Download Lecture 2 - The Local Group

Lecture 2 — [Chapter 21]
Tuesday, Jan 17th
Administrative Items
Assignments this week:
— read Ch 21 and Ch 22 in the textbook
— complete “Pre-Lecture Ch22 HW” assignment
— complete “Ch 21 HW” assignment
[Pre-Lecture Ch 22 HW will be available today and due Thursday by 8am.]
[Ch 21 HW should be available today and due Friday.]
No Discussion Sections this week.
First quiz will be next week in Discussion (Jan 25th).
Where were we?…
Coulomb’s Law:
1
F=
4π𝝐0
q 1q 2
r2
r^12
Electric forces are much stronger than gravity
Compare gravitational and electric forces between an electron and a
proton for a given separation r:
mpme
Fg = G 2
r
2
Fe = k
qeqp
r2
ke2
= 2
r
2
(9 ×10 9 )(1.6 ×10 −19 )
Fe
ke
=
=
Fg Gmpme (6.67 ×10 −11 )(1.67 ×10 −27 )(9.11×10 −31 )
Fe
≈ 2 ×10 39
Fg
The electric force is almost 40 orders of magnitude stronger
than the gravitational force.
So why don’t we usually notice electric forces ?
Electric forces are much stronger than gravity
Compare gravitational and electric forces between an electron and a
proton for a given separation r:
mm
Fg = G p2 e
r
Fe = k
(
qeqp
r2
ke2
= 2
r
)(
2
)
9 ×10 9 1.6 ×10 −19
Fe
ke2
=
=
Fg Gmpme
6.67 ×10 −11 1.67 ×10 −27 9.11×10 −31
(
)(
)(
)
Fe
≈ 2 ×10 39
Fg
The electric force is almost 40 order of magnitude stronger
than the gravitational force.
So why don’t we usually notice electric forces ?
➡ Gravity is always attractive.
➡ Ordinary matter contains almost exactly equal amount of
positive and negative charges, and thus electric forces tend to
cancel out.
Principle of Superposition: 3 or more charges
Suppose we want to find the net force exerted by charge #1 AND #2 on
charge #3.
Charge #2
Charge #1
Charge #3
The principle of superposition says that we can just work out
the force q1 exerts on q3 and add it vectorially to the force that
q2 exerts on q3.
Principle of Superposition Example: What’s total force on q3?
Two pt. charges are located on the x-axis: q1 = 1.0 nC at x = +2.0 cm and
q2 = -3.0 nC at x = +4.0 cm. What is the total force exerted by q1 and q2
on a charge q3 = 5.0 nC at x=0?
Principle of Superposition Example: What’s total force on q3?
Two pt. charges are located on the x-axis: q1 = 1.0 nC at x = +2.0 cm and
q2 = -3.0 nC at x = +4.0 cm. What is the total force exerted by q1 and q2
on a charge q3 = 5.0 nC at x=0?
Principle of Superposition Example: What’s total force on q3?
Two pt. charges are located on the x-axis: q1 = 1.0 nC at x = +2.0 cm and
q2 = -3.0 nC at x = +4.0 cm. What is the total force exerted by q1 and q2
on a charge q3 = 5.0 nC at x=0?
Principle of Superposition Example: What’s total force on q3?
Two pt. charges are located on the x-axis: q1 = 1.0 nC at x = +2.0 cm and
q2 = -3.0 nC at x = +4.0 cm. What is the total force exerted by q1 and q2
on a charge q3 = 5.0 nC at x=0?
F1→3 = 112µN (in the negative x direction)
FF2→3
1→2 = 84µN (in the positive x direction)
ke = 9 x 109 N.m2/C2
Vector sum:
(i.e. negative x direction)
Principle of Superposition: 3 or more charges
Charge #2
Charge #1
Charge #3
Remember: add electrical forces using vector
addition
➡Split in x and y components for each force.
➡Sum the x(y)-components together.
➡Get the magnitude using Pythagorean Thm.
and the direction angle using tanθ = y/x (where θ
is measured counter-clockwise from the +x-axis).
Quiz: Superposition principle I
Three point charges lie at the vertices of an equilateral triangle as shown. All three
charges have the same magnitude, but charges #1 and #2 are positive (+q) and
charge #3 is negative (–q).
The net electric force that charges #2 and #3 exert on charge #1 is in
A.the +x-direction
B.the -x-direction
y
Charge #2
+q
C.the +y-direction
D.the -y-direction
E.None of the above
x
Charge #1
+q
–q
Charge #3
Quiz: Superposition principle I
Three point charges lie at the vertices of an equilateral triangle as shown. All three
charges have the same magnitude, but charges #1 and #2 are positive (+q) and
charge #3 is negative (–q).
The net electric force that charges #2 and #3 exert on charge #1 is in
A.the +x-direction
B.the -x-direction
y
Charge #2
+q
C.the +y-direction
D.the -y-direction
E.None of the above
x
Charge #1
+q
–q
Charge #3
Quiz: Superposition principle II
Three point charges lie at the vertices of an equilateral triangle as shown. All three
charges have the same magnitude, but charge #1 is positive (+q) and charges #2
and #3 are negative (–q).
The net electric force that charges #2 and #3 exert on charge #1 is in
A.the +x-direction
B.the -x-direction
Charge #2
–q
y
C.the +y-direction
D.the -y-direction
E.None of the above
x
Charge #1
+q
–q
Charge #3
Quiz: Superposition principle II
Three point charges lie at the vertices of an equilateral triangle as shown. All three
charges have the same magnitude, but charge #1 is positive (+q) and charges #2
and #3 are negative (–q).
The net electric force that charges #2 and #3 exert on charge #1 is in
A.the +x-direction
B.the -x-direction
C.the +y-direction
D.the -y-direction
E.None of the above
Charge #2
–q
y
x
Charge #1
+q
–q
Charge #3
Quiz: Electric Force I
Two balls with charges +Q and +4Q are separated by 3R. Where
should you place another charge ball Q0 on the line between the two
charges such that the net force on Q0 will be zero?
+4Q
+Q
A
B
C
D
2R
R
3R
E
Quiz: Electric Force I
Two balls with charges +Q and +4Q are separated by 3R. Where
should you place another charge ball Q0 on the line between the two
charges such that the net force on Q0 will be zero?
+4Q
+Q
A
B
C
D
E
2R
R
3R
QQ
F on Q0 due to +Q is: F = k 0 2
R
F on Q0 due to +4Q is: F = k Q0 4Q
(2R)2
Since +4Q is 4 times bigger than +Q, Q0
needs to be farther from +4Q.
If Q0 is 2x as far from +4Q to have net
zero force - as the force depends on the
distance squared in Coulomb’s law.
Coulomb’s law problem solving strategy
Use Coulomb’s law whenever you need to find the electric force between two (or more) charges.
1.Make a diagram of the location of all charges, their signs, and magnitudes
2.Setup the (x,y) coordinate system
3.Identify the particle(s) (eg Q) that you want to find the electric force on
4.For each particle that exerts a force on Q, find the magnitude using Coulomb’s law
5.Draw the FBD of all forces acting on Q, showing the magnitude and direction of the force vectors.
✓ Direction will depend on whether the force is attractive or repulsive (i.e. the two charges
involved)
6.Use the principle of superposition to find the resultant force on Q by adding the vectors in the FBD.
✓ You may need to break the force vectors into (x,y) components to do this.
Tips:
✓Use symmetries whenever possible to simplify the problem.
✓Watch out for the units !
✓ r: meter
✓ q: Coulomb
µC=10-6 C nC=10-9 C
✓ Check that your numerical answer makes sense !!!
Example: Electric force in a plane
Two equal and positive charges q1=q2=2.0µC are located at x=0, y=0.3m and x=0,
y=-0.3m, respectively. What are the magnitude and direction of the total electric
force that q1 and q2 exert on a third charge Q=4.0µC at x=0.4m and y=0?
y
Charge #1
q1
Q
x
q2
Charge #2
Example: Electric force in a plane
Two equal and positive charges q1=q2=2.0µC are located at x=0, y=0.3m and x=0,
y=-0.3m, respectively. What are the magnitude and direction of the total electric
force that q1 and q2 exert on a third charge Q=4.0µC at x=0.4m and y=0?
y
Charge #1
q1
Q
x
q2
Charge #2
First, look for symmetries:
๏ q1 and q2 have the same magnitude, the
same charge, and are at the same distance
from Q
➡ magnitude of the forces F1 and F2 will
be equal.
✓ q1 and q2 are located symmetrically along
the y-axis.
➡ y-components of F1 and F2 will cancelout.
➡ Force will be in +x direction.
Example: Electric force in a plane
Two equal and positive charges q1=q2=2.0µC are located at x=0, y=0.3m and x=0,
y=-0.3m, respectively. What are the magnitude and direction of the total electric
force that q1 and q2 exert on a third charge Q=4.0µC at x=0.4m and y=0?
Example: Electric force in a plane
Two equal and positive charges q1=q2=2.0µC are located at x=0, y=0.3m and x=0,
y=-0.3m, respectively. What are the magnitude and direction of the total electric
force that q1 and q2 exert on a third charge Q=4.0µC at x=0.4m and y=0?
Example: Electric force in a plane
Two equal and positive charges q1=q2=2.0µC are located at x=0, y=0.3m and x=0,
y=-0.3m, respectively. What are the magnitude and direction of the total electric
force that q1 and q2 exert on a third charge Q=4.0µC at x=0.4m and y=0?
x-component of F1: F1cosα
x-component of F2: F2cosα
Since F1=F2, the total force on Q is:
(y-components cancel)
Balloon Bending Water Trick
(Balloon has just been rubbed on hair.)
Balloon Bending Water Trick
(Balloon has just been rubbed on hair.)
How do balloons bend water?
Electrons from hair
give balloon extra
negative (-) charge.
The H2O molecule is a “dipole”: => has a positive (+) and negative (-) side.
The (+) sides are attracted to the balloon.
Molecules orient themselves w/ (+) side (where H’s are) pointing towards balloon.
While the (-) side is repelled, it’s farther from the balloon, so the (+) tug wins.
Electric dipoles
•Nature is FULL of dipoles
✓
Many molecules are essentially dipoles
•
A dipole is two equal magnitude, opposite sign point
charges q, separated by a distance d.
•
Dipole has no net charge. The electric field arises from
the slight separation of the two opposite charges.
•
Eg: Dipole moment of H2O is responsible for some of its
important properties (i.e. almost a universal solvent)
In water, Na+ & Cl- (salt)
can be pulled apart due to
the H20 dipole moment
Electric forces on neutral objects
We’ve seen that charged objects exert forces on each other.
A charge object can also exert a force on an uncharged object.
Induced charge effect: electric charges shift (polarization) leading to a
stronger attractive force than repulsion force.
Electric Fields
Idea of Electric field
q2
q1
Force on q2 from q1:
Idea of Electric field
q2
Force on q2 from q1:
q3
Force on q3 from q1:
q1
Idea of Electric field
q2
Force on q2 from q1:
q3
Force on q3 from q1:
q1
Force on q4 from q1:
q4
Idea of Electric field
Force from q1 on any charge qi at position r:
q1
qi
Idea of Electric field
Force from q1 on any charge qi at position r:
q1
qi
Electric field: Depends only on
the charge q1 and spatial
location.
Idea of Electric field
Force from q1 on any charge qi at position r:
q1
qi
!"!
!"
F0 = q0 E
Idea of Electric field
Force from any charge q on any other charge q0 at position r:
q’
q
!"!
!"
F0 = q0 E
Electric field from point charge q:
Electric Field vs. Electric Force
Electric field:
➡
➡
Exists in all points around a charged object
When another charged object (charge q0) enters this electric field, an electric force acts on it.
!"!
!"
F0 = q0 E
Electric Field vs. Electric Force
Electric field:
➡
➡
Exists in all points around a charged object
When another charged object (charge q0) enters this electric field, an electric force acts on it.
!"!
!"
F0 = q0 E
A charge distribution creates an electric field, which exerts a force on
any charge that is present in the field.
Quiz (show of hands, no clickers)
What is the direction of the force felt by a negative
charge in the presence of this electric field?
E
a)
b)
F
F
Quiz
What is the direction of the force felt by a negative
charge in the presence of this electric field?
E
!"!
!"
F0 = q0 E
a)
b)
F
F
Electric Field of a Point Charge
Quiz (show of hands, no clickers)
Two point charges and a point P lie at the vertices of an equilateral triangle as
shown. Both point charges have the same amount of charge (q), but are of
opposite sign. There is nothing at point P.
The net electric field that charges #1 and #2 produce at point P is in
A.the +x-direction
B.the -x-direction
C.the +y-direction
D.the -y-direction
E.None of the above
Charge #1
–q
y
P
+q
x
Charge #2
Quiz (show of hands, no clickers)
Two point charges and a point P lie at the vertices of an equilateral triangle as
shown. Both point charges have the same amount of charge (q), but are of
opposite sign. There is nothing at point P.
The net electric field that charges #1 and #2 produce at point P is in
A.the +x-direction
B.the -x-direction
C.the +y-direction
D.the -y-direction
E.None of the above
Charge #1
–q
y
P
+q
x
Charge #2
Example: Electric field of a point charge
A point charge q=-8.0 nC is located at the origin. Find the electric field
vector at (x=1.2m, y=-1.6m).
Need magnitude & direction
-ve
r
(1.2m, -1.6m)
Example: Electric field of a point charge
A point charge q=-8.0 nC is located at the origin. Find the electric field
vector at (x=1.2m, y=-1.6m).
Need magnitude & direction
-ve
r
(1.2m, -1.6m)
(1.2 m)2 + (-1.6 m)2
Example: Electric field of a point charge
A point charge q=-8.0 nC is located at the origin. Find the electric field
vector at (x=1.2m, y=-1.6m).
Need magnitude & direction
-ve
E
(1.2m, -1.6m)
(1.2 m)2 + (-1.6 m)2
Electric field: continuous charge distribution
The system of closely spaced charges is
equivalent to a total charge that is continuously
distributed alone some line, over a surface, or
throughout some volume.
Procedure:
1. Divide the charge distribution into small
elements, each of which contains dq
2. Calculate the electric field, dE, due to one of
these elements at point P
3. Evaluate the total field by summing the
contributions of all the charge elements.
dE =
1 dQ
r̂
2
4πε 0 r
Charge Densities
Linear charge density: when a charge is distributed along a line
✓
λ ≡ Q / ℓ with units C/m
Surface charge density: when a charge is distributed evenly over a surface area
✓
σ ≡ Q / A with units C/m2
Volume charge density: when a charge is distributed evenly throughout a volume
✓
ρ ≡ Q / V with units C/m3
If the charge is uniformly distributed over a volume, surface, or line, the
amount of charge, dq, is given by
✓
For the length element: dq = λ dℓ
✓
For the surface: dq = σ dA
✓
For the volume: dq = ρ dV
Example: Ring of charge
Suppose we have a total charge Q distributed uniformly around a thin ring of radius a. We want to find the
electric field at point P, located at a distance x from the center of the ring, along the axis normal to the ring.
dQ
Q
=
ds 2π a
1.Break charge distribution into small segments
ds, with charge dQ.
Example: Ring of charge
Suppose we have a total charge Q distributed uniformly around a thin ring of radius a. We want to find the
electric field at point P, located at a distance x from the center of the ring, along the axis normal to the ring.
dQ
Q
=
ds 2π a
1.Break charge distribution into small segments
ds, with charge dQ.
2.Evaluate the electric field dE due to dQ at P. Because P is on the symmetry axis of the ring:
✓We can see that dEy for a segment at the top of the ring will cancel-out with dEy’ for a segment at the bottom of the
ring. Thus Ey=0 and only Ex contributes.
✓Every point on the ring is at the same distance from point P:
r = x2 + a2
dE =
1 dQ
1 Q
ds
=
4πε 0 r 2 4πε 0 2π a x2 + a2
Example: Ring of charge (cont.)
dE =
1 dQ
1 Q
ds
=
4πε 0 r 2 4πε 0 2π a x2 + a2
The magnitude of dE and the angle α
are the same for every point on the ring,
the x-component of dE is the same for
every point on the ring:
dEx = dEcosα
dEx =
1 Q
ds
4πε 0 2π a x2 + a2
x
x2 + a2
Example: Ring of charge (cont.)
dE =
1 dQ
1 Q
ds
=
4πε 0 r 2 4πε 0 2π a x2 + a2
The magnitude of dE and the angle α are
the same for every point on the ring, the xcomponent of dE is the same for every
point on the ring:
dEx = dEcosα
dEx =
1 Q
ds
4πε 0 2π a x2 + a2
x
x2 + a2
Example: Ring of charge (cont.)
dE =
1 dQ
1 Q
ds
=
4πε 0 r 2 4πε 0 2π a x2 + a2
The magnitude of dE and the angle α are
the same for every point on the ring, the xcomponent of dE is the same for every
point on the ring:
dEx = dEcosα
dEx =
1 Q
ds
4πε 0 2π a x2 + a2
x
x2 + a2
Now we can apply the principle of
superposition and integrate over the whole
ring:
1 Q
x
Ex = ∫ dEx =
4πε 0 2π a x2 + a2
(
See book example 21.11: line of charge
See book example 21.12: uniform charge disk
Ex =
Q
x
4πε 0 (x2 + a2 )3/2
2π a
3/2
)
∫ ds
0
Example: Field of uniformly charged disk of radius R
Think of many thin rings…
Field of each ring of radius a & charge q:
σ ≡ Q / A with units C/m2
Example: Field of uniformly charged disk of radius R
q of each ring
Field of each ring of radius a & charge q:
σ ≡ Q / A with units C/m2
at any radius r:
Example: Field of uniformly charged disk of radius R
q of each ring (radius r)
Field of each ring of radius a & charge q:
σ ≡ Q / A with units C/m2
Example: Field of uniformly charged disk of radius R
σ ≡ Q / A with units C/m2
Field of each ring of radius a & charge q:
at any radius r:
Quiz
Limit of an infinitely large plate (R —> infinity)?
Quiz
Limit of an infinitely large plate (R —> infinity)?
A.
B.
C.
Quiz
Limit of an infinitely large plate (R —> infinity)?
A.
B.
C.
Example: Field of uniformly charged disk of radius R
Limit of an infinitely large plate (R —> infinity)
Example: Field of uniformly charged disk of radius R
Limit of an infinitely large plate (R —> infinity)
If P is to the left of the plane, all the same except -ve sign:
Infinite plane sheet
2 infinite plane sheets
2 infinite plane sheets
2 infinite plane sheets
2 infinite plane sheets
2 infinite plane sheets
The field between oppositely changed sheets is uniform as
long as the sheets are much larger than their separation.
Field outside is zero.
Example: Electron in a uniform field
-e = -1.6×10-19C
me=9.11 ×10-31 kg
E=1×104 N/C
a) What is the acceleration of the electron ?
b) What speed & kinetic energy does the electron acquire while traveling 1cm?
c) How much time does it take the electron to travel the distance
Example: Electron in a uniform field
-e = -1.6×10-19C
me=9.11 ×10-31 kg
E=1×104 N/C
a) What is the acceleration of the electron ?
Example: Electron in a uniform field
-e = -1.6×10-19C
me=9.11 ×10-31 kg
E=1×104 N/C
c) How much time does it take the electron to travel the distance
Example: Electron in a uniform field
1 eV = the work done by electric field when a charge e is moved through a potential of 1 Volt
-e = -1.6×10-19C
1 eV = 1.602 ×10-19 Joule
me=9.11 ×10-31 kg
E=1×104 N/C
b) what is speed and KE of e after traveling 1cm?
Electric field lines: Visualization of E-Field
Electric field lines: Visualization of E-Field
Electric field lines: Visualization of E-Field
An E-field line is an imaginary line or curve such as the direction of the E-field is
tangent to the curve at every point.
Number of field lines ending/starting on a charge is proportional to the magnitude of the
charge.
For a dipole (two equal and opposite charges), the field lines start on the positive charge
and end on the negative charge.
The electric field can only have one direction at any single point and thus never intersect.
The density of the field lines near a particular point is an indication of the strength
(magnitude) of the electric field there.
•Field lines close together where the field is strong
•Field lines are farther apart where the field is weak
For a uniform E-field (eg 2 large parallel charged planes), the field lines are parallel and
evenly space. The field as the same direction and the same magnitude everywhere.
Electric field lines: Visualization of E-Field