Lecture02 Electric Field; Electric Forces;

8/3/2019 Lecture02 Electric Field; Electric Forces;

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adapted from http://www.lab-initio.com/


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Make sure you have a copy ofSpecial Homework #2, which is

due tomorrow.


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Announcements

Homework for tomorrow:

21: 21: 25, 33a, 54a, 78, Special Homework #2(reminder: all solutions must start with OSEs)

You only need to do part a of problems 33 and 54.You dont need to be paranoid about using OSEs, but dontstart a problem with a random equation from your book.

Labs begin next Monday.If you have lab next week, dont miss it!

http://campus.mst.edu/physics/courses/24lab/


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Todays agenda:

The electric field.You must be able to calculate the force on a charged particle in an electric field.

Electric field due to a point charge.You must be able to calculate electric field of a point charge.

Motion of a charged particle in a uniform electric field.You must be able to solve for the trajectory of a charged particle in a uniform electric field.

The electric field due to a collection of point charges.

You must be able to calculate electric field of a collection of point charges.

The electric field due to a continuous line of charge.You must be able to calculate electric field of a continuous line of charge.


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Coulomb's Law quantifies the interaction between chargedparticles.

+ -

Coulombs Law:The Big Picture

Q1 Q2

r121 22

0 12

q q1F = ,12 4 r

Coulombs Law was discovered through decades of experiment.

By itself, it is just useful." Is it part of something bigger?


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You experienced gravitational fields in Physics 23.

Gravitational Fields

1 2

2

12

m mF =G , attractiveG

r

&

GFg(r) =m

&& &

is the local gravitational field. On earth, it is about 9.8N/kg, directed towards the center of the earth.g(r)& &

Units of g areactually N/kg!

If the last equationlooks like this, youhave missing fonts.


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Could you measure glocal with yourself and a bathroom scale?

Click here for an animated viewof the earths gravitational field.

Wouldn't your mass, which contributes to the local gravitationalfield, introduce a perturbation into the measurement?

Your mass is of order magnitude 102 kg and the earth's mass is about 6x1024 kg.

Something to think about: if you move, how long does it take the earth to realize it?

Gravity Recovery and Climate Experiment model 01.


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The Electric Field

Coulomb's Law (demonstrated in 1785) shows that chargedparticles exert forces on each other over great distances.

How does a charged particle "know" another one is there?

Electric, magnetic, and gravitational forces result from directand instantaneous interaction between particles.

Action At A Distance Viewpoint

Relativity theory shows why this viewpoint is wrong. Theconcept of electric field gives the correct explanation.


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Faraday, beginning in the 1830's, was the leader in developingthe idea of the electric field. Here's the idea:

yA charged particle propagates (sendsout) a "field" into all space.

yAnother charged particle senses the field,and knows that the first one is there.

+

+-

likechargesrepel

unlikecharges

attract

F12

F21

F31

F13


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The idea of an electric field is good for a number of reasons:

y

It makes us feel good, like weveactually explained something.

y

We can develop a theory based on thisidea. From this theory may springunimagined inventions.

+

+-

likechargesrepel

unlikecharges

attract

F12

F21

F31

F13

OK, that was a flippant remark. There are serious reasonswhy the idea is good.

If the theory explains past observations and leads to newpredictions, the idea was good.


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We define the electric field by the force it exerts on a testcharge q0:

0

0

FE =q

&

&

This is your second starting equation. By convention the direction of the electric fieldis the direction of the force exerted on a POSITIVE test charge. The absence ofabsolute value signs around q0 means you must include the sign of q0 in your work.

If the test charge is "too big" it perturbs the electric field, so thecorrect definition is

0

0

q 00

FE = lim

qp

&&

The subscript 0 reminds you the force is on thetest charge. I wont require the subscripts whenyou use this equation for boardwork or on exams.

Any time you know the electric field, you can use this equation to calculate the force

on a charged particle in that electric field.

You wont be required to usethis version of the equation.

F = qE& &

Im not mad, I tell you, not mad. Thelittle voices tell me Im quite sane.


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The units of electric field are

newtons/coulomb. ? A

0

0

F NE = =

q C

-

-

&&

In chapter 23, you will learn that the units of electric field can

also be expressed as volts/meter:

? AN V

E = =C m

The electric field exists independent of whether there is acharged particle around to feel it.


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+Remember: the electric field direction is the

direction a + charge would feel a force.

A + charge would be repelled by another + charge.

Therefore the direction of the electric field is away frompositive (and towards negative).

http://regentsprep.org/Regents/physics/phys03/afieldint/default.htm


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Todays agenda:








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The Electric FieldDue to a Point Charge

Coulomb's law says

... which tells us the electric field due to a point charge q is

1 2

2

12

q qF =k ,

12 r

This is your third starting equation.

q 2qE =k , away from +r

&

or just 2

qE=kr


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2

qE=k

r

A physics 24 equation is not a toaster!

You cant expect to just shove the numbers in and out pops thecorrect answer.

To experience the optimum user satisfaction from your physics24 toaster equations you need to understand what they meanand think about what you are doing with them.


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The equation for the electric field of a point charge then

becomes:

2

qE=k r

r

&

We define as a unit vector from the source point to the fieldpoint:

r

+source point

field point

r

You may start with either equationfor the electric field (this one or theone on the previous slide). Butdont use this one unless youREALLY know what you aredoing! (So dont use it!)


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Example: calculate the electric field at the electrons distance

away from the proton in a hydrogen atom (5.3x10-11

m).

To be worked at the blackboard.

For comparison, air begins to break down and conductelectricity at about 30 kV/cm, or 3x106V/m.


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Quiz time (maybe for points, maybe just for practice!)


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Todays agenda:



Motion of a charged particle in a uniform electricfield.You must be able to solve for the trajectory of a charged particle in a uniform electric field.

The electric field due to a collection of point charges.You must be able to calculate electric field of a collection of point charges.



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Motion of a Charged Particlein a Uniform Electric Field

A charged particle in an electric field experiences a force, and ifit is free to move, an acceleration.

- - - - - - - - - - - - -

+ + + + + + + + + + + + +

-

F

If the only force is due to theelectric field, then

F ma qE.! !& &&

If E is constant, then a is constant, and you can use theequations of kinematics* (remember way back to the beginningof physics 23?).

*If you get called to the board, you can use the Physics 23 starting equations. They are posted.

E


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Example: a proton and an electron enter a region of uniformelectric field. Describe their motion.

Direction of forces?

Magnitudes of accelerations?

Shape of trajectories?

Well work these out in the following example.


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Example: an electron moving with velocity v0 in the positive xdirection enters a region of uniform electric field that makes a

right angle with the electrons initial velocity. Express theposition and velocity of the electron as a function of time.

- - - - - - - - - - - - -

+ + + + + + + + + + + + +

E

x

y

-

v0

To be worked at the blackboard.

What would be different for a proton?


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Todays agenda:




The electric field due to a collection of point charges.You must be able to calculate electric field of a collection of point charges.



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Example: calculate the electric field at position P due to the two

protons shown.

+ +

DD

PQ1=+e Q2=+e

xE1

E2

1 2

P 1 2 22

k Q k Q E E E i iD 2D

! ! & & &

P 2 2 2

ke ke 5ke E i i iD 4D 4D

! ! &


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Todays agenda:








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Electric Field Due To A Line of Charge

+ + + + + + + + + + + + + + + +

Matter is made of discrete atoms, but appears"continuous" to us, and in Physics 23 we treated

matter as being a continuous entity.

Similarly, a charge distribution is made of individual

charged particles, but we can treat it as if the chargewere continuous.


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Think of a line of charge as a collection of very very tiny pointcharges all lined up. The official term for very very tiny isinfinitesimal.

So we get the electric field for the line of charge by adding theelectric fields for all the infinitesimal point charges.

What happens in calculus when you add infinitesimals?

+ + + + + + + + + + + + + + + +


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The electric field due to a small "chunk" (q of charge is

2

0

1 qE = r4 r

((&

The electric field due to collection of "chunks" of charge is

i ii 2i i0 i

1 qE = E = r 4 r

((

& &

unit vector from (qto wherever you

want to calculate(E

unit vector from (qito wherever youwant to calculate E

As (qpdqp0, the sum becomes an integral.


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If charge is distributed along a straight line segment parallel tothe x-axis, the amount of charge dq on a segment of length dxis Pdx.

P is the linear density of charge (amount of charge per unitlength). P may be a function of position.

Think P length. P times the length of line segment is thetotal charge on the line segment.

"

x

dx

P Pdx

Consider charge distributed along a line (e.g., electrons on athread).


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The electric field at point P due to the charge dq is

x

dq

P

2 2

0 0

1 dq 1 dxdE = r' = r'

4 r' 4 r'P&

r

r'

dE

Im assuming positively charged objectsin these distribution of charges slides.

Absolute value signs not needed around dqbecause Im assuming positively charged objects.


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The electric field at P due to the entire line of charge is

2

0

1 (x) dxE = r' .

4 r'&

The integration is carried out over the entire length of the line, which neednot be straight. Also, P could be a function of position, and can be takenoutside the integral only if the charge distribution is uniform.

x

dq

P

r

r'

E


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Example: A rod of length L has a uniform charge per unit lengthP and a total positive charge Q. Calculate the electric field at a

point P along the axis of the rod a distance d from one end.

P x

y

d L

Lets put the origin at P. The linear charge density and Q are

related by Q= and Q = L

LP P

Note that the problem statement says Q is positive, so the

electric field points away from the rod.


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P x

y

d L

The electric field points away from the rod. The electric field onthe axis of the rod has no y-component (why?). dE from thecharge on an infinitesimal length dx of rod is

dEx dx dQ = P dx

2 2

dq dxdE = k k

x x

P!

Note: dE is in the x direction. dE is the magnitude of dE. Iveused the fact that Q>0 (so dq>0) to eliminate the absolutevalue signs in the starting equation.

This is a legalstarting equation. Infact, this is the best

way to start a problem

like this one.


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P x

y

d L

dEx dx dQ = P dx

d L

d+L d+L d+L2 2d d d

d

dx dx 1 E = dE = -k i = -k i = -k ix x x

P P P

& &

1 1 d d L L kQ E = -k i = -k i= -k i= - i

d L d d d L d d L d d L P P P

&


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Homework Hints (may not apply every semester)

There are two kinds of problems from todays lecture:

1. Given an electric field, calculate the force on a

charged particle.

2. Given one or more charged particles, calculatethe electric field they produce.

Make sure you understand which kind of problem you areworking on!

2

qE=k

r

F = qE& &


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Symmetry is your friend. Use it when appropriate. Dont use itwhen not appropriate.

G,pair 2

GmMF , attractive

r!

&

The above equation is on the Physics 23 Starting Equation

Sheet, which is posted in the recitation classrooms. You are freeto use Physics 23 starting equations at any time.


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Your starting equations so far are:

(plus Physics 23 starting equations).

1 2

2

12

q qF k12 r

! 0

0

FE =

q

&&

2

qE=k

r

2

dqdE=k .

r

This is a legal variation (use it if you are assigned 21.54a or

21.90!):

You can remove the absolute value signs if you know that dq isalways positive.


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Electric field at center of semicircular line of positive charge.

By symmetry, Ex=0 (why?).RN

dNds

dE

2 2 2

dq ds Rd ddE = k k k k

R R R R

P P N P N! ! !

y

d sin( )dE = dE sin( ) k

R

P N NN !

y y0

d sin( )E = dE = k R

T P N N Magnitude comes from above integral, direction is y.

y

x

There is dq of chargeon the ds of the line.That dq of charge gives

rise to dE of field.


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32 2 2

dx

x a

The integrals below are on page A-4 (appendix B) of your text.

32 2 2

x dx

x a

Your recitation instructor will supply you with needed integrals.

The above integrals may or may not be needed this semester.


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Lecture02 Electric Field; Electric Forces;

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