Electric Field - Eleanor Roosevelt High School

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Lesson 17
Electric Fields
and Potential
Eleanor Roosevelt High School
Chin-Sung Lin
Electric Fields
Gravitational & Electric
Forces
 What are the formulas for the following physics laws?
 Law of Universal Gravitation
 Coulomb’s Law
Gravitational & Electric
Forces
 Law of Universal Gravitation
Fg = G
m1 m2
r2
 Coulomb’s Law
Fe = k
q1 q2
r2
Gravitational Field
 What’s the definition of gravitational field?
Gravitational Field
 Gravitational Field: Force per unit mass
g=
Fg
m
• Fg:
gravitational force (N)
• m:
mass (kg)
• g:
gravitational field strength (N/kg, or m/s2)
Electric Field
 Electric Field: Force per unit charge
E=
Fe
q
• Fe:
electric force (N)
• q:
charge (C)
• E:
electric field strength (N/C)
Gravitational & Electric Fields
 Gravitational Field
g=
Fg
m
 Electric Field
E=
Fe
q
Electric Field
Source Charge
+Q
Test Charge
+q
r
Fe
E=
Fe
q
Electric Field
Source Charge
–Q
Test Charge
r
+q
Fe
E=
Fe
q
Electric Field
 Electric field is a vector
 A vector includes ___________ and ____________
Source Charge
Test Charge
+Q
–Q
+q
r
r
Fe
Fe
+q
E=
Fe
q
Electric Field
 Electric field is a vector
 A vector includes direction and magnitude
Source Charge
Test Charge
+Q
–Q
+q
r
r
Fe
Fe
+q
E=
Fe
q
Electric Field
 Can you apply Coulomb’s law to this formula and
then simplify it?
E=
Fe
q
= ???
Source Charge
+Q
Test Charge
r
+q
Fe
Electric Field
 Electric Field: Force per unit charge
E=
•
•
•
•
•
•
Fe:
q:
Q:
E:
r:
k:
Fe
q
qQ
=k
r2
q
Q
= k
electric force (N)
test charge (C)
source charge (C)
electric field strength (N/C)
distance between charges (m)
electrostatic constant (N m2/C2)
r2
Electric Field Example
 What is the magnitude of the electric field strength
when an electron experiences a 5.0N force?
Electric Field Example
 What is the magnitude of the electric field strength
when an electron experiences a 5.0N force?
E = Fe / q
E = 5 N / (1.6 x 10-19 C)
= 3.13 x 1019 N/C
Electric Field Example
 What are the magnitude and direction of the
electric field 1.5 m away from a positive charge of
2.1*10-9 C?
Electric Field Example
 What are the magnitude and direction of the
electric field 1.5 m away from a positive charge of
2.1*10-9 C?
E = k Q / r2
E = (8.99 x 109 N m2/C2) (2.1 x 10-9 C) / (1.5 m)2
= 8.4 N/C
Direction: away from the positive charge
Electric Field Exercise
 There is a negative charged particle of 0.32 C in the
free space. (a) What are the magnitude and
direction of the electric field 2.0 m away from the
particle? (b) What are the magnitude and direction
of the electric force when an electron is placed 2.0
m away from this particle?
[3 minutes]
– 0.32 C
2.0 m
e–
Electric Field Exercise
 There is a negative charged particle of 0.32 C in the
free space. (a) What are the magnitude and
direction of the electric field 2.0 m away from the
particle?
E = k Q / r2
E = (8.99 x 109 N m2/C2) (0.32 C) / (2.0 m)2
= 7.2 x 108 N/C
Direction: toward the negative charge
Electric Field Exercise
 There is a negative charged particle of 0.32 C in the
free space. (b) What are the magnitude and
direction of the electric force when an electron is
placed 2.0 m away from this particle?
E = Fe / q
Fe = q E
Fe = (1.6 x 10-19 C) (7.2 x 108 N/C)
= 1.15 x 10-10 N
Aim: Electric Field
DoNow:
(4 minutes)
 Write down the definition of Electric Field in words
 Write down the formulas of Electric Field in two
different forms
 Define every symbol in the formula and identify
their units
 Identify the relationships between Electric Field
and other variables
Aim: Electric Field
Electric Field
 Electric Field: Force per unit charge
E=
•
•
•
•
•
•
Fe:
q:
Q:
E:
r:
k:
Fe
q
Q
= k
r2
electric force (N)
test charge (C)
source charge (C)
electric field strength (N/C)
distance between charges (m)
electrostatic constant (N m2/C2)
Electric Field
 Electric Field: Force per unit charge
E ~ Fe E ~ Q
•
•
•
•
•
•
Fe:
q:
Q:
E:
r:
k:
1
E~ q
electric force (N)
test charge (C)
source charge (C)
electric field strength (N/C)
distance between charges (m)
electrostatic constant (N m2/C2)
1
E~ 2
r
Electric Field
 If you shift the test charge around, where can you
find the electric field with the same magnitude?
Source Charge
Test Charge
+Q
+q
Fe
E=
Fe
q
Electric Field
Source Charge
+Q
Test Charge
+q
E
Fe
Electric Field
 What will happen if you move the test charges
away from the source charge?
Source Charge
+Q
Test Charge
+q
E
Fe
Electric Field
Source Charge
+Q
Test Charge
+q
E
Fe
Electric Field
Source Charge
+Q
Test Charge
E
+q
Fe
E
Fe
Electric Field
Source Charge
+Q
Test Charge
+q
E
Test Charge
+q
Fe
E
Fe
Electric Field Representation
Vector representation
+
-
Electric Field Representation
Line-of-Force representation
+
-
Electric Field Representation
 How do you decide the strength of electric field?
+
-
Electric Field Representation
 When the field lines are denser, the field is stronger
+
-
Electric Field Representation
 Where can you find the the strongest electric field?
E
C
A
B
D
Electric Field: Point Charge
Line-of-Force representation
+
-
Electric Field: Pair of Charges
Line-of-Force representation
Electric Field: Pair of Charges
 Sketch the electric field for like charges?
+
+
Electric Field: Pair of Charges
Line-of-Force representation
+
+
Electric Field: Pair of Charges
Line-of-Force representation
Electric Field: Parallel Plates
Line-of-Force representation
Electric Field: Parallel Plates
 Anything special for the electric field between the
parallel plates charged with opposite charges?
Electric Field: Parallel Plates
 The electric field between the parallel plates is
uniform except at both ends
Electric Field Example
 A charged droplet of mass 5.87 x 10-10 kg is
hovering motionless between two parallel plates.
The parallel plates have a electric field of 1.2 x 107
N/C and are 2.00 mm apart. (a) What is the charge
on the particle? (b) By how many electrons is the
particle deficient?
Electric Field Example
 A charged droplet of mass 5.87 x 10-10 kg is
hovering motionless between two parallel plates.
The parallel plates have a electric field of 1.2 x 107
N/C and are 2.00 mm apart. (a) What is the charge
on the particle?
E = Fe / q
Fe = E q
Fg = m g
Fe = F g
Eq=mg
(1.2 x 107 N/C) q = (5.87 x 10-10 kg) (9.81 m/s2)
q = 4.80 x 10-16 C
Electric Field Example
 A charged droplet of mass 5.87 x 10-10 kg is
hovering motionless between two parallel plates.
The parallel plates have a electric field of 1.2 x 107
N/C and are 2.00 mm apart. (b) By how many
electrons is the particle deficient?
e - = 1.6 x 10-19 C
number of e - = 4.80 x 10-16 C / 1.6 x 10-19 C = 3000 e –
Electric Field Exercise
 A charged droplet of mass 5.87 x 10-10 kg is
hovering motionless between two parallel plates.
The parallel plates have a electric field of 9.6 x 106
N/C and are 2.00 mm apart. What is the charge on
the particle?
Electric Field Exercise
 A charged droplet of mass 5.87 x 10-10 kg is
hovering motionless between two parallel plates.
The parallel plates have a electric field of 9.6 x 106
N/C and are 2.00 mm apart. What is the charge on
the particle?
E = Fe / q
Fe = E q
Fg = m g
Fe = F g
Eq=mg
(9.6 x 106 N/C) q = (5.87 x 10-10 kg) (9.81 m/s2)
q = 6.0 x 10-16 C
Electric Field Exercise
 A positively charged ball with mass 20 g is hanging
between two charged parallel plates from the
ceiling through an insulating wire with length 0.1 m.
The electric field strength of the charged parallel
plates is 4.2 x 109 N/C. When the ball is in balance,
the wire and the vertical line form an angle of 60o.
What is the charge of the ball?
Electric Fields
Electric Fields
Electric Fields
Electric Fields
Electric Fields
Electric Fields and Shielding
E=0
Electric Shielding
Electric Fields and Shielding
 Cancellation of electric force
 The electric forces of area A
and area B on P completely
cancel out
A
P
B
Electric Potential
Aim: Electric Potential
DoNow:
(3 minutes)
 Write down the formulas of Electric Field
 Draw the electric field surrounding a pair of
opposite charge
 Draw the electric field surrounding a pair of
charged parallel plates
Aim: Electric Potential
Electric Field
 Electric Field: Force per unit charge
E=
•
•
•
•
•
•
Fe:
q:
Q:
E:
r:
k:
Fe
q
Q
= k
r2
electric force (N)
test charge (C)
source charge (C)
electric field strength (N/C)
distance between charges (m)
electrostatic constant (N m2/C2)
Electric Field: Pair of Charges
Line-of-Force representation
Electric Field: Parallel Plates
Line-of-Force representation
Gravitational Potential Energy (GPE)
Gravitational Potential Energy (GPE)
 What happens in the
picture?
 What types of energy
have been converted?
Gravitational Potential Energy (GPE)
 What happens in the
picture?
 What types of energy
have been converted?
 If we want to pull the
weight up back to its
original position, what
should we do?
Gravitational Potential Energy (GPE)
 What happens in the
picture?
 What types of energy
have been converted?
 If we want to pull the
weight up back to its
original position, what
should we do?
 How much work do we
need?
Electric Potential Energy (EPE)
What are they in common?
+
+
-
+
+
--
Electric Potential Energy (EPE)
 What happens in the
picture?
 What types of energy
have been converted?
+
+
 If we want to pull the
weight up back to its
original position, what
should we do?
 How much work do we
need?
-
+
+
--
Electric Potential Energy (EPE)
What are they different?
+
+
-
+
+
--
Electric Potential Energy
What did the monkey do in order to bring a positively
charged ball toward a positively charged object?
Electric Potential Energy
What did the monkey do in order to bring a positively
charged ball toward a positively charged object?
When the ball was released, what happened to the ball?
Why?
Gravitational Potential Energy
 The work performed in taking a mass from height A
to height B does not depend on the path
B
A
A
Electric Potential Energy
 The work performed in taking a charge from A to B
does not depend on the path
B
A
Gravitational Potential Energy
 What happens when we
double the mass?
x2
Gravitational Potential Energy
 When mass is doubled, the
gravitational potential energy
is also doubled
x2
 PE2 = (2m)gh = 2(mgh) = 2PE1
Electric Potential Energy
 What did we double here?
+ +
+ +
x2
-
Electric Potential Energy
 We doubled the charge.
+ +
+ +
x2
-
 What happens when we
double the charge?
Electric Potential Energy
+ +
+ +
x2
-
 When charge is doubled, the
electric potential energy is
also doubled
Electric Potential Energy
 Potential Energy – Capability to Do Work
W
• W:
or
EPE
electric (potential) energy
aka work (J)
W is a scalar (not a vector)
Electric Potential Energy
 Potential Energy – Capability to Do Work
W
• W:
or
EPE
electric (potential) energy
aka work (J)
W
~
q
Electric Potential
 Electric potential energy per unit charge
V=
W
q
• W:
electric (potential) energy, aka work (J)
• V:
electric potential, aka potential difference,
aka voltage (V, Volts)
• q:
charge (C)
Electric Potential
W = qV
 Electric potential (V) is based on a zero reference
point
 Only the potential difference matters
 Electric potential (Voltage, V) is the work (W)
required to bring a unit charge (1 C) from the zero
reference point
 V is a scalar (not a vector)
Electric Potential Example
 How much work is required to move 3.0 C of
positive charge from the negative terminal of a 12volt battery to the positive terminal?
Electric Potential Example
 How much work is required to move 3.0 C of
positive charge from the negative terminal of a 12volt battery to the positive terminal?
V=W/q
W=qV
W = (3.0 C) (12.0 V) = 36.0 J
Electric Potential Example
 If an electron loses 1.4 x 10-15 J of energy in
traveling from the cathode to the screen of a
computer monitor, across what potential
difference must it travel?
Electric Potential Example
 If an electron loses 1.4 x 10-15 J of energy in
traveling from the cathode to the screen of a
computer monitor, across what potential
difference must it travel?
V=W/q
W=qV
V = (1.4 x 10-15 J) / (1.6 x 10-19 C) = 8750 V
Electric Potential Example
 Can you make up a question using the definition of
electric potential?
Electric Potential Example
Electric Potential: Parallel Plates
 What is special about the electric field between the
charged parallel plates?
Electric Potential: Parallel Plates
 Electric Field (E): is uniform due to uniform density of
the electric field lines
E
E
E
Electric Potential: Parallel Plates
 If we place test charge at different locations between
the charged parallel plates, compare the forces
experienced by these test charges
E
E
E
Electric Potential: Parallel Plates
 Electric Force (Fe): experienced by a test charge is
constant due to Fe = qE
E
E
Fe
E
Fe
Fe
Electric Potential: Parallel Plates
 Compare the work required to move test charges
from the negative plate to the positive plate
E
Fe
E
Fe
Fe
E
d
Electric Potential: Parallel Plates
 Electric potential energy / work (W): required to
move a test charge from negative plate to positive
plate is constant due to W = Fe d
E
E
Fe
Fe
E
Fe
d
Electric Potential: Parallel Plates
 Given the following formulas, can you derive the
formula for Electric potential (V) and Electric field
(E)?
W = Fe d
Fe = q E
W
V= q
Electric Potential: Parallel Plates
 Electric potential (V):
W = Fe d
and
Fe = q E
W=qEd
qEd
W
V= q = q =Ed
V
V = Ed
or
E= d
Electric Potential: Parallel Plates
 Electric potential (V): relative to the negative plate
is proportional to the distance to it due to V = E d
E
E
E
V
d
Equipotential Lines: Parallel Plates
 Equipotential Lines: on an equipotential line,
voltages are all the same
 Equipotential lines are perpendicular to field lines
V2
E
V1
V2
V2
E d2
E
V1
V1
d1
Equipotential Lines: Point Charge
 Equipotential Lines for a point charge:
Equipotential Lines: Point Charge
 Equipotential Lines for a charge pair
Electric Field and Potential
 Electric field
V
E= d =
• E:
• V:
• d:
• q:
Fe
q
electric field (N/C or V/m)
electric potential / potential difference /
voltage (V, Volts)
distance between parallel plates (m)
charge (C)
Electric Potential Example
 A charged droplet of mass 5.87x 10-10 kg is hovering
motionless between two parallel plates. The
parallel plates have a potential difference of 24000
V and are 2.00 mm apart. What is the charge on the
particle?
Electric Potential Example
 A charged droplet of mass 5.87x 10-10 kg is hovering
motionless between two parallel plates. The
parallel plates have a potential difference of 24000
V and are 2.00 mm apart. What is the charge on the
particle?
E = Fe / q
Fe = E q
Fg = m g
Fe = F g
Eq=mg
E=V/d
Vq/d=mg
(24000 V) q / ((0.002 m) = (5.87 x 10-10 kg) (9.81 m/s2)
q = 4.80 x 10-16 C
The End

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