Electric Charge and the Electric Field (Ch 26)

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Electric Charge and the Electric Field (Ch 26)

• Four Fundamental Forces
• Electromagnetic
• Gravity
• Strong Nuclear Force
• Weak Nuclear Force

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Electric Charge

• Electricity Elektron (amber)
• Amber (tree resin) gets a charge when rubbed with
  cloth
• Ben Franklin
• Positive Charge on a rubber rod
• Negative Charge on an amber/plastic rod

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(No Transcript)
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Electric Charge

• Law of conservation of charge Charge is never
  created or destroyed

Cloth
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Rubber rod
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Insulators/Conductors
Insulators Semiconductors Conductors
Do not conduct electricity Conduct electricity, but only at a higher voltage Conduct electricity well
Small atoms Electrons tightly held Non-Metals Medium atoms Electrons held medium Metalloids Large atoms Electrons not tightly held Metals
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(Discuss movement of the electrons)
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Conduction

• Transfer of charge by touching
• Electrons can flow between substances
• Can produce a net charge on the substance

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Induction

• Transfer of charge without touching
• Electrons migrate within the substance to cause a
  separation of charge
• Net charge on substance is still zero

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Induction Grounding

• Only way to produce a net charge by induction
• Earth can easily absorb or donate electrons
• Electrons can leave a substance, then break the
  ground

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Electroscope
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Charging the electroscope
by induction by conduction (charge
separation) (net charge)
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Using the electroscope

• Can be used to detect the charge on a substance
• (Must charge electroscope first)

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Coulombs Law

• Coloumb experimented to determine magnitude of
  electromagnetic force
• Measured the forces between charged spheres
  (angle of deflection).

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• F k Q1Q2
• r2
• F Force
• k 9.0 X 109 N-m/C2 (proportionality constant)
• Q Charge (C)
• r Radius (m)
• Varies with inverse-square of the radius(distance)

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The Coulomb

• 1 Coulomb 1 Amperesecond
• Unit of charge
• Point charges small objects (charge doesnt get
  distributed much)
• Elementary Charge
• Charge on the electron and proton
• Quantized (cant have ½ an electron)
• e 1.602 X 10-19 C

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• Determine the magnitude of the force between a
  proton and an electron in a hydrogen atom.
  Assume the distance from the electron to the
  nucleus is 0.53 X 10-10 m. (8.2 X 10-8 N)

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• Calculate the force between an electron and the
  three protons in a Li atom if the distance is
  about 1.3 X 10-10 m. (3.9 X 10-8 N)

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• Three charged particles are arranged in a
  straight line as shown in the diagram. Calculate
  the net force on particle 3. (-1.5 to the left)

0.30 m
0.20 m
-
-

Q1 -8.0 mC Q2 3.0 mC Q3 -4.0 mC
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• F k Q1Q2
• r2
• F31 (9.0 X 109 N-m2/C2 )(4.0X 10-6 C)(8.0X10-6
  C)
• (0.50 m)2
• F31 1.2 N (Repulsive to the right)
• F32 (9.0 X 109 N-m2/C2 )(4.0X 10-6 C)(3.0X10-6
  C)
• (0.20 m)2
• F32 2.7 N (Attractive to the left)

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• Fnet F31 - F32
• Fnet 1.2 N 2.7 N -1.5 to the left

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• Three charges are in a line. The first is 2.00
  mC. The second is -2.50 mC at 25 cm. The third
  charge is 2.00 mC at the 40 cm mark.
• Calculate the net force on the center charge.
• Will the center charge move to the left or right?

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• Three charges are in a line. The first is 3.00
  mC. The second is -2.00 mC at 5 cm. Where
  should a the third charge (4.0 mC) be placed so
  the middle charge does not move?

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• Three charges are in a line. The first is -10.00
  mC. The second is -15.00 mC at 100 cm. Where
  should a middle charge (20.0 mC) be placed so it
  does not move?

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Coulombs Law Ex 4

• What is the resultant force on charge q3 if the
  charges are arranged as shown below. The
  magnitudes of the charges are
• q1 6.00 X 10-9 C
• q2 -2.00 X 10-9 C
• q3 5.00 X 10-9 C

25
4.00 m

-
q2
q3
37o
3.00 m
5.00 m

q1
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• First calculate the forces on q3 separately
• F13 k Q1Q3
• r2
• F13 (9.0 X 109 N-m2/C2)(6.00 X 10-9 C)(5.00 X
  10-9 C)
• (5.00m )2
• F13 1.08 X 10-8 N

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• F23 k Q2Q3
• r2
• F23 (9.0 X 109 N-m2/C2)(2.00 X 10-9 C)(5.00 X
  10-9 C)
• (4.00m )2
• F23 5.62 X 10-9 N

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F13 1.08 X 10-8 N
F23 5.62 X 10-9 N
37o

-
q2
q3
37o

q1
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• F13x F13cos37o (1.08 X 10-8 N)cos37o
• F13x 8.63 X 10-9 N
• F13y F13sin37o (1.08 X 10-8 N)sin37o
• F13y 6.50 X 10-9 N
• Fx F23 F13x
• Fx -5.62 X 10-9 N 8.63 X 10-9 N 3.01 X
  10-9 N
• Fy F13y 6.50 X 10-9 N

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• Fx 3.01 X 10-9 N
• Fy 6.50 X 10-9 N
• FR \/ (3.01 X 10-9 N)2 (6.50 X 10-9 N)2
• FR 7.16 X 10-9 N
• sin q Fy/FR
• sin q (6.50 X 10-9 N)/ (7.16 X 10-9 N)
• q 64.7o

FR
Fy
Fx
q
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Coulombs Law Ex 5

• Calculate the net electrostatic force on charge
  Q3 as shown in the diagram

Q3 65 mC

60 cm
30 cm
30o

-
Q2 50 mC
Q1 -86 mC
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• First calculate the forces on Q3 separately
• F13 k Q1Q3
• r2
• F13 (9.0 X 109 N-m2/C2)(65 X 10-6 C)(86 X 10-6
  C)
• (0.60 m )2
• F13 140 N
• F23 (9.0 X 109 N-m2/C2)(65 X 10-6 C)(50 X 10-6
  C)
• (0.30 m )2
• F23 330 N

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F23

Q3
F13
30o

-
Q2
Q1
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• F13x F13cos30o (140 N)cos30o
• F13x 120 N
• F13y -F13sin30o (140 N)sin30o
• F13y -70 N
• Fx F13x 120 N
• F7 330 N - 70 N 260 N

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• Fx 120 N
• Fy 260 N
• FR \/ (120 N)2 (330 N)2
• FR 290 N
• sin q Fy/FR
• sin q (260 N)/ (290 N)
• q 64o

FR
Fy
Fx
q
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• A small plastic bead has a mass of 15 mg and a
  charge of -10 nC (nano 10-9). A glass rod of
  charge 10 nC is held 1.0 cm above the bead.
• Calculate the electric field strength of the rod
  at the position of the bead.
• Calculate the force on the bead.
• Will the bead leap off the table?

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Calculus Example 1

• Calculate the force on charge q from the charged
  rod shown below. The charge per unit length of
  the rod is l Q/l

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• F kqQ
• x2
• dF kqdQ dQ ldx
• x2
• dF kqldx
• x2
• F kql ? dx (from a to al)
• x2
• F -kql 1 a l
• x a

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• F -kql 1 a l
• x a
• F -kql 1 - l
• a1 a
• F kqll
• a(a1)

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Calculus Ex 2

• A total positive charge of Q is evenly
  distributed on a semicircular ring of radius R.
  Calculate the force felt by the charge q at the
  center of the semicircle.

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• F kqQ
• R2
• However, x-components cancel
• Fy Fsinq
• Fy kqQ sinq
• R2
• Break into a small unit of force
• dFy kqdQ sinq
• R2

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• l Q/pR
• dQ lds
• s Rq
• dQ l Rdq
• dFy kq lR sinq dq
• R2

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• Fy kq l - cosq p
• R 0
• Fy 2kq l
• R

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Electric Field

• Contact forces
• Friction
• Pushes and pulls
• Forces at a distance
• Gravity
• Electromagnetism
• Field Invisible lines that extend from a body

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• Proton

A positive test charge would be repelled by the
field


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• Electron

A positive test charge would be attracted by the
field

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• Opposite charges attract

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• Like Charges repel

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• E in terms of a test charge
• E F
• q
• Vector quantity
• Force that the test charge q would feel. The
  smaller the charge, the larger the Force

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• E in terms of a point charge
• E kQ
• r2
• Vector quantity
• Point charge is the source charge producing the
  electric field
• k 1/ 4pe0
• e0 8.85 X 10-12 C2/Nm2 (permittivity constant)

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Electric Field Example 1

• Find the electric force on a proton placed in an
  electric field of 2.0 X 104 N/C
• E F
• q
• F qE
• F (1.602 X 10-19 C)(2.0 X 104 N/C)
• F 3.2 X 10-15N

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Electric Field Example 2

• Calculate the magnitude and direction of an
  electric field at a point 30 cm from a source
  charge of Q -3.0 X 10-6 C.
• E kQ
• r2
• E (9.0 X 109 N-m2/C2)(3.0 X 10-6 C)
• (0.30 m)2
• E 3.05 X 105 N/C towards the charge

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• What electric field is required for a copier to
  carry toner particles of mass 9.0 X 10-16 kg.
  Each particle carries 20 electrons to provide the
  test charge.
• Assume the copier must overcome twice the weight
  of each particle.

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• E F
• q
• F 2mg (twice the weight)
• E 2mg (2)(9.0 X 10-16 kg)(9.8m/s2)
• q (20)(1.602X10-19C)
• E 5500 N/C

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Electric Field Example 3

• Two point charges are separated by a distance of
  10.0 cm. What is the magnitude and direction of
  the electric field at point P, 2.0 cm from the
  negative charge?

2 cm
8 cm
P
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Q1 -25 mC
Q2 50 mC
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E2
E1
2 cm
8 cm
P
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Q1 -25 mC
Q2 50 mC
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• E E1 E2 (both point to the left)
• E kQ
• r2
• E1 (9.0 X 109 N-m2/C2)(25 X 10-6 C) (0.020
  m)2
• E1 5.625 X 108 N/C
• E2 (9.0 X 109 N-m2/C2)(50 X 10-6 C) (0.080
  m)2
• E2 7.031 X 107 N/C
• E E1 E2 6.3 X 108 N/C

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Electric Field Example 3a

• What acceleration would an electron feel if it
  were placed at point P? Would it move to the
  right or the left? An electron has a mass of 9.1
  X 10-31 kg.

E2
E1
2 cm
8 cm
P
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Q1 -25 mC
Q2 50 mC
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• E F
• q
• F ma
• E ma
• q
• a Eq/m
• a (6.3 X 108 N/C)(1.602 X 10-19 C)
• (9.1 X 10-31 kg)
• a 1.1 X 1020 m/s2

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Electric Field Example 4

• Charge Q1 7.00 mC is placed at the origin.
  Charge Q2 -5.00 mC is placed 0.300 m to the
  right. Calculate the electric field at point P,
  0.400 m above the origin.

P
0.400 m
0.300 m
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Q1 7.00 mC
Q2 -5.00 mC
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P
0.400 m
c 0.500 m
q
0.300 m
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Q1 7.00 mC
Q2 -5.00 mC
c2 a2 b2 c2 (0.400 m)2 (0.300 m)2 c
0.500 m tan q opp/adj 0.400/0.300 q 53.1o
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• E kQ
• r2
• E1 (9.0 X 109 N-m2/C2)(7.00 X 10-6 C) (0.400
  m)2
• E1 3.94 X 105 N/C
• E2 (9.0 X 109 N-m2/C2)(5.00 X 10-6 C) (0.500
  m)2
• E2 1.80 X 105 N/C

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E1
E2x
P
E2
E2y
q
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E2x E2cosq (1.80 X 105 N/C)(cos 53.1o) E2x
1.08 X 105 N/C (to the right) E2y E2sinq
(1.80 X 105 N/C)(sin 53.1o) E2y -1.44 X 105 N/C
(down)
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• Ex E2x
• Ex 1.08 X 105 N/C (to the right)
• Ey E1 E2y
• Ey 3.94 X 105 N/C -1.44 X 105 N/C
• Ey 2.49 X 105 N/C
• ER2 (1.08 X 105 N/C )2 (2.49 X 105 N/C)2
• ER 2.72 X 105 N/C
• tan f Ey/Ex 2.49 X 105/ 1.08 X 105
• f 66.6o

f
P
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Electric Field Example 5

• Calculate the electric field at point A, as shown
  in the diagram

A
30 cm
52 cm
-

Q1 50.0 mC
Q2 -50.0 mC
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• Ans E 4.5 X 106 N/C at an angle of 76o

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Electric Field Example 6

• Calculate the electric field at point B, as shown
  in the diagram.

B
30 cm
26 cm
26 cm
-

Q1 50.0 mC
Q2 -50.0 mC
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• Ans E 3.6 X 106 N/C along the x direction