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


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


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


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


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


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

⎛

⎝⎜⎜⎜

⎞

⎠⎟⎟⎟⎟

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


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


F1→3 =−k q1q3

x3−x1( )2 (−x̂) F2→3 =−k q2q3

x3−x2( )2 (−x̂)Fnet→3 =

F1→3 +

F2→3


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.


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?


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


F2→1 +

F4→1 =

F1,net

F2→1

F4→1




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



!  Key Idea: Evaluate the x and y components of the F4"1 vector


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 …


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


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:


T cosθ−Fg = 0

Fe = k q2

d2

Fg = mg

sinθ=

d / 2

=d2

Fe = k q2

2sinθ( )2 = k q2

42 sin2 θ


Charged Balls Simplify

!  We divide two force component equations to get:

!  Substitute in our equations for Fe and Fg:

Calculate


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

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟

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:


m = 0.764 kg

T sinθ≈mgθ= Fe = k q2

d2 ≈ k q2

2θ( )2

m =kq2

4g2θ3 =8.99 ⋅109 N m2

C2

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟

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?





!  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)




!  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

electric fields and electrostatics gauss’s law · electric fields and gauss’s law...

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