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Lecture 3: Chapter 19 Cont Electric Charges, Forces and Electric Fields
Lecture 3: Chapter 19 Cont
Electric Charges, Forces and
Electric Fields
Agenda
Compare the electric force to Gravitational force
Superposition of forces
Electric Field
Superposition of Electric Field
Shielding and Charging by Induction
Electric Flux and Gauss’s Law
What did we study last lesson?
Electric Charges: positive and neg.
Electron (neg) and proton (pos)
Charges are conserved
Polarization
Insulators & Conductors
Coulomb’s Law
Recap
Two rods and a cat fur:
Glass Rod (positive charges)
Rubber Rod (negative)
Charging by Conduction
Recap
Coulomb’s Law
(electrostatic force between point charges)
2
21
er
qqkF
7
Recap
Polarization is realignment of charge within individual molecules.
Produces induced charge on the surface of insulators.
how e.g. rubber or glass can be used to supply electrons.
8
Student Discussion
A positively charged object hanging from a string is brought near a non conducting object (ball). The ball is seen to be attracted to the object.
1.Explain why it is not possible to determine whether the object is negatively charged or neutral.
2.What additional experiment is needed to reveal the electrical charge state of the object?
? +
9
Cont
Two possibilities:
Attraction between objects of unlike charges.
Attraction between a charged object and a neutral object subject to polarization.
- +
+ - - - -
+ + + +
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What additional experiment is needed to reveal the electrical charge state of the object?
Two Experiments:
Bring known neutral ball near the object and observe whether there is an attraction.
Bring a known negatively charge object near the first one. If there is an attraction, the object is neutral, and the attraction is achieved by polarization.
? 0
- + + + +
+- -+-
+ -
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Question
Name the first action at a distance force you have encountered in physics so far.
Electric Force
Electrical Forces are Field Forces
This is the second example of a field force Gravity was the first
Remember, with a field force, the force is exerted by one object on another object even though there is no physical contact between them
There are some important similarities and differences between electrical and gravitational forces
Electrical Force Compared to Gravitational Force
Both are inverse square laws
The mathematical form of both laws is the same Masses replaced by charges
G replaced by ke
Electrical forces can be either attractive or repulsive
Gravitational forces are always attractive
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Consider a proton (mp=1.67x10-27 kg; qp=+1.60x10-19 C) and an electron (me=9.11x10-31 kg; qe=-1.60x10-19 C) separated by 5.29x10-11 m. The particles are attracted to each other by both the force of gravity and by Coulomb’s law force. Which of these has the larger magnitude? A. Gravitational force B. Coulomb’s law force
Example: Student Participation
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Which of these has the larger magnitude? A. Gravitational force B. Coulomb’s law force
N102.8
1029.5
1060.11060.11099.8 :Coulomb
N106.31029.5
1011.91067.11067.6 :Gravity
8
211
19199
2
47
211
312711
2
r
qqkF
r
mGmF
epe
ep
We will mostly ignore gravitational effects when we
consider electrostatics.
The steps will be found on Pg. 659. By what factor the electric force is greater than the gravitational force?
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Problem solving steps
1. Visualize problem – labeling variables
2. Determine which basic physical principle(s) apply
3. Write down the appropriate equations using the variables
defined in step 1.
4. Check whether you have the correct amount of
information to solve the problem (same number of
knowns and unknowns).
5. Solve the equations.
6. Check whether your answer makes sense (units, order of
magnitude, etc.).
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Superposition Principle
From observations: one finds that whenever multiple charges are present, the net force on a given charge is the vector sum of all forces exerted by other charges.
Electric force obeys a superposition principle.
The Superposition Principle
How to work the problem?
Find the electrical forces between pairs of charges separately
Then add the vectors
Remember to add the forces as vectors
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F12 F13
Superposition Principle Example: 3 charges in a line
q1 = 6.00 uC q2 = 1.50 uC q3 = -2.00 uC d1 = 3.00 cm d2 = 2.00 cm K = 8.99 x 109 N.m2 /C2
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Example: Superposition Principle (Not along a line)
Consider three point charges at the corners of a triangle, as shown in the next slide. Find the resultant force on q3 if
q1 = 6.00 x 10-9 C
q2 = -2.00 x 10-9 C
q3 = 5.00 x 10-9 C
Superposition Principle Example
The force exerted by q1 on q3 is
The force exerted by q2 on q3 is
The total force exerted on q3 is the vector sum of
and
13F
13F
23F
23F
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Consider three point charges at the corners of a triangle, as shown below. Find the resultant force on q3.
Solution:
2
2
2
2
9 9
3 2 9 9
32 22
9 9
3 1 9 8
31 22
9
32 31
9
31
2 2 9
5.00 10 2.00 108.99 10 5.62 10
4.00
5.00 10 6.00 108.99 10 1.08 10
5.00
cos37.0 3.01 10
sin 37.0 6.50 10
7.16 10
Nme C
Nme C
o
x
o
y
x y
C Cq qF k N
r m
C Cq qF k N
r m
F F F N
F F N
F F F N
65.2o
Spherical Charge Distribution
Students need to read and work on pg 663/664
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Electric Field - Discovery
Electric forces act through space even in the absence of physical contact.
Suggests the notion of electrical field (first introduced by Michael Faraday (1791-1867).
Electric Field
A charged particle, with charge Q, produces an electric field in the region of space around it
A small test charge, qo, placed in the field, will experience a force.
“The electric field is the force per charge at a given location”
Electric Field
Mathematically,
SI units are N / C
o
FE
q
Given: One finds:
2
o
e
q qF k
r
2e
qE k
r
Electric Field
Use this for the magnitude of the field
The electric field is a vector quantity
The direction of the field is defined to be the direction of the electric force that would be exerted on a small positive test charge placed at that point
o
FE
q
Direction of Electric Field
The electric field produced by a negative charge is directed toward the charge
A positive test charge would be attracted to the negative source charge
Direction of Electric Field, cont
The electric field produced by a positive charge is directed away from the charge
A positive test charge would be repelled from the positive source charge
Electric Field
More About a Test Charge and The Electric Field
The test charge is required to be a small charge
It can cause no rearrangement of the charges on the source charge
The electric field exists whether or not there is a test charge present
The Superposition Principle can be applied to the electric field if a group of charges is present
Electric Fields and Superposition Principle
The superposition principle holds when calculating the electric field due to a group of charges
Find the fields due to the individual charges
Add them as vectors
Use symmetry whenever possible to simplify the problem
Problem Solving Strategy, cont
Apply Coulomb’s Law
For each charge, find the force on the charge of interest
Determine the direction of the force
Sum all the x- and y- components
This gives the x- and y-components of the resultant force
Find the resultant force by using the Pythagorean theorem and trig
Examples of Electric Field
a) E will be constant (a charged plane)
b) E will decrease with distance as (a point charge)
c) E in a line charge, decrease with a distance 1/r
2
1
r
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Example:
An electron moving horizontally passes between two horizontal planes, the upper plane charged negatively, and the lower positively.
- vo
- - - - - - - - - - - - - - - - - - - - - -
+ + + + + + + + + + + + + + + + + + + + + +
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Example:
A uniform, upward-directed electric field exists in this region. This field exerts a force on the electron. Describe the motion of the electron in this region.
- vo
- - - - - - - - - - - - - - - - - - - - - -
+ + + + + + + + + + + + + + + + + + + + + +
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- vo
- - - - - - - - - - - - - - - - - - - - - -
+ + + + + + + + + + + + + + + + + + + + + +
Observations:
Horizontally:
No electric field
No force
No acceleration
Constant horizontal velocity
0
0
0
x
x
x
x o
o
E
F
a
v v
x v t
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- vo
- - - - - - - - - - - - - - - - - - - - - -
+ + + + + + + + + + + + + + + + + + + + + +
Observations:
Vertically:
Constant electric field
Constant force
Constant acceleration
Vertical velocity increase linearly with time.
2
/
/
1/
2
y o
y o o
y o o o
y o o o
o o o
E E
F q E
a q E m
v q E t m
y q E t m
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-
- - - - - - - - - - - - - - - - - - - - - -
+ + + + + + + + + + + + + + + + + + + + + +
Conclusions:
The charge will follow a parabolic path downward.
Motion similar to motion under gravitational field only except the downward acceleration is now larger.
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Example: Electric Field Due to Two Point Charges (Superposition of Fields)
Question:
Charge q1=7.00 mC is at the origin, and charge q2=-10.00 mC is on the x axis, 0.300 m from the origin. Find the electric field at point P, which has coordinates (0,0.400) m.
x
y
0.300 m q1 q2
0.4
00 m
P
E1
E2
E
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Question: Charge q1=7.00 mC is at the origin, and charge q2=-10.00 mC is on the x axis, 0.300 m from the origin. Find the electric field at point P, which has coordinates (0,0.400) m.
Observations: First find the Electric field at point P due to
charge q1 and q2.
Field E1 at P due to q1 is vertically upward.
Field E2 at due to q2 is directed towards q2.
The net field at point P is the vector sum of E1 and E2.
The magnitude is obtained with
2e
qE k
r
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Question: Charge q1=7.00 mC is at the origin, and charge q2=-10.00 mC is on the x axis, 0.300 m from the origin. Find the electric field at point P, which has coordinates (0,0.400) m.
Set up the problem:
q1=7.00 mC
q2=-10.00 mC
K = 8.99 x 109 N. m2 /C2
r1 = 0.400m
r2= ? What do we do?
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Question: Charge q1=7.00 mC is at the origin, and charge q2=-10.00 mC is on the x axis, 0.300 m from the origin. Find the electric field at point P, which has coordinates (0,0.400) m.
1) Calculate r2
2) Calculate the electric field at P
3) I will set it up, but you will need to finish (it looks like example in pg. 669)
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Question: Charge q1=7.00 mC is at the origin, and charge q2=-10.00 mC is on the x axis, 0.300 m from the origin. Find the electric field at point P, which has coordinates (0,0.400) m.
Solution:
Electric Field Lines
A convenient aid for visualizing electric field patterns is to draw lines pointing in the direction of the field vector at any point
These are called electric field lines and were introduced by Michael Faraday
Electric Field Line Patterns
Point charge
The lines radiate equally in all directions
For a positive source charge, the lines will radiate outward
Electric Field Line Patterns
For a negative source charge, the lines will point inward
Electric Field Line Patterns
An electric dipole consists of two equal and opposite charges
The high density of lines between the charges indicates the strong electric field in this region
Electric Field Line Patterns
Two equal but like point charges
At a great distance from the charges, the field would be approximately that of a single charge of 2q
The bulging out of the field lines between the charges indicates the repulsion between the charges
The low field lines between the charges indicates a weak field in this region
Electric Field Patterns
Unequal and unlike charges
Note that two lines leave the +2q charge for each line that terminates on -q
Rules for Drawing Electric Field Lines
The lines for a group of charges must begin on positive charges and end on negative charges In the case of an excess of charge, some
lines will begin or end infinitely far away
The number of lines drawn leaving a positive charge or ending on a negative charge is proportional to the magnitude of the charge
No two field lines can cross each other
Electric Field Lines, final
The electric field lines are not material objects
They are used only as a pictorial representation of the electric field at various locations
They generally do not represent the path of a charged particle released in the electric field
Question:
Is it safe to stay inside an automobile during a lightning storm? Why?
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Mini-Discussion
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Question:
Is it safe to stay inside an automobile during a lightning storm? Why?
Answer:
Yes. It is. The metal body of the car carries the excess charges on its external surface. Occupants touching the inner surface are in no danger.
SAFE
Shielding and Charging by Induction
When no net motion of charge occurs within a conductor, the conductor is said to be in electrostatic equilibrium
An isolated conductor has the following properties: The electric field is zero everywhere inside the
conducting material
Any excess charge on an isolated conductor resides entirely on its surface
Properties, cont
The electric field just outside a charged conductor is perpendicular to the conductor’s surface
On an irregularly shaped conductor, the charge accumulates at locations where the radius of curvature of the surface is smallest (that is, at sharp points)
Property 1
The electric field is zero everywhere inside the conducting material Consider if this were not true
If there were an electric field inside the conductor, the free charge there would move and there would be a flow of charge
If there were a movement of charge, the conductor would not be in equilibrium
Property 2
Any excess charge on an isolated conductor resides entirely on its surface
A direct result of the 1/r2 repulsion between like charges in Coulomb’s Law
If some excess of charge could be placed inside the conductor, the repulsive forces would push them as far apart as possible, causing them to migrate to the surface
Electric Flux
Field lines penetrating an area A perpendicular to the field
The product of EA is the flux, Φ
In general:
ΦE = E A cos θ
Electric Flux, cont.
ΦE = E A cos θ The perpendicular to the area A is at an
angle θ to the field
SI unit: N.m2 /C
If the surface is close, the sign is The flux is positive for field lines that leave
the enclosed volume of the surface
The flux is negative for field lines that enter the enclosed volume of the surface.
Gauss’ Law
Gauss’ Law states that the electric flux through any closed surface is equal to the net charge Q inside the surface divided by εo
εo is the permittivity of free space and equals 8.85 x 10-12 C2/Nm2
The area in Φ is an imaginary surface, a Gaussian surface (spherical surface) (derivation in pg. 676)
insideE
o
Q
SI unit;
N.m2 /C
Electric Field of a Charged Thin Spherical Shell
The calculation of the field outside the shell is identical to that of a point charge
The electric field inside the shell is zero
2e
o
2 r
Qk
r4
QE
Electric Field of a Nonconducting Plane Sheet of Charge
Use a cylindrical Gaussian surface
The flux through the ends is EA, there is no field through the curved part of the surface
The total charge is Q = σA
Note, the field is uniform
o2E
Electric Field of a Nonconducting
Plane Sheet of Charge, cont.
The field must be perpendicular to the sheet
The field is directed either toward or away from the sheet
Parallel Plate Capacitor
The device consists of plates of positive and negative charge
The total electric field between the plates is given by
The field outside the plates is zero
o
E
What did we study?
Electric Force (Coulomb’s force): Superposition of forces
l Electric Field (point charge):
Definition of electric Flux, :
Gass’s Law electric Flux (q is enclosed by a surface):
2
21
er
qqkF
2e
qE k
r
ΦE = E A cos θ
insideE
o
Q
