Giancoli- chapter 16

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Electric Forces and Fields

• Giancoli- chapter 16

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Electric Charge

• Electric charge is conserved
• Electric Charge is quantized
• One unit of charge e 1.60219 x 10-19 C
• C stands for Coulomb, the unit of electric charge
• A proton has a charge of 1.60 x 10-19 C
• An electron has a charge of -1.60 x 10-19 C

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Conductors Insulators

• Conductors Materials in which electric charges
  move freely
• Examples most metals
• Insulators Materials in which electric charges
  do not move freely
• Examples Plastic, glass, silk, rubber

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Charging by contact

• The two objects are rubbed together and electrons
  are transferred from one to the other
• electrons from the fur are transferred to the rod

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Charging by Induction

• To charge by induction, a charged object is
  brought close to (not touching!) a conductor and
  then a conducting wire connects the conductor to
  the ground and the electrons travel to the ground

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Charging by Polarization

• Charging by polarization creates a surface charge
• A charged object is brought close to an insulator
  and the electrons and protons realign themselves
  to create one side that is more positive and one
  that is more negative

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Coulombs Law

• Coulombs Law describes the mathematical
  relationship between electric force, distance and
  electric charge for two objects

kC 8.99 x 109 Nm2 C2
Electric force Coulombs Constant x (charge
1)(charge 2) distance2
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Coulombs Law

• Remember that force is a vector!
• For problems involving two charges, the direction
  is either attractive or repulsive
• the direction of the force between a positive
  charge and negative charge is attractive a
• the direction of the force between two negative
  charges is repulsive

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Example Problem

• Two identical conducting spheres are placed with
  their centers 0.30 m apart. One is given a charge
  of 12 x 10-9 C and the other is given a charge
  of -18 x 10-9 C
• A. Find the electric force exerted on one sphere
  by the other
• B. The spheres are connected by a conducting
  wire. After equilibrium has occurred, find the
  electric force between the two spheres.

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Use Coulombs Law
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• What does it mean after equilibrium has
  occurred?
• The charge on each sphere is the same

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Three charges in a line

• Three charges are arranged in a line as shown
  below. The distance between Q1 and Q2 is 0.35 m
  and the distance between Q2 and Q3 is 0.45 m.
• What is the net electrostatic force acting on Q3?

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• Find the force acting on Q3 from Q1
• Find the force acting on Q3 from Q2
• Net force acting on Q3 1.33 N Left

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Two Dimensional Problem

• Three charges are arranged in a triangular
  pattern as shown below. The distance between Q1
  and Q2 is 0.35 m and the distance between Q2 and
  Q3 is 0.45 m.
• Find the net electrostatic force on Q2

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Solving the Problem

• We already have the numerical values for the
  force acting on Q2from Q3but the direction is
  different
• F32 1.78 N Right
• Find the force from Q1 on Q2

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Finish the Problem

• Use vector addition (Pythagorean theorem) to find
  the net force acting on Q2.
• Direction?

1.47 N
1.78 N
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Electric Field

• Electric force, like gravitational force, is a
  field force
• Remember Field forces can act through space even
  when there is no physical contact between the
  objects involved
• A charged object has an electric field in the
  space around it

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Electric Field Lines

• Electric Field Lines point in the direction of
  the electric field
• The number and spacing of field lines is
  proportional to the electric field strength
• The electric field is strong where the field
  lines are close together and weaker when they are
  far apart

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Electric Field Lines

• The lines for a positive charge point away from
  the charge
• The lines for a negative charge point towards the
  charge

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Electric Field Lines

• This diagram shows the electric field lines for
  two equal and opposite point charges
• Notice that the lines begin on the positive
  charge and end on the negative charge

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Electric Field Lines

• This diagram shows the electric field lines for
  two positive point charges
• Notice that the same number of lines emerges from
  each charge because they are equal in magnitude

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Electric Field Lines

• If the charges are unequal, then the number of
  lines emerging from them will be different
• Notice that the positive charge has twice as many
  lines

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Calculating Electric Field Strength

• The equation for the electric field produced by a
  point charge is
• Kc9x109 Nm2/C2 ,r is the distance from the
  charge and q is the charge producing the field
• The unit for E is N/C
• Electric field strength is a vector!!
• If q is positive, then E is directed away from q
• If q is negative, then E is directed toward q

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Calculating the force from an electric
field

• If a charged object is placed in an electric
  field, we can calculate the force acting on it
  from the electric field
• Remember that F is a vector!!

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Sample Problem p. 647 3

• An electric field of 2.0 x 104 N/C is directed
  along the positive x-axis
• a. What is the electric force on an electron in
  this field?
• b. What is the electric force on a proton in
  this field?

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Sample Problem p. 647 3

• E 2.0 x 104 N/C , q 1.6 x 10-19 C
• FqE 3.2 x 10-15 N for both the electron and the
  proton
• What about the direction?
• The electric field is pointing along the positive
  x axis (to the right) which means theres a
  positive charge to the left

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For the proton

• Since the electric field is pointing to the
  right, if you put a proton in it, the proton will
  want to move away towards the right and the
  direction of the force on it will be to the right
• Answer 3.2 x 10-15 N along the positive x axis
  (to the right)

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For the electron

• Since theres a positive charge causing the
  electric field to point towards the right, an
  electron would feel attracted to the positive
  charge. Therefore, the force acting on it is
  toward the left
• Answer 3.2 x 10-15 N along the negative x axis
  (to the left)

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Sample Problem

• Find the electric field at a point midway between
  two charges of 30 nC and 60 nC separated by a
  distance of 30.0 cm

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• For the 30 nC charge
• Direction of the E-field for both charges is
  away since theyre both positive

• For the 60 nC charge

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Which one will win?

• At the midway point, the 30nC charges field
  strength is 12000 N/C toward the 60 nC charge and
  the 60 nC charges field strength is 24,000 N/C
  toward the 30 nC charge.
• The 60 nC charge will win. Since the fields
  point in opposite directions, you have to
  subtract
• Answer 12,000 N/C toward the 30 nC charge

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Sample Problem

• A constant electric field directed along the
  positive x-axis has a strength of 2.0 x 103 N/C.
• Find the electric force exerted on a proton by
  the field
• Find the acceleration of the proton

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Answer ?

• FqE(1.6x10-19 C)(2.0 x 103 N/C)
• 3.2 x 10-16 N
• Direction?
• Answer 3.2 x 10-16 N along the positive x-axis
  (to the right)

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Answer ?

• B. What is the acceleration?
• Ask Newton!
• Fma
• a F/m 3.2 x 10-16 N/1.6x10-27 kg
• a 2 x 1011 m/s2 along the positive x axis