50 meter from a point charge of
  5.0 X 10-8 coulomb. The intensity of the electric
  field at point P is most nearly
• (A) 2.5 x 10-8 N/C
• (B) 2.5 x 101 N/C
• (C) 9.0 x 102 N/ C
• (D) 1.8 x 103 N/C
• (E) 7.5 x 108 N/C

30
iClicker

• Charges Q and - 4Q are situated as shown above.
  The net electric field is zero nearest which
  point?
• (A) A
• (B) B
• (C) C
• (D) D
• (E) E

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Superposition of Fields

• When more than one charge contributes to the
  electric field, the resultant electric field is
  the vector sum of the electric fields produced by
  the various charges.
• Again, as with force vectors, this is referred to
  as superposition.

32
iClicker

• The diagram above shows an isolated, positive
  charge Q. Point (B) is twice as far away from Q
  as point A. The ratio of the electric field
  strength at point A to the electric field
  strength at point B is
• (A) 8 to 1
• (B) 4 to 1
• (C) 2 to 1
• (D) 1 to 1
• (E) 1 to 2

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Electric Field Lines

• Electric field lines are NOT VECTORS, but may be
  used to derive the direction of electric field
  vectors at given points.
• The resulting vector gives the direction of the
  electric force on a positive charge placed in the
  field.

34
Problem

• A particle bearing -5.0 µC is placed at -2.0 cm,
  and a particle bearing 5.0 µC is placed at 2.0
  cm. What is the field at the origin?

35
iClicker

• An electron is accelerated from rest for a time
  of 10-9 second by a uniform electric field that
  exerts a force of 8.0 x 10-15 Newton on the
  electron.
• What is the magnitude of the electric field?
  
• (A) 8.0 x 10-24 N/C
• (B) 9.1 x 10-22 N/C
• (C) 8.0 x 10-6 N/C
• (D) 2.0 x 10-5 N/C
• (E) 5.0 x 104 N/C

36
iClicker

• An electron is accelerated from rest for a time
  of 10-9 second by a uniform electric field that
  exerts a force of 8.0 x 10-15 Newton on the
  electron.
• The speed of the electron after it has
  accelerated for the 10-9 second is most nearly
• (A) 101 m/s
• (B) 103 m/s
• (C) 105 m/s
• (D) 107 m/s
• (E) 109 m/s

37
Problem

• A particle bearing 10.0 mC is placed at the
  origin, and a particle bearing 5.0 mC is placed
  at 1.0 m. Where is the field zero?

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Electric Potential
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Electric Potential Energy

• Electrical potential energy is the energy
  contained in a configuration of charges.
• Like all potential energies, when it goes up the
  configuration is less stable when it goes down,
  the configuration is more stable.
• The unit is the Joule.

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Electric Potential Energy

• Electrical potential energy increases when
  charges are brought into less favorable
  configurations

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-
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Electric Potential Energy

• Electrical potential energy decreases when
  charges are brought into more favorable
  configurations.

-
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Work

• Work must be done on the charge to increase the
  electric potential energy.

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Electric Potential

• Electric potential is hard to understand, but
  easy to measure.
• We commonly call it voltage, and its unit is
  the Volt.
• 1 V 1 J/C
• Electric potential is easily related to both the
  electric potential energy, and to the electric
  field.

44
Electric Potential

• The change in potential energy is directly
  related to the change in voltage.
• ?U q?V
• ? U change in electrical potential energy (J)
• q charge moved (C)
• ? V potential difference (V)
• All charges will spontaneously go to lower
  potential energies if they are allowed to move.

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Electric Potential

• Since all charges try to decrease U, and ?U q
  ?V, this means that spontaneous movement of
  charges result in negative ?U.
• V U / q
• Positive charges like to DECREASE their potential
  (?V lt 0)
• Negative charges like to INCREASE their
  potential. (?V gt 0)

46
Problem

• A 3.0 µC charge is moved through a potential
  difference of 640 V. What is its potential energy
  change?

47
E-field Potential

• The electric potential is related in a simple way
  to a uniform electric field.
• ?V -Ed
• ?V change in electrical potential (V)
• E Constant electric field strength (N/C or V/m)
• d distance moved (m)

48
Problem

• An electric field is parallel to the x-axis. What
  is its magnitude and direction of the electric
  field if the potential difference between x 1.0
  m and x 2.5 m is found to be 900 V?

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More Electric Field Stuff
50
iClicker

• Two large, flat, parallel, conducting plates are
  0.04 m apart, as shown above. The lower plate is
  at a potential of 2 V with respect to ground.
  The upper plate is at a potential of 10 V with
  respect to ground. Point P is located 0.01 m
  above the lower plate. The electric potential at
  point P is
• a. 10 V
• b. 8 V
• c. 6 V
• d. 4 V
• e. 2 V

51
iClicker

• The electron volt is a measure of
• (A) Charge
• (B) Energy
• (C) Impulse
• (D) Momentum
• (E) velocity

52
iClicker

• A point P is 0.50 meter from a point charge of
  5.0 X 10-8 coulomb. The electric potential at
  point P is most nearly
• (A) 2.5 x l0-8 V
• (B) 2.5 x 101 V
• (C) 9.0 x 102 V
• (D) 1.8x 103 V
• (E) 7.5x 103 V

53
iClicker

• The figure above shows two particles, each with a
  charge of Q, that are located at the opposite
  corners of a square of side d. What is the
  direction of the net electric field at point P ?
• (A) Up and to the Right
• (B) Up and to the Left
• (C) Down and to the Right
• (D) Down and to the Left
• (E) Down

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Charge Location

• Electric field lines are more dense near a sharp
  point, indicating the electric field is more
  intense in such regions.
• All lightning rods take advantage of this by
  having a sharply pointed tip.
• During an electrical storm, the electric field at
  the tip becomes so intense that charge is given
  off into the atmosphere, discharging the area
  near a house at a steady rate and preventing a
  sudden blast of lightning.

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Excess Charges

• Excess charges reside on the surface of a charged
  conductor.
• If excess charges were found inside a conductor,
  they would repel one another until the charges
  were as far from each other as possible, thus the
  surface.

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E-Field of a Charged Sphere

• The electric field inside a conductor must be
  zero.

E0












57
Conductor in an E-Field

• The electric field inside a conductor must be
  zero.

58
iClicker

• A positive charge of 10-6 coulomb is placed on an
  insulated solid conducting sphere. Which of the
  following is true?
• The charge resides uniformly throughout the
  sphere
• (B) The electric field inside the sphere is
  constant in magnitude, but not zero.
• (C) The electric field in the region surrounding
  the sphere increases with increasing distance
  from the sphere.
• (D) An insulated metal object acquires a net
  positive charge when brought near to, but not in
  contact with, the sphere.
• (E) When a second conducting sphere is connected
  by a conducting wire to the first sphere, charge
  is transferred until the electric potentials of
  the two spheres are equal.

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Electric Potential Energy
60
iClicker

• Of the following phenomena, which provides the
  best evidence that particles can have wave
  properties?
• (A) The absorption of photons by electrons in an
  atom
• (B) The alpha-decay of radioactive nuclei
• (C) The interference pattern produced by neutrons
  incident on a crystal
• (D) The production of x-rays by electrons
  striking a metal target
• (E) The scattering of photons by electrons at
  rest

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Conservation of Energy

• In a conservative system, energy changes from one
  form of mechanical energy to another.
• When only the conservative electrostatic force is
  involved, a charged particle released from rest
  in an electric field will move so as to lose
  potential energy and gain an equivalent amount of
  kinetic energy.
• The change in electrical potential energy can be
  calculated by
• ?U q?V.

62
Problem

• If a proton is accelerated through a potential
  difference of -2,000 V, what is its change in
  potential energy?
• How fast will this proton be moving if it started
  at rest?

63
iClicker

• Two parallel conducting plates are connected to a
  constant voltage source. The magnitude of the
  electric field between the plates is 2,000 N/C.
  If the voltage is doubled and the distance
  between the plates is reduced to 1/5 the original
  distance, the magnitude of the new electric field
  is
• (A) 800 N/C
• (B) 1,600 N/C
• (C) 2,400 N/C
• (D) 5,000 N/C
• (E) 20,000 N/C

64
Problem

• A proton at rest is released in a uniform
  electric field. How fast is it moving after it
  travels through a potential difference of -1200
  V? How far has it moved?

65
Electric Potential Energy

• Electric potential energy is a scalar, like all
  forms of energy.
• U kq1q2/r
• U electrical potential energy (J)
• k 9 109 N m2 / C2
• q1, q2 charges (C)
• r distance between centers (m)
• This formula only works for spherical charges or
  point charges.

66
Electric Potential

• For a spherical or point charge, the electric
  potential can be calculated by the following
  formula
• V kq/r
• V potential (V)
• k 9 109 N m2 / C2
• q charge (C)
• r distance from the charge (m)

67
Problem

• How far must the point charges q1 7.22 µC and
  q2 -26.1 µC be separated for the electric
  potential energy of the system to be -126 J?

68
Problem

• The electric potential 1.5 m from a point charge
  q is 2.8 x 104 V. What is the value of q?

69
iClicker

• The hollow metal sphere shown above is positively
  charged. Point C is the center of the sphere and
  point P is any other point within the sphere.
  Which of the following is true of the electric
  field at these points?
• a. It is zero at both points.
• b. It is zero at C, but at P it is not zero and
  is directed inward.
• c. It is zero at C, but at P it is not zero and
  is directed outward.
• d. It is zero at P, but at C it is not zero.
• e. It is not zero at either point.

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E-Field and Potential
71
Electric Field and Potential

• E - V / d
• Two things about E and V
• The electric field points in the direction of
  decreasing electric potential.
• The electric field is always perpendicular to the
  equipotential surface.

72
E-Field and Potential

• The electric field points in the direction of
  decreasing electric potential.

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E-Field is Perpendicular

• Zero work is done when a charge is moved
  perpendicular to an electric field
• W Fd cos 0
• If zero work is done, and V - W/q, then there
  is no change in potential.
• Thus, the potential is constant in a direction
  perpendicular to the electric field.

74
Problem

• Draw a negative point charge of -Q and its
  associated electric field.
• Draw 4 equipotential surfaces such that ?V is the
  same between the surfaces, and draw them at the
  correct relative locations.
• What do you observe about the spacing between the
  equipotential surfaces?