Transcript q 2

```Electric Charge and the Electric Field
(Ch 26)
Four Fundamental Forces
1. Electromagnetic
2. Gravity
3. Strong Nuclear Force
4. Weak Nuclear Force
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
Electric Charge
• Law of conservation of charge – Charge is never
created or destroyed
Cloth
- + +++ +
Rubber rod
Insulators/Conductors
Insulators
Do not conduct
electricity
•Small atoms
•Electrons tightly
held
•Non-Metals
Semiconductors
Conduct
electricity, but
only at a higher
voltage
•Medium atoms
•Electrons held
medium
•Metalloids
Conductors
Conduct electricity
well
•Large atoms
•Electrons not
tightly held
•Metals
(Discuss movement of the electrons)
Conduction
• Transfer of charge by touching
• Electrons can flow between substances
• Can produce a net charge on the substance
Induction
• Transfer of charge without touching
• Electrons migrate within the substance to cause a
separation of charge
• Net charge on substance is still zero
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
Electroscope
Charging the electroscope
by induction
by conduction
(charge separation) (net charge)
Using the electroscope
• Can be used to detect the charge on a substance
• (Must charge electroscope first)
Coulomb’s Law
• Coloumb experimented to determine magnitude
of electromagnetic force
• Measured the forces between charged spheres
(angle of deflection).
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)
The Coulomb
• 1 Coulomb = 1 Ampere•second
• Unit of charge
• Point charges – small objects (charge doesn’t get
distributed much)
• Elementary Charge
– Charge on the electron and proton
– Quantized (can’t have ½ an electron)
– e = 1.602 X 10-19 C
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)
+
-
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)
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
Q1= -8.0 mC
0.20 m
+
Q2= +3.0 mC
Q3 = -4.0 mC
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)
Fnet = F31 - F32
Fnet = 1.2 N – 2.7 N = -1.5 to the left
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.
a) Calculate the net force on the center charge.
b) Will the center charge move to the left or right?
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?
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?
Coulomb’s 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
q2
q3
=
=
=
+6.00 X 10-9 C
-2.00 X 10-9 C
+5.00 X 10-9 C
q2
-
+
37o
3.00 m
q1
4.00 m
+
5.00 m
q3
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
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
F13 = 1.08 X 10-8 N
F23 = 5.62 X 10-9 N
q2
-
q1
+
+
37o
37o
q3
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
FR
Fx = 3.01 X 10-9 N
Fy = 6.50 X 10-9 N
Fy
- q
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
Fx
Coulomb’s Law: Ex 5
Calculate the net electrostatic force on charge Q3 as
shown in the diagram:
+
Q3 = +65 mC
60 cm
30 cm
+
Q2 = +50 mC
30o
-
Q1 = -86 mC
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
F23
Q3
+
+
Q2
F13
30o
Q1
-
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
FR
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
Fy
- q
Fx
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.
a. Calculate the electric field strength of the rod at
the position of the bead.
b. Calculate the force on the bead.
c. Will the bead leap off the table?
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
F = kqQ
x2
dF = kqdQ
x2
dF = kqldx
x2
F = kql ∫ dx
x2
F = -kql 1 a+ l
x
a
dQ = ldx
(from “a” to “a+l”)
F = -kql 1
x
a+ l
a
F = -kql 1 - l
a+1 a
F = kqll
a(a+1)
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.
F = kqQ
R2
However, x-components cancel
Fy = Fsinq
Fy = kqQ sinq
R2
Break into a small unit of force
dFy = kqdQ sinq
R2
l = Q/pR
dQ = lds
s = Rq
dQ = l Rdq
dFy = kq lR sinq dq
R2
Fy = kq l - cosq
R
Fy = 2kq l
R
p
0
Electric Field
• Contact forces
– Friction
– Pushes and pulls
• Forces at a distance
– Gravity
– Electromagnetism
• Field – Invisible lines that extend from a body
Proton
+
+
A positive
test charge
would be
repelled by
the field
Electron
+
-
A positive
test charge
would be
attracted
by the field
• Opposite charges attract
• Like Charges repel
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
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)
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
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
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.
E= F
q
F = 2mg (twice the weight)
E = 2mg
q
= (2)(9.0 X 10-16 kg)(9.8m/s2)
(20)(1.602X10-19C)
E = 5500 N/C
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
Q1 = -25 mC
8 cm
P
+
Q2 = +50 mC
E2
E1
2 cm
Q1 = -25 mC
8 cm
P
+
Q2 = +50 mC
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
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
Q1 = -25 mC
8 cm
P
+
Q2 = +50 mC
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
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
Q1 = 7.00 mC
Q2 = -5.00 mC
P
0.400 m
c = 0.500 m
+
0.300 m
Q1 = 7.00 mC
q
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
-
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
E1
P
+
E2x
E2
E2y
q
-
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)
Ex = E2x
Ex = 1.08 X 105 N/C (to the right)
P
f
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
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
Ans: E = 4.5 X 106 N/C at an angle of 76o
Electric Field: Example 6
Calculate the electric field at point B, as shown in
the diagram.
B
30 cm
+
26 cm
Q1 = +50.0 mC
26 cm
-
Q2 = -50.0 mC
Ans: E = 3.6 X 106 N/C along the +x direction
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