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Electric Fields and Electrostatics Gauss’s Law · Electric Fields and Gauss’s Law...
Transcript of Electric Fields and Electrostatics Gauss’s Law · Electric Fields and Gauss’s Law...
January 13, 2014 Physics for Scientists & Engineers 2, Chapter 22 1
Electric Fields and Gauss’s Law Electrostatics
Announcements ! If you are not registered for this course yet,
section 1 has openings • The PA Undergraduate Secretary, Kim Crosslan, in 1312 BPS can
register you ! Homework set 0 is due 1/13 (tonight) ! Homework set 1 is due 1/20 ! See book for example calculations ! Clicker questions starting today
• Clicker points only for the section in which you are registered ! Lecture notes: linked from lon-capa, or directly at
http://www.pa.msu.edu/~schwier/courses/2014SpringPhy184/ ! Section 1 lecture notes:
http://www.pa.msu.edu/~nagy_t/phy184/lecturenotes.html
January 13, 2014 Physics for Scientists & Engineers 2, Chapter 21 2
Preliminary Helproom hours ! Strosacker learning center ! BPS 1248 ! Starting today ! Schedule is flexible – let me know if I should change
something
! Mo: 10am – noon, 1pm – 9pm ! Tue: noon – 6pm ! We: noon – 2pm ! Th: 10am – 1pm, 2pm – 9pm
January 13, 2014 Physics for Scientists & Engineers 2, Chapter 21 3
Electrostatic Force – Coulomb’s Law ! The law of electric charges is evidence of a force
between any two charges at rest ! Experiments show that for the electrostatic force exerted by
charge 2 (q2) on charge 1 (q1), the force on q1 points toward q2 if the charges have opposite signs and away from q2 if the charges have like signs
January 13, 2014 Physics for Scientists & Engineers 2, Chapter 21 4
Electrostatic Force – Coulomb’s Law ! Coulomb’s Law gives the magnitude of this force as
! k is Coulomb’s constant given by ! We can relate Coulomb’s constant to the
electric permittivity of free space ε0
January 13, 2014 Physics for Scientists & Engineers 2, Chapter 21 5
F = k q1q2
r2
q1 and q2 are electric chargesr =r1−r2 is the distance between the charges
k = 8.99⋅109 N m2
C2
k =
14πε0
, ε0 = 8.85⋅10−12 C2
N m2
January 13, 2014 Physics for Scientists & Engineers 2, Chapter 21 7
Force between Two Charges ! PROBLEM ! What is the magnitude of the force between two 1.00 C
charges 100.0 cm apart? ! SOLUTION
! This force is approximately the same as the gravitational force of 450 loaded Space Shuttles!
F = k q1q2
r2
F = 8.99⋅109 N m2
C2
⎛
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⎞
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1.00 C( )2
1.00 m( )2
F = 8.99⋅109 N
Electrostatic Force Vector ! Coulomb’s Law can be written in vector form as
January 13, 2014 Physics for Scientists & Engineers 2, Chapter 21 8
F2→1 =−k q1q2
r3r2−r1( )=−k q1q2
r2 r̂21
Physics for Scientists & Engineers 2 9
Math Reminder (1)
a
a
d
a
b
c x
y
January 13, 2014
Superposition Principle ! Consider the force exerted on charge q3 by two other
charges q1 and q2
January 13, 2014 Physics for Scientists & Engineers 2, Chapter 21 10
F1→3 =−k q1q3
x3−x1( )2 (−x̂) F2→3 =−k q2q3
x3−x2( )2 (−x̂)Fnet→3 =
F1→3 +
F2→3
January 13, 2014 Physics for Scientists & Engineers 2, Chapter 21 12
Example - Forces between Electrons
! What is relative strength of the force of gravity compared with the electric force for two electrons?
! So the electric force is always very much larger than the gravitational force
! Macroscopic objects are usually uncharged so only gravity plays a role • Motion of the planets
! Gravity is irrelevant for sub-atomic processes
Felectric = kqe2
r2
Fgravity = Gme2
r2
FelectricFgravity
=kqe
2
Gme2 =
(8.99 ⋅109 N ⋅m2 / C2 )(1.602 ⋅10−19 C)2
(6.67 ⋅10-11 N ⋅m2 /kg2 )(9.109 ⋅10-31 kg)2 = 4.2 ⋅1042
Three Charges ! Two charged particles of q1=1.6
10-19C and q2=3.2 10-19C are fixed on the x axis with a separation of R=0.02m. Particle 3 with charge q3=-3.2 10-19C is placed at a distance of 3/4R from particle 1. What is the net force on particle 1 due to particles 2 and 3?
! Key idea: The electrostatic force on q1 is the vector sum of the forces resulting from its interactions with the other two charges q2 and q3.
January 13, 2014 Physics for Scientists & Engineers 2, Chapter 21 13
F2→1 +
F3→1 =
F1,net
Three Charges ! Two charged particles of q1=1.6
10-19C and q2=3.2 10-19C are fixed on the x axis with a separation of R=0.02m. Particle 3 with charge q3=-3.2 10-19C is placed at a distance of 3/4R from particle 1. What is the net force on particle 1 due to particles 2 and 3?
January 13, 2014 Physics for Scientists & Engineers 2, Chapter 21 14
F3→1
F2→1: repulsive
: attractive
F2→1 = k q1q2
R2 = 8.99 ⋅109 N m2 /C2( ) 1.6 ⋅10−19C( ) 3.2 ⋅10−19 C( )0.022 =1.15 ⋅10−24 N
F3→1 = k q1q3
(34
R)2= 8.99 ⋅109 N m2 /C2( ) 1.6 ⋅10−19C( ) 3.2 ⋅10−19 C( )
(34
0.02)2= 2.05 ⋅10−24 N
F1,net = −F2→1 + F3→1 = −1.15 ⋅10−24 N + 2.05 ⋅10−24 N = 9.0 ⋅10−25 N pointing to the right
F2→1
F3→1
F2→1
Three Charges ! Two charged particles of q1=1.6 10-19C
and q2=3.2 10-19C are fixed on the x axis with a separation of R=0.02m. Particle 4 with charge q4=-3.2 10-19C is placed at a distance of 3/4R from particle 1 in the xy plane with an angle of θ=60 degrees relative to the x axis. What is the net force on particle 1 due to particles 2 and 4?
! Key idea: Superposition principle
January 13, 2014 Physics for Scientists & Engineers 2, Chapter 21 16
F2→1 +
F4→1 =
F1,net
F2→1
F4→1
Three Charges ! Two charged particles of q1=1.6 10-19C
and q2=3.2 10-19C are fixed on the x axis with a separation of R=0.02m. Particle 4 with charge q4=-3.2 10-19C is placed at a distance of 3/4R from particle 1 in the xy plane with an angle of θ=60 degrees relative to the x axis. What is the net force on particle 1 due to particles 2 and 4?
January 13, 2014 Physics for Scientists & Engineers 2, Chapter 21 17
F2→1 = kq1q2
R2 = 8.99 ⋅109 N m2 /C2( ) 1.6 ⋅10−19C( ) 3.2 ⋅10−19 C( )0.022 =1.15 ⋅10−24 N
F4→1 = kq1q4
(34
R)2= 8.99 ⋅109 N m2 /C2( ) 1.6 ⋅10−19C( ) 3.2 ⋅10−19 C( )
(34
0.02)2= 2.05 ⋅10−24 N
F2→1
F4→1
Three Charges ! Two charged particles of q1=1.6 10-19C
and q2=3.2 10-19C are fixed on the x axis with a separation of R=0.02m. Particle 4 with charge q4=-3.2 10-19C is placed at a distance of 3/4R from particle 1 in the xy plane with an angle of θ=60 degrees relative to the x axis. What is the net force on particle 1 due to particles 2 and 4?
! Key Idea: Evaluate the x and y components of the F4"1 vector
January 13, 2014 Physics for Scientists & Engineers 2, Chapter 21 18
x y
F4→1,x = F4→1 cos(60°)= 2.05 ⋅10−24 cos(60°)=1.025 ⋅10−24 NF4→1,y = F4→1 sin(60°)= 2.05 ⋅10−24 sin(60°)=1.775 ⋅10−24 N
F2→1
F4→1
Three Charges ! Two charged particles of q1=1.6 10-19C and
q2=3.2 10-19C are fixed on the x axis with a separation of R=0.02m. Particle 4 with charge q4=-3.2 10-19C is placed at a distance of 3/4R from particle 1 in the xy plane with an angle of θ=60 degrees relative to the x axis. What is the net force on particle 1 due to particles 2 and 4?
! Almost done …
January 13, 2014 Physics for Scientists & Engineers 2, Chapter 21 19
x y
F1,net ,x = F2→1,x + F4→1,x = −1.15 ⋅10−24 N+1.025 ⋅10−24 =-1.25 ⋅10−25 NF1,net ,y = F2→1,y + F4→1,y = 0+1.775 ⋅10−24 N =1.775 ⋅10−24 N
F = Fx2 + Fy
2 tanθ = Fy / Fx
F1,net =1.78 ⋅10−24 N
θ = tan−1(F1,net ,y
F1,net ,x)= −86°
θ =180°+ tan−1(F1,net ,y
F1,net ,x)=180°− 86°
F2→1
F4→1
Charged Balls
PROBLEM: ! Two identical charged balls hang
from the ceiling by insulated ropes of equal length, l = 1.50 m.
! A charge q = 25.0 μC is applied to each ball.
! Then the two balls hang at rest, and each supporting rope has an angle of 25.0° with respect to the vertical.
! What is the mass of each ball?
January 13, 2014 Chapter 21 21
Copyright © The McGraw-Hill Companies. Permission required for reproduction or display.
Problem solving strategy ! Use this approach to solve problems, in particular if at first you have no clue.
Recognize the problem What’s going on?
Think Describe the problem
in terms of the field What does this have to do with…?
Sketch
Plan a solution How do I get out of this?
Research Execute the plan
Let’s get an answer! Simplify, Calculate, Round
Evaluate the solution Can this be true?
Double-check
! Draw a picture ! Phrase the question in your own
words ! Relate the question to something
you just learned ! Identify physics quantities, forces,
fields, potentials,… ! Find a physics principle
(symmetry, conservation, …) ! Write down the equations ! Solve equations, starting with
intermediate steps ! Check units, order-of-magnitude,
insert into original question, …
Step 1 Think
Step 2 Sketch
Step 3 Research
Step 4 Execute
Step 5 Double-check
1/13/14 22 Physics for Scientists & Engineers 2
SOLUTION: Think
! Each ball has three forces acting on it: • Force of gravity • Repulsive electrostatic force • Tension in the supporting rope
! The forces must sum to zero. Sketch
! A free body diagram of one of the balls is shown. Research
! The sum of the x-components of the forces gives us:
Charged Balls
January 13, 2014 Chapter 21 23 T sinθ−Fe = 0
Copyright © The McGraw-Hill Companies. Permission required for reproduction or display.
Charged Balls ! The sum of the y-components of the forces gives us:
! The electrostatic force is given by:
! The force of gravity is given by: ! The distance between the balls is given by:
! The electrostatic force between the balls is:
January 13, 2014 Chapter 21 24
T cosθ−Fg = 0
Fe = k q2
d2
Fg = mg
sinθ=
d / 2
=d2
Fe = k q2
2sinθ( )2 = k q2
42 sin2 θ
Copyright © The McGraw-Hill Companies. Permission required for reproduction or display.
Charged Balls Simplify
! We divide two force component equations to get:
! Substitute in our equations for Fe and Fg:
Calculate
January 13, 2014 Chapter 21 25
T sinθT cosθ
=Fe
Fg
⇒ tanθ= Fe
Fg
tanθ=
k q2
42 sin2 θmg
⇒ m =kq2
4g2 sin2 θ tanθ
m =8.99 ⋅109 N m2
C2
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25.0 ⋅10−6 C( )2
4 9.81 m/s2( ) 1.50 m( )2sin2 25.0° tan25.0°
= 0.764116 kg
Charged Balls Round
Double-check
! To double-check our results, let’s make the small angle approximation that sinθ ≈ tanθ ≈ θ and cosθ ≈ 1:
January 13, 2014 Chapter 21 26
m = 0.764 kg
T sinθ≈mgθ= Fe = k q2
d2 ≈ k q2
2θ( )2
m =kq2
4g2θ3 =8.99 ⋅109 N m2
C2
⎛
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⎞
⎠⎟⎟⎟⎟
25.0 ⋅10−6 C( )2
4g 1.50 m( )20.436 rad( )3 = 0.768 kg
Two Spheres ! Two identical, conducting, isolated
spheres A and B are separated by a distance a that is large compared to the radius of the spheres. Sphere A is positively charged, +Q, and sphere B is neutral.
! a) There is no electrostatic force since q=0 for sphere B (also no induced charge since distance a large)
! b) For a moment, the spheres are connected by a conducting wire. What happens?
January 13, 2014 Physics for Scientists & Engineers 2, Chapter 21 27
Two Spheres ! Two identical, conducting, isolated
spheres A and B are separated by a distance a that is large compared to the radius of the spheres. Sphere A is positively charged, +Q, and sphere B is neutral.
January 13, 2014 Physics for Scientists & Engineers 2, Chapter 21 28
! b) For a moment, the spheres are connected by a conducting wire. What happens?
! Key Idea: The negatively charged electrons on sphere B are attracted by the positive charge of sphere A and some move along the wire onto sphere A. As sphere B loses negative charge, it becomes positively charged; A gains negative charge and becomes less positive. This happens until the charges on A and B are equal (+Q/2) – see (c)
Two Spheres ! Two identical, conducting, isolated
spheres A and B are separated by a distance a that is large compared to the radius of the spheres. Sphere A is positively charged, +Q, and sphere B is neutral.
January 13, 2014 Physics for Scientists & Engineers 2, Chapter 21 29
! c) The wire is removed. What is the force between the spheres?
! Key Idea: The force is repulsive (both spheres are positively charged).
FAB = k (Q / 2)(Q / 2)
a2 =14
k Q2
a2