Electrostatics and electric field review
January 24, 2014 Physics for Scientists & Engineers 2, Chapter 21 1
Exam 1 ! Wait outside lecture hall before the exam ! As you come in, I will assign seating and hand out bubble
sheets ! Equation sheet:
• One 8” by 5” sheet or index card • hand-written
! Exams will then be handed out by row ! Coded exams
• Put the code on the bubble sheet • Put your name on the handout
! You will have 45 minutes
January 24, 2014 Physics for Scientists & Engineers 2, Chapter 21 2
Coded exams
! Put the code on the bubble sheet • No code → no grade
! Put your name on the handout ! Hand back both exam handout and bubble sheet
January 24, 2014 Physics for Scientists & Engineers 2, Chapter 21 3
Put this code on the bubble sheet!
Put your name here
January 24, 2014 Physics for Scientists & Engineers 2 4
Law of Charges ! This result leads to the Law of Charges
• Like charges repel and opposite charges attract
! The unit of charge is defined as
+ -
+
-
-
+
1 C = 1 A s
January 24, 2014 Physics for Scientists & Engineers 2, Chapter 21 5
Insulators and Conductors ! Materials that conduct electricity well are called conductors
• Electrons can move freely (some of the electrons) • Metals
! Materials that conduct electricity poorly are called insulators • Electrons cannot move freely
• Glass • Plastic • Cloth
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 24, 2014 Physics for Scientists & Engineers 2, Chapter 21 6
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 24, 2014 Physics for Scientists & Engineers 2, Chapter 21 7
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
General Charge Distributions ! Line of charge: charge dq distributed over length dx:
charge density
! Sheet of charge: charge dq distributed over area dx × dy = dA: charge density
! Volume of charge: charge dq distributed over volume dx × dy × dz = dV: charge density
January 24, 2014 Physics for Scientists & Engineers 2, Chapter 22 8
λ = dq
dx
σ = dq
dA
ρ = dq
dV
January 24, 2014 Physics for Scientists & Engineers 2, Chapter 22 9
The Electric Field ! The electric force on a charge is parallel or antiparallel to the
electric field at that point
! The magnitude of the force is
! The magnitude of the field is F = q E
points toward charge 2, as shown in Figure 21.13b.
F
F
positive charge
negative charge
E
E
E = F / q
Superposition of electric fields ! The total electric field at any point is the vector sum of all
individual electric fields
! Note that this means adding the x-, y-, and z- components separately
January 24, 2014 Physics for Scientists & Engineers 2, Chapter 21 10
Enet =
Ei
i∑
Enet ,X = Ei ,Xi∑
Enet ,Y = Ei ,Yi∑
Enet ,Z = Ei ,Zi∑
Field Lines ! To draw an electric field line, we imagine placing a small
positive test charge at each point in the electric field. ! Electric field lines always originate on positive charges and
terminate on negative charges ! The higher the field line density, the larger the field ! Field lines never cross ! Field lines are perpendicular to the surface of a conductor
January 24, 2014 Chapter 22 11
Electric Field due to Point Charges ! The magnitude of the electric force on a point charge q0 due
to another point charge q is
! We can take q0 as a test charge so can express the magnitude of the electric field due to q as
January 24, 2014 Physics for Scientists & Engineers 2, Chapter 22 12
F =
14πε0
qq0
r2
E =
Fq0
=1
4πε0
qr2
September 8, 2008 Physics for Scientists & Engineers 2, Lecture 8 13
Review – Gauss’s Law
q = net charge enclosed by S
!E ⋅d!A
S"∫ = q
ε0
Homework hint ! A cubic box of side a = 0.390 m is
placed so that its edges are parallel to the coordinate axes, as shown in the figure. There is NO net electric charge inside the box, but the space in and around the box is filled with a nonuniform electric field of the following form: E(x,y,z) = Kz j + Ky k, where K = 3.70 N/(Cm) is a constant.
! What is the electric flux through the top face of the box?
January 24, 2014 Physics for Scientists & Engineers 2, Chapter 21 14
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
1/24/14 15 Physics for Scientists & Engineers 2
Homework hint ! A cubic box of side a = 0.390 m is
placed so that its edges are parallel to the coordinate axes, as shown in the figure. There is NO net electric charge inside the box, but the space in and around the box is filled with a nonuniform electric field of the following form: E(x,y,z) = Kz j + Ky k, where K = 3.70 N/(Cm) is a constant.
! What is the electric flux through the top face of the box?
! Think, sketch: ! what is the field?
i, j, k are unit vectors in the x, y, z directions
January 24, 2014 Physics for Scientists & Engineers 2, Chapter 21 16
!E
Φ =!E id!A∫
Homework hint 1. Research: Components of
electric field: • Ex = 0 • Ey = Kz (depends on z) • Ez = Ky (depends on y)
2. Surface normal vector? In z direction " Only z component of field
matters 3. Execute: Flux
January 24, 2014 Physics for Scientists & Engineers 2, Chapter 21 17
Φ =!E id!A∫
!E
!E
= Ez dA∫ = Ez dx dy∫ = Kydx dy∫ = Kydy∫ dx∫ = 12 K(y 2 −0)(x −0)
Homework hint ! Result:
• For x = 0.39 m, y = 0.39 m, K=3.70 N/(Cm)
! Check: units ok • Order-of-magnitude estimate: field in the center times total area
• Agrees perfectly because field is a linear function of position
January 24, 2014 Physics for Scientists & Engineers 2, Chapter 21 18
Φ = 12 Ky 2x
!E
!E
Φ = 0.110 Nm2 /C
Φ ∼ Ezcenter ∗area = K 1
2 y xy = 12 Ky 2x = 0.110Nm2 /C
January 24, 2014 Physics for Scientists & Engineers 2, Chapter 22 19
Shielding ! Two important consequences of Gauss’s Law
• The electrostatic field inside any isolated conductor is always zero. • Cavities inside conductors are shielded from electric fields.
! To examine these consequences, let’s suppose a net electric field exists at some moment at some point inside an isolated conductor
! Charges will move to cancel the field inside the conductor
September 8, 2008 Physics for Scientists & Engineers 2, Lecture 8 20
Review - Electric Fields from Charge Distributions ! The electric field E at distance r from a
wire with charge density λ is
! The electric field E produced by an
infinite non-conducting plate with charge density σ is
E = λ
2πε0r= 2kλ
r
02E σ
ε=
September 8, 2008 Physics for Scientists & Engineers 2, Lecture 8 21
Review - Electric Fields from Charge Distributions (2) ! The electric field E produced by an
infinite conducting plane with charge density σ is
! The electric field inside a spherical shell of charge q is zero
! The electric field outside a spherical
shell of charge q is the same as the field from a point charge q.
0
E σε
=
2
qE kr
=
September 8, 2008 Physics for Scientists & Engineers 2, Lecture 8 22
Review - Spherical Charge Distributions
r R
+
+
+
+ + +
+ +
R Q Q
Conducting sphere
E
+ +
+ +
+
+ +
+
+
+ +
+
Non-conducting sphere
E=0
r R
E
September 8, 2008 Physics for Scientists & Engineers 2, Lecture 8 23
Electric Fields from a Ring of Charge ! The electric field E resulting from a ring of charge
(radius R, charge density λ=q/(2πR)) on the axis ! Strategy: Imagine the ring is divided into differential
elements of charge dq=λds. Use the electric field of a point charge for every one of them.
ds
( ) ( )
2 2 2
2 2
3/ 2 3/ 22 2 2 2
cos and cos
( )
z
z
dq dsdE k kr R z
zdE dER z
ds z kqzE z kR z R z
λ
θ θ
λ
= =+
= =+
= =+ +
∫λ kq/z2 for large z
September 8, 2008 Physics for Scientists & Engineers 2, Lecture 8 24
Example - Charge in a Cube ! Q=3.76 nC is at the center of a
cube. What is the electric flux through one of the sides?
! Gauss’ Law: 0/Q εΦ =
! Since a cube has 6 identical sides and the point charge is at the center
Q
one face 6ΦΦ =
2Nm70.8C
=0
16Qε
=
September 8, 2008 Physics for Scientists & Engineers 2, Lecture 8 26
Example - E Field and Force ! The figure shows the deflecting plates of an ink-jet printer. A negatively
charged ink drop (q=1.5 x 10-13 C) enters the region between the plates with a velocity of v=18 m/s along x. The length L of each plate is 1.6 cm. The plates are charged to produce an electric field at all points between them (E=1.4 x 106 N/C). The vertical deflection of the drop at x=L is 0.64 mm. What is the mass of the ink drop?
! Idea: A constant electrostatic force
of magnitude qE acts upward on the drop.
yF qEam m
= =
… constant acceleration
September 8, 2008 Physics for Scientists & Engineers 2, Lecture 8 27
Example - E Field and Forces (2) ! What is the
mass of the ink drop?
! Idea: Let t be the time required to pass through the plates. Then…
212
2
2 2
10
and
2 2which implies ( / )
So... (substitute the numerical values) 1.3 10 kg
y at L vt
y yvaL v L
qEma
−
= =
= =
= = = ×
Exam 1 ! Wait outside lecture hall before the exam ! As you come in, I will assign seating and hand out bubble
sheets ! Equation sheet:
• One 8” by 5” sheet or index card • hand-written
! Exams will then be handed out by row ! Coded exams
• Put the code on the bubble sheet • Put your name on the handout
! You will have 45 minutes
January 24, 2014 Physics for Scientists & Engineers 2, Chapter 21 28