Electrostatics and electric field review - Michigan State...

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


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


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


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


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

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


λ = dq

dx

σ = dq

dA

ρ = dq

dV


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


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


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?


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


!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


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


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


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


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=


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

=


Review - Spherical Charge Distributions

r R

+

+

+

+ + +

+ +

R Q Q

Conducting sphere

E

+ +

+ +

+

+ +

+

+

+ +

+

Non-conducting sphere

E=0

r R

E


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


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

=


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


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


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


Put this code on the bubble sheet!

Put your name here

Electrostatics and electric field review - Michigan State...

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Transcript of Electrostatics and electric field review - Michigan State...