Assignment #1: Where Do Charges Come From...c.) Polarization involves the use of a charged object to...

Assignment 1 Where Do Charges Come From To see where charges come from you need to zoom in to the atomic model In the atomic representation given in the picture below the wood box and the surface are composed of millions of atoms What is an atom You may recall some atomic models you have learned in previous science classes

A Simple Atomic Model Scientists usually look for the simplest model as possible For our current physics understanding a neutral atom is simply represented as a positive center (the atomrsquos nucleus) matched with negative electrons surrounding it

The previous charge model that was constructed using the tapes does not incorporate the motions of the particles or the coexistence of both positive and negative particles

Atomic Models of Neutral Material and Charged Material View a computer simulation of atomic models with your teacher

In the charge model (your tapes) you did not consider the motion of particles The atomic model of a neutral material takes this into consideration Think about the following questions while looking at the modelsimulation of neutral tapes

1) How can you tell if a material is neutral from its atomic representation

It has equal numbers of positive and negative so that they cancel each other out

2) Which part of the atomic model is more dynamic (is more readily able to move) the positive center or the electron

Only the electrons are able to move

Your teacher will now click the button Charge the Tape to simulate the charged tapes (peeling off the tapes from each other) The top shows the interaction between charged tapes using the charge model The bottom illustrates the atomic model as if you can zoom in to see the particles

3) Record your observations of the zoomed in atomic model of the charged tapes

The left piece of tape has less electrons and is positive The right piece of tape has more electrons and is negative

4)

a) From an atomic perspective how do materials become charged through friction

Electrons leave one material (making it positive) and go to another (making it negative)

b) As shown in the atomic model two oppositely charged tapes attract each other Since each

tape contain both + and - particles why dont they repel each other due to the repulsion between + amp + particles and repulsion between minus amp minus particles Force between overall charges must be greater than that between individual charges

Assignment 2 Charge Polarization

1) Determine whether the following statement is true or false based on your observations in the introduction (magic experimental tape) activity

ldquoA neutral object and a charged object will attract each otherrdquo True (True or False)

2) Wersquoll now look closer at this statement Rub a balloon with some fur (or in your hair if you donrsquot mind it getting messy) Place the rubbed side of the balloon against the wall

a) What happens to the balloon when you put it near the wall It sticks to it (the balloon is attracted to the wall)

b) In this scenario what is the charged object What is the neutral object The balloon The wall c) Polarization involves the use of a charged object to cause the rearrangement of electrons

within a neutral object This rearrangement causes one side of the object to be positive and the other side to be negative There is no gain or loss of electrons in the neutral object so the object remains neutral

Once again wersquore brought back to the rule of opposite charges attract and (repel or attract)

like charges repel (repel or attract)

d) When you rubbed the balloon the balloon became negatively-charged Based on the idea of polarization (ldquorearranging chargesrdquo) label the wallrsquos molecules to show them being polarized

3) Rub some fur across a rubber rod Turn the water in the faucet on a very low steady stream Place the rod near to BUT NOT IN the stream of water

a) What happens to the water when you put the rod next to it The stream bends toward the rod b) In this scenario what is the charged object What is the neutral object The rod The water (H2O)

c) When you rubbed the rubber rod the rod became negatively-charged So why did it bend the waterrsquos stream Read the last paragraph of p 513 in your textbook Draw a picture of a H2O molecule below Label the dipoles

d) In terms of polarization explain what happened when the rod was

brought near the stream of water The water molecules rearrange themselves so that their positive end attracts the negative rubber rod

ndash ndash ndash

+ minus + minus + minus

Assignment 3 Creating Charge by Friction amp Contact Directions Answer each of the following questions followed by an explanation of why the correct

answer is correct and why the incorrect answers are incorrect 1) A positively charged glass rod attracts object X The net charge of object X

(1) must be positive (3) may be zero or positive (2) may be zero or negative (4) must be negative Explanation

bull Positive repels positive (ldquolike charges repelrdquo) so the answers 1 amp 3 are incorrect bull Positive attracts negative (ldquoopposite charges attractrdquo) bull However 4 is not correct because neutral objects (objects with an overall charge of zero) can

polarize (positives and negatives separate within the material) and will therefore attract to any charged object Therefore answer 2 is correct

2) When a neutral metal sphere is charged by contact with a positively charged glass rod the sphere (1) loses protons (2) gains protons (3) loses electrons (4) gains electrons Explanation

bull A positive glass rod is electron deficient (has less electrons than protons) bull When the two objects come in contact the electrons will want to be evenly spread between

the two materials bull The neutral sphere will lose electrons to the glass rod (resulting in a positively-charged

sphere) This is why answer 3 is correct bull Answers 1 amp 2 are incorrect because protons are never transferred

3) A negatively charged plastic comb is brought close to but does not touch a small piece of paper If the comb and the paper are attracted to each other the charge on the paper (1) must be positive (3) may be positive or neutral (2) must be negative (4) may be negative or neutral

Explanation

bull Negative repels negative (ldquolike charges repelrdquo) so the answers 2 amp 4 are incorrect bull Positive attracts negative (ldquoopposite charges attractrdquo) bull However 1 is not correct because neutral objects (objects with an overall charge of zero) can


4) A glass rod becomes positively charge when it is rubbed with silk This net positive charge accumulates because the glass rod (1) loses protons (2) gains electrons (3) gains protons (4) loses electrons Explanation

bull Only electrons can move (So answers 1 amp 3 are incorrect) bull A positive charge is created because an object has less electrons than it has protons bull If the rod had gained electrons it would be negatively-charged (So answer 2 is incorrect) bull Therefore the rod must have lost electrons amp the correct answer is 4

Assignment 4 Creating Charges by Induction

Induction charging is a method used to charge an object without actually touching the object to any other charged object An understanding of charging by induction requires an understanding of charging by conduction and an understanding of the polarization process If you are not already comfortable with these topics you might want to familiarize yourself them prior to reading further (See Assignments 1 amp 2) View the following diagrams being sure to read each of the descriptions below each one Then answer the questions that follow 1 - Charging a Two-Sphere System Using a Negatively Charged Object

2 - Charging a Two-Sphere System Using a Positively Charged Object

Question Before viewing the next diagrams it is important to know what ldquogroundingrdquo is Define grounding (p 511 of the text may be helpful)

A way to neutralize an object by allowing electrons to move to or from the object to the ground (an infinite reservoir of charges)

3 - Charging a Single Sphere Using a Negatively Charged Object

4 - Charging a Single Sphere Using a Positively Charged Object

Questions The following were all upheld in each of the examples of charging by induction 1) The charged object is never touched to the object being charged by induction (touched or never touched)

2) The charged object does not transfer electrons to or receive electrons from the (does or does not)

object being charged

3) The charged object serves to polarize the object being charged (conduct frictionize or polarize)

4) The object being charged is sometimes touched by a ldquogroundrdquo Electrons are (Electrons Protons or Neutrons)

transferred between the ground and the object being charged (either into the object or out of it)

5) The object being charged ultimately receives a charge that is opposite that of (opposite or the same as)

the charged object which is used to polarize it

Assignment 5 Whatrsquos a Charge 1) The Atom - Fill in the subatomic particle chart below based on your knowledge of chemistry and

your Physics Reference Tables

Particle Symbol Location in Atom

Relative Charge (elementary

charges) Charge

(Coulombs) Mass (kg)

proton p+ inside nucleus +1e +16 x 10minus19 C 167 x 10minus27 kg

neutron n0 inside nucleus 0 0 167 x 10minus27 kg

electron e- outside nucleus minus1e minus16 x 10minus19 C 911 x 10minus31 kg

write in the total amount of charge in Coulombs not just +1 or -1

2) Based on the chart above what are the three types of charge positive negative and no charge (neutral) 3) A lithium atom is made up of three protons four neutrons and three electrons The electrons are

found outside of the nucleus

a) Draw a picture of the lithium atom Be sure to label the p+ n0 and e- 3 p+ amp 4 n0 in the nucleus 3 e- outside the nucleus

b) What is the charge in elementary charges (ldquorelativerdquo units) of the lithium nucleus

+3 e

c) What is the charge in Coulombs of the lithium nucleus +3e x 16 x 10-19C = +48 x 10-19C 1 e

d) What is the charge in elementary charges (ldquorelativerdquo units) of the lithiumrsquos electrons minus3 e

e) What is the charge in Coulombs of the lithiumrsquos electrons minus3e x 16 x 10-19C = minus48 x 10-19C 1 e

f) What is the charge of the lithium atom 0 C ((+48 x 10-19 C) + (minus48 x 10-19 C) = 0 C) 4) You learned in chemistry that some atoms have a greater desire for a certain subatomic particle

than others This particle is sometimes even gained or lost to make the atom ldquostablerdquo Which subatomic particle is this

(1) proton (2) neutron (3) electron

5) Hopefully you chose ELECTRON (choice 3) in question 4 ELECTRONS do a LOT of moving in the world around us It is the imbalance of electrons that

cause charges to occur Letrsquos see if you remember this - Atoms that gain or lose electrons are called ions

If an ion is negatively-charged that means the atom gained electrons (gained or lost) If an ion is positively-charged that means the atom lost electrons (gained or lost) 6) A lithium atom likes to lose one electron to become stable a) What would be the charge of the lithium ion in elementary (relative) units +1 e

b) What would be the charge of the lithium ion in Coulombs +16 x 10-19 C 7) One chlorine atom has 17 protons 18 neutrons and 17 electrons Chlorine atoms like to gain

electrons to become stable a) An atom of chlorine will gain one electron What will the relative charge be for the ion (in

elementary charges) minus1 e b) What will be the charge of the chloride ion in Coulombs minus16 x 10-19 C 8) Which of the following is true When objects become electrically charged it is because (1) Both protons and electrons may be moving on to or off of the object (2) Only protons are moving on to or off of the object (3) Only electrons are moving on to or off of the object (4) Becoming electrically charged has nothing to do with the movement of protons or electrons In physics we use Coulomb units for our charges It is also important to be able to convert Coulombs to and from elementary charges (aka ndash ldquorelativerdquo units)

Assignment 6 Conserving amp Converting Charge

1) In the last unit we discussed the conservation of energy In the unit before that we discussed the conservation of momentum For our current unit there is conservation of charge Read Section 322 in the textbook (p 502-504) and answer the questions that follow

a) In terms of subatomic particles how does an object become electrically charged By gaining or losing electrons

When a rubber rod is rubbed with a piece of furhellip

b) What subatomic particle is transferred electrons c) What happens to the charge of the rubber rod Why

Becomes negative ndash it gains electrons

d) What happens to the charge of the fur Why Becomes positive ndash it loses electrons

e) Summarize what ldquoconservation of chargerdquo means The net electrical charge will always be the same ndash electrons are not created or

destroyed just transferred

You may have noticed in Assignment 5 that you will often need to convert between elementary charges and Coulomb units Follow along with the example below then answer the questions that follow Example Neutral Object A is rubbed against neutral object B transferring 20 x 109 electrons to it What is the charge on each object in units of elementary charges and in units of

Coulombs

Object A Object B -Lost 20 x 109 electrons (got rid of ndash) -Gained 20 x 109 electrons (got more ndash) -Will now be positive (more + than ndash ) -Will now be negative (more ndash than + ) Charge +20 x 109 e = +32 x 10-10C Charge ndash20 x 109 e = ndash32 x 10-10C Two ways to convert e to C Method 1 20 x 109 e x 16 x 10-19 C = 32 x 10-10C 1 e Method 2 20 x 109 e x 1C = 32 x 10-10C 625 x 1018 e 2) Explain how charge was conserved when Object A was rubbed against Object B

Electrons lost by Object A were gained by B The net electrical charge did not

change (still 0 C)

3) What is the net charge of an object possessing an excess of 60 x 106 electrons

-6 x 106 e

-6 x 106 e x 16 x 10-19 C = -96 x 10-13 C 1 e 4) An alpha particle consists of two protons and two neutrons The alpha particles charge of +2

elementary charges is equivalent to

(1) 80 x 10-20 C (2) 32 x 1019 C (3) 32 x 10-19 C (4) 12 x 1019 C

+2 e x 16 x 10-19 C = +32 x 10-19 C 1 e

5) What is the smallest electric charge that can be put on an object

(1) 625 x 1018 C (2) 160 x 10-19 C (3) 911 x 10-31 C (4) 900 x 109 C 6) An object cannot have a charge of

(1) 32 x 10-19 C (2) 45 x 10-19 C (3) 96 x 10-19 C (4) 80 x 10-19 C

= 2 e =28 e =6 e =5 e canrsquot have a fractional value

7) What is the net static electric charge on a metal sphere having an excess of +3 elementary charges

(1) 480 x 1019 C (2) 160 x 10-19 C (3) 480 x 10-19 C (4) 300 x 100 C

+3 e x 16 x 10-19 C = +48 x 10-19 C 1 e 8) Which of the following is a true statement

(1) Two positively-charged objects will attract one another

(2) Two negatively-charged objects will attract one another

(3) A positively-charged object and a negatively-charged object will repel one another

(4) A positively-charged object and a negatively-charged object will attract one another

9) The following particles are linked together +13e -10e +27e -44e and +19e

a) What will be the net charge when they are all linked together in units of ldquoerdquo

+13 ndash 10 + 27 ndash 44 + 19 = + 5 e

b) What will be the net charge when they are all linked together in units of Coulombs

+5 e x 16 x 10-19 C = +8 x 10-19 C 1 e

Assignment 7 Creating Charges by Contact

When two identical (as in size) charged objects come into contact the net charge of the two objects together is shared equally between them Sometimes this process is also called charging by conduction

Example

emdash move from the more negative

object to a more The end result is an even positive one distribution of charge

Now itrsquos your turn

+100 microC +300 microC +200 microC +200 microC Which way do the

electrons flow LEFT TO RIGHT

-5 C neutral -25 C -25 C Which way do the electrons flow LEFT TO RIGHT

+1800 microC -600 microC +600 microC +600 microC Which way do the

electrons flow RIGHT TO LEFT

Assignment 8 Electrical Charge Problems ID 1) What is the net charge on the nucleus of any atom lDiit

2) What is the standard unit of charge CouCIINo ie) 3) Like charges Ceye each other while opposite charges ~oc-a

4) When an object loses electrons it becomes o-ht~ charged

5) As a positive and negative charge are brought together a force acts on

(1) mostly the positive charge OnIY the negative charge (2) only the positive charge 4) both charges equally Ja I~ o~ -reoCt~~CN s 3 ~~

6) If a negatively charged rod is brought close t9 a neutral pithball

(1) the pith ball will repel due to transfer of charge the pithball will attract due to transfer of charge he pithball will become polarized and attract ~ the pithball will become polarized and repel

7) If the rod (from question 6) touches the pithball the pithball will

~acquire a charge opposite the rod (3) not acquire a charge but remain polarized ampcquire the same charge as the rod (4) explode upon gaining a charge

8) After rubbing on a students hair a balloon that was initially neutral has 5000 excess electrons What is the charge on the balloon in coulombs

- 5ot()e ~ pIC)f( ~

10) How does the charge of an electron differ from the charge of a proton I ~

~-4t OS) TIt _-e -c bull 1)-( +amp +c)I 0 C

11) If you scuff electrons onto your feet while walking across a rug are you negatively or positively charged Is the rug negatively or positively charged -aL ~

lw wOacl amp- ~~t- C~ ~) t )e~ yK~c

1

r 12) Look at Figure 328 (below)

o c Figure 328 i Charging by induction

Would the charges induced on spheres A and B be exactly equal and opposite Explain why or

why not feS--n~ ~- o- =1 ~W A~t ~ewe6 ~ ~~tr

13) Why does the negative rod in Figure 328 (question 12) have the same charge before and after the spheres are charged (look carefully at number of electrons present) - shy

~ nO ne~ c~ ~ cooJT

A)~ ~ St~S~ 14) Look at Figure 329 (below)

~ NEl CHAR6pound IS lERO NET OIAlcbE S +

middot ~f L~ t ~+ffmiddot-~~ bull middotmiddotmiddot~hmiddot-middot U6 b c u d

Figure 329 i Chllrge induction by grounding

In this case why does the negative rod NOT have the same charge before and after (look carefully at number of electrons present)

1k tcgt4 C-rL Qow-~ ~i~ ~ bol So eec~Ott ~ ~s~trtd

r Base your answers to questions 15-22 on the information and diagram below

The diagram below shows two identical metal spheres A and B on insulated stands Sphere A has a charge of - 4 x 103 C and sphere B has a charge of +6 x 103 C

15) What is the total (net) charge of the 2 spheres E ~ (+c-WO-C) (-LtOcJ ItO shy

16) If the 2 spheres are brought into contact with each other and placed so they are touching what will the total charge be Why -_ -i

+ ~OOOe ~otNL 0 GJaoampJ ~cou 0 cO$UV-n of ~ - no oS 0 ~o ~ tc+tr~ poundrO c~ fI fe+ iL4t(~middotca cNat1r~)

17) When the spheres are in contact what will the electrons do Explain -I

poundec-h-~~ w- ~NLr lie w) MOve 1VAJO 4Q

~ MOtt -pol itt ~fC ~

18) If the spheres are now separated what will be the charge on sphere A on B

+lOOO C _ + 000 ~ O~ e~ Sp~c-2 19 W~article is transferred between sphere A and B

~Iectron (2) proton (3) neutron

20) Did the particles move from A to B or from B to A Explain )

~ A - ~ (se e)(t--~~ -I oq

21) How much charge (in coulombs) was transferred UANlt shy ooot)S)l t)~~ (SfNt ~~ CI~ --vw-A

St~ ~ ~en~ ~Ofc ~e --0 COOc 22) How many electrons were transferred Show all Wl~ )

S~ f)- ~ e J ~ Oe I- Jt tt) laquo

Assignment 9 Coulomb Force Internet Activity (PART 1)

Open your web browser to wwwexplorelearningcom Sign in using the Username rh student 1 and Password 1 Search Coulomb Force -7 Run the gizmo Coulomb Force (Static)

1) Keep the Charge on A and on B at 100 x1 04 C

Would you expect the charges to repel or attract each other r ~

2) Click on the show force vector for charge A and B Draw the arrows below

( A BIIIi

q = 1100) x 10 C

3) Change the value on Charge B to -100 X 104 C

Do the charges repel or attract each other _fA=k~-tCt=lPl~----____ Draw the arrows below

Abull B I

4) Change the value on Charge A to -100 x10middot4 C Keep Charge B at -100 x 104 C

Do the charges repel or attract each other __rll3IIIU~p-t--I--____ Draw the arrows below

A Bbull l( P

bull

5) Which of the following images represents the Coulomb force between two positively charged particles

(1) Image A A B(2) Image B Imagec A B A B

~ ~ --+ -+ lt_Image]gt

C 0

A B A B ~ ~~ -

Check your answer with the simulation

6) Define a vector quantity ~~o Nt bcyenr =uA(tt) ~ e~ (~fMhM~ eOu-al -)

7) Change the value of Charge A to a value that is smaller than 100 x1 0-4 C and keep Charge B at 100 x10-4 C How does the magnitude of the vector change ba~ oIroW~

~tt ~tyU 1Mtf~At

8) Change both the values of Charge A and Charge B to values that are smaller than 100 x1 0-4 C

How does the magnitude of the vector change ~ 10gt Gtc~~ 1- tu ~Nw-er I)er ~~

9) Change both the values of Charge A and Charge B to values that are larger than 100 x10-4 C How does the magnitude of the vector change~ It dior- ~~ fN)~

~u- 1 fte~~d

10) In each of the above scenarios how did the force of Charge A on Charge B compare to the force of Charge B on Charge A (ie - how do the lengths of the vectors compare)

~~ w~ o_~S ~ ~(~ ~cL

11) Try dragging Charge A closerto Charge what happens to the force vector Describe how b th the directi n and the magnitude of the vector are affected

Assignment 10 The Inverse Square Law

In the realm of nature the inverse square law shows Lip over and over again Heres an example of it from something you are sort of familiar with having your picture taken

The light from a flash falls off with distance When you double the distance you get one quarter as much light This relationship is

called the inverse s uare law

When the flash fires the beam of light expands as it moves father from the camera so its intensity falls off with distance As a result subjects nearer the flash will be illuminated with a more intense light than subjects farther away The rate at which the light falls off is described by the inverse square law The law states that if the distance between the flash and subject is doubled only one quarter the amount of light will reach the subject because the same amount of light is spread over a larger area Conversely when the distance is halved four times as much light falls on a given area

THINK Based on this info can you predict how much weaker the light intensity in the picture above will be if the distance from the camera flash was tripled instead of doubled

Earlier in the year you learned that the law of universal gravitation works this way as well As the distance between objects with mass triples for example the force of gravitational attraction becomes 19 as much as it was originally

Notice that the relationship in both cases is inverse meaning as distance increases the other variable decreases (light intensity and gravity in our two examples) Further the relationship in inverse squared meaning that not only does the other variable decrease but it decreases by a factor that is the square of the other variables increase As distance became 2 times larger in the camera flash example the light intensity at that distance became not V2 as much but rather as much This same relationship can be seen in Newtons Law of Gravitation

Note that when distance (r) =1 below the denominator has a value of 1 but if distance is doubled so r =2 then the denominator value is 4

In the study of static electricity the inverse square law again pops up as it describes way that electrically charged objects exert force on each other

So think about it

If two protons are sitting 1 meter apart lets say they exert F amount of repelling force on each other

o 0 I~ ~I

r = 1 meter

So

1 If the distance between the protons doubles to 2 meters what will happen to the repelling force between the protons

a) Would it get weaker or stronger

b) By what factor

(1) One times weaker ~hree times weaker (2) Two times weaker ~our times weaker

Explain your choice in terms of the inverse square relati~nship between F and r l shy

~c~ C -z ~s StampfO-cl () 1Wt ~~QW~~ becM~ L 1) ~ ~ t~M (~op o~ C- middotc-ltflIA~S F cS (~ker)

2) If the protons are pushed closer together will the repelling force between them increase or decrease

3 If the protons are pushed 4 times closer to each other (025 m apart now) what is likely to happen to the force they exert on each other

(1) The force will become 4 times weaker (2) The force will become 4 times stronger ~he force will become 16 times weaker ~ he force will become 16 times stronger

4) A student moves the protons some distance farther apart than 1 meter She then measures the ~ between the protons to be 136th of its original value How many meters apart are the

protons now J r ]_fz~ r1 _r - -M _

5 Which graph below do you think best depicts the inverse squared relationship between F and r

w-_ bullbulll ~0

ra~( bull ~ rtVl middot jiI ___ b~~)

=

KE~ Assignment 11 Coulomb Force Internet Activity (PART 2)

1) In each of the scenarios from PART 1 (Assign 9) how did the force of Charge A on Charge B compare to the force of Char e B on Charge A (ie - how do the lengths th vectors compare)

2) What happened to the magnitude of the vectors when the two charges were brought closer together ~ ~

3) What happened to the magnitude of the vectors when the two charges were brought further apart

~ ~AtL Click the show distance toggle at the top of the screen Click the show vector notation at the bottom of the screen The vector notation shows us the value of the force We will use only the magnitude of the force (in absolute value signs) do not worry about the i and j notation

4) Keep both Charge A and Charge B at 100 x 10-4 C a What is the force when they are 20 meters apart

b Would you expect the force to increase or decrease when the charges are brought closer together

c What is the force when they are 10 meters apart

d What is the force when they are 5 meters apart

e Would you expect the force to increase or decrease when the charges are brought further apart

f What is the force when they are 30 meters apart

g What is the force when they are 40 meters apart

h How is the force between charges and the distance between them related

t~ SC)6I - 5) Two charged particles are separated by a distance of 12 m The Coulomb force between them is

20 N What will the Coulomb force be if the same particles are separated by a distance of 6 m f la

~o N (2) 40 N (3) 10 N (4) 5 N ramp~ I 6) What happened to the magnitude of the vectors when the values of Charge A and Charge B were

decreased

0 N

7) What happened to the magnitude of the vectors when the values of Charge A and Charge B were increased

rLteayd 8) Keeping the distance constant

a What is the force when Charge A and Charge B each have a value of 100 x 10-4 C

l ~6 r 0) qo N

b What is the force when Charge A stays at 100 x 10-4 C and Charge B is decreased to 4

50 x 10- C 5 N

c What is the force when Charge A stays at 100 x 10-4 C and Charge B is decreased to 20 x 10-4 C

ttO N

d What is the force when Charge A stays at 100 x 10-4 C and Charge B is increased to 150 x 10-4 C

~S N

e What is the force when both Charge A and Charge B have a value of 150 x 10-4 C

f How are charge and torce related G rec+ At c~ middotuoS4 F laquoamp1

( l s+-cA w shy9) Keeping Charge A constant which impacts the fo~re 0 champ~cof

(a) dividing Charge B in half OR ~iViding the distance in half IS J IO-eJbo N) (amp0 )

Explain your choice

~nee- F d ~t -c cIrt~ ce~ ~~ ~~ -1 1 hlalN

t~ F hAlf ~ca F Atcl Qlew+ ~ II bull

middot S ~_l rcdcc1I cliv~middotdr r f h ~ttH eo

re ~ F be~ adf

r

Assignment 12 Coulombs Law Problems

1) Given the mathematical representation of Coulombs Law F =k ql2 describe in words the r

relationship betwee=tric force and charge as well as electric force and distance

bull EedTmiddot CAl - - S ciirc+ty r-ohO ~ trocLado oc ~c 2 CNL~S

middot~W~ ~ _~~ul -~ r(uon ~ ~ cl~u ~~ ~O ~y

2) Why does electrical force between a pair of charged objects diminish~ they are moved farther

apart oeCAU -C -tNL U (suorecw recdC)fshy

3) By how much does the electric force between a pair of charged bodies diminish when their separation is (a) doubled (b) tripled

~middot1U~ 4~ (~-yen ~ (-hamp~ )

4) By what factor doe~ t~e force between two charged bodies increase if the separating distance between them is (a) reduced to V2 of its original (b) reduced to 14 its original

Lt ~ ~~ur- ~ -shy(F~li) ~) (lt)~ampltt 1-)

5) Two positive charges of 60 x 10 C are separated by 050 m Draw a force diagram for each of the charges considering only electrostatic forces What is the magnitude of the force between the

6

t(a Oe)shycharges Is this force repulsive or attractive ~Ill-baWe ~ l~O tlmiddot at a bull r 010 (01$0-)

~ ~ ~ t ~amp amp~ O ~ ~t14 t-l) ~bull ~~~ ~

C ~ LIr to~4L Q G)I=- l aM ~ 1= +)

4 4 6) A negative charge of 20 x 10 C and a positive charge of 80 x 10 C are separated by 030 m

What is the magnitude of the force between the charges Is thii t~rce repulsi e or attractiVe)

-1-0-( ~t ~ (8~O -1 Omiddott Omiddot~ + ~ crt Camp 0 0-) ~ 1 ~

~ oHWampCN CCAlir-C I 0lO -aD at 0 ~ - 8CftIt ~~ ~~- ~ - c 6) ~ c w ~o feMit~

r 7) A force of 225 x 10

3 N exists between two identical point charges of 50 x 10

6 C What distance

separates the charges

~ 11~ bull 0middot ~ Fe ~ ~ ~ kq~ Co (1 rIt S 0middot C

68) A force of 30 x 10 N exists between two charged objects whose centers are 30 m apart

7One of the objects has a charge of 15 x 10 C Find the charge on the other object

-I tA ~ ~ ~t a ~~ C) M ~ J ) Q ~ f 1 kca ~~~

-C -- Ishy $-0 G (JC 0 NA~~J shy~~ ~ c~d ~)~IDC)

~ (i y Omiddotsl

9) What is the magnitude of two identical point charges which are 50 meters apart when the force between them is 144 N

1= - ~ =) ~ r=S~

=

Assignment 13 Graphing Electric Force

Directions -Determine the force between the two charges listed at the following distances -USING A PENCIL - Graph force vs distance using the graph on the following page

and connect the data points with a smooth curved line -Use a pencil for the graph -Show all work for the force calculations (include equation substitution and final

answers with units)

Distance =1m

F gttJtOmiddot M

Distance = 2m

O~F ~ O

Distance =3m

F Ct 0 N

Distance =4m

F 3 ~ O~ N

Distance = 5m

f~ ~ ON

Distance =6m

F IS~OC~

Distance = 7m

f ~ 1 ~ 0 ~

Distance =8m

F ~ ~ loN

Distance =9m

F laquo1- ~ tOea N

Distance = 10m

F S~ ~ O~

i~ f(1G-nc Of el 41 ~ middotsanC_

rr ~d

bull

Z wft

bull-tftJ ~~

1pound J~ ~ d CG f 1

1lltW2 UlI~

~

~

~

C~

~~

1- 5 bull

~

-

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I

g ffim ++ ~

~ g1 llJ

shy

~ i

E iiiL +shy J I Ishy

I

-

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mEI J t m~ ~ I

~ U

II t=R=H IiIIr j

EBi~ ~

BEI11III ~ n

-000II ~ jiiiI ~

rIshy --bull J

IIIIIIII

m ffim II IJ

bull 100

I ~

bull 0

Assignment 14 Putting It All Together Electrical Charges amp Electric Force Problems

1) Gravitational force depends on mass What property determines electrical force

c~ 2) When the distance between two charges is cut in half the electric force is b )

pound-- tc1 3 a) If an electron at a certain distance from a charged particle is attracted with a certain force

how will the force compare at twice the distance

~ A ~~ (F - tC-~) b) Is the charged particle in this case positive or negative PWyen-

4) Metal sphere A has a charge of +12 elementary charges and identical sphere B has a charge of +16 elementary charges After the two spheres are brought into contact what is the charge on sphere A

Sphere ~ Sphere 8

5) An electroscope is a device with a metal knob a metal stem and freely hanging metal leaves used to detect charges The diagram below shows a positively charged leaf electroscope

As a positively charged glass rod is brought near the knob of the electroscope the separation of the electroscope leaves will

(1) remain the same (2) decrease (sect)increase

6) Which graph best represents the relationship between the magnitude of the electrostatic force and the distance between two 0 positely charged particles

CDQ) CD eee o ooo lJ LLLLlJ

Distance DistanceDistance D~e (1 ) (2) (4)amp

r 7) Two electrically neutral metal spheres A and B on insulating stands are placed in contact with

each other A negatively charged rod is brought near but does not touch the spheres as shown in the diagram below

NEGATIVELYHow are the spheres now charged CHARGED ROD

(1) A is negative and B is negative is negative and B is positive

3 is positive and B is negative ~ A is positive and B is positive

INSULATING STANDS

8) The diagram to the ri~ht shows two metal spheres +10 x 106 C +30 x 106 C charged to +10 x 10- coulomb and +30 x 10-6 coulomb respectively on insulating stands separated by a distance of 010 meter

The spheres are touched together and then returned to their original positions

As a result the magnitude of the electrostatic force between the spheres changes from 27 N to ___

Show all work including equation(s) and substitution with O10m 1 units~ _ --r ~Qot~ 111 10middot Co c1

~~cgt~o- JF~ ~~(bullmiddothdN-i tCi J 9C1 0 ~ f r (0 0 I

C at--o M 1 9) The diagram to the right shows two negatively charged

balloons suspended from nonconducting strings being held by a student

What occurs as the student brings the balloons closer to each other without allowing them to touch

amphe magnitude of the electrostatic force between the balloons increases and they repel each other

(2) The magnitude of the electrostatic force between the balloons increases and they attract each other

(3) The magnitude of the electrostatic force between the balloons decreases and they repel each other

(4) The magnitude of the electrostatic force between the balloons decreases and they attract each other

J-tf _- I - -) -~

10) Two protons are located one meter apart Compared to the gravitational force of attraction between the two protons the electrostatic force between the protons is

(1) weaker and attractive ampstronger and attractive (2) weaker and repulsive ~tronger and repulsive

11) In the diagram below a positive test charge is located between two charged spheres A and B Sphere A has a charge of +2q and is located 02 meter from the test charge Sphere B has a charge of -2q and is located 01 meter from the test charge

A Test charge

~ ~ 02 rn Olm

If the magnitude of the force on the test charge due to sphere A is F what is the magnitude of the force on the test charge due to sphere B cS=amp) ~ f t ~ ~F (2) 2F (3) f (4) F ~ 4 2

12) Four small metal spheres R S T and U on insulating stands act on each other by means of electrostatic forces

It was known that sphere S is negatively charged The following observations were made 1l T U cC b4I -t O~ ~~~

bull Sphere S attracts all the other spheres I hN lDT ~4ln bull Spheres Tand U repel each other -~ ~o~ ~ ~ J bull Sphere R attracts all the other spheres - ~ ~ fI4~

Determine the charge on each sphere and complete the table below noting for each sphere if it is positive (+) negative (-) or neutral (0)

Sphere Charge

R 0 T 4shyU

r Base your answers to questions 13 amp 14 on the following information

A lightweight sphere hangs by an insulating thread A student wishes to determine if the sphere is neutral or electrostatically charged She has a negatively charged rubber rod and a positively charged glass rod She does not touch the sphere with the rods but runs tests by bringing them near the sphere one at a time

I u -- n I Rubber Rod w -- u

r-------~ Glass ROd

13) Describe the test result that would prove that the sphere in the given situation is neutral

1M ~~e ~~cl ~ o~~ ~ Ju~ fOcS

14) Describe the test result that would prove that the sphere in the given situation is positively charged

-1ht -St~ -S -k ~t

r

Assignment 15 Electric Field Simulation

This activity is designed to demonstrate the appearance of electric fields around various charge configurations

bull Go to httpphetcoloradoedusimulationssimsphpsim=Charges_and_Fields bull Click the green Run Now button to begin

1 Take a positive 1 nanD-Coulomb charge from the box of positive charges and drag it to the middle of the screen Now take a field sensor and move it around the positive charge you placed to investigate the field around it (For the field sensor to work you must release it for a moment then pick it up again)

a Does the field get stronger or weaker as you get further away from the charge How can you tell th 1)

lA)tpctr ~~ ~Cl~ ri6- o~- n~ eC ~ (~1udat) ~e+s ~rNM

b As you mov9he sensor around the positive charge do you see any pattern to the direction of the electric field Describe the direction of the electric field around the positive charge + is w~S fO~ Qa)~w~ ~ lO~~ ~~~

Electric field direction at any location is DEFINED to be the same as the direction of the force a positive charge would feel at that point In other words the field sensor you were playing with was just another positive charge and the arrow on it was just showing the strength and direction of the force it was feeling

2 Click on Show E - Field to show the electric field Notice that the electric field at any point always points away from the positive charge This is because any positive test charge would always feel a force in that direction (since like charges repel)

3 Knowing that electric field direction is defined by the direction of the force a positive charge would feel how do you think the field around a single negative charge would look

1) ~-eel ~G ~ ~~~ ~t With the box for Show E-Field still checked hit the Clear All button and this time drag a ~ngle negative charge to the middle of the screen to see if your prediction was correct ~P ~

4 Create and draw the Electric Field lines for the charge configurations on the following page (Instead of drawing a million tiny arrows however just draw some smooth lines to show the basic shape of the field Dont forget to put arrowheads on your lines to indicate direction0 As an example heres what the field around your positive charge in 1 would look like

r

a)

b)

d)

r

Assignment 16 GraQhing Electric Field Strength ~

Electric Field Strength (NC) Radius (m) 780 x 10~ OS 200 x 10~ 10 090 X 10) 15 060 x 103 20 050 x 1O~ 25 030 x 103 30 026 x 10 35 020 x 1 O~ 40 010 X 10) 45

The data above reflects electric field strength measurements at specific radii as a test charge is moved at varying distances from a large positive charge as shown below

o Large Positive Charge Test Charge is a proton

radius

Directions 1) USING A PENCIL - Graph Electric Field Strength vs Radius on p 45 2) Label the x and yaxes Plot data points and best fit curve through your points 3) Give the graph a title

Questions 1) What happens to the electric field strength as the positive test charge is moved further away

from the large positive charge

If Jets uc ~ I sYNatr

2) Explain how this can be similar to gravitational field strength Ishye ~r~1 ()J objw- s ~ otO~~~~ (Sueh as PcrSOf sel~ -A-~ Eor ~ lcS -- ~ 0- G~~~ it re

3) What type of mathematical relationship exists between electric field strength and distance separating charges e t f +~c

IUt ~ middot ~ or~ +004amp E f _ 11C1a Itt

-t

- C ~ ~ct l - fl f

shy y

I r

4) a) Calculate the force on the proton at a distance of 15 m from the large positive charge SHOW ALL WORK BELOW INCLUDING EQUATIONS SUBSTITUTION AND INCLUSION OF UNITS

r SIf E ~ F Eq ~) ~ 0 ItI)~~c Cl -= (O~ to N~( lIS 4+-~ Co 134 O-N

b) Calculate the force on the proton at a distance of 35 m from the large positive charge SHOW ALL WORK BELOW INCLUDING EQUATIONS SUBSTITUTION AND INCLUSION OF UNITS

r- ~i~ F E (02 0 MXw yen C)-e)E o Le~C)MI~ Co l +In IOmiddotC ~ bull IIgt If U)It N ]

c) Write a statement comparing Electric Field Strength Force on a charge and distance separating charges

A~ C __~~ be~4tr ~es middott-~es

~ ~ecm-c ~~ Stt~~ otc ~Ctc OW cr Ct-x~e lI)i ell 0 ~dIa-c-0- ~ d~CL ~u0rt6 middot

~ or-4 C - o~ noampfSt tvlre ~ c-)

ceo-~~h~ ~

~ ~ CIO +-+-+-I-+shy

~ 1--+-+Ishy2 l-a

~ t-+-+I-I-+-+--I--+-+--I----l-----Ishy

f~ cOl++t+--+-+--H---+-I-I-I---+-+--t-Ishy

I-t-H-I-J-+-J-+-lshyIS- ~+I+--I----l-l-l-

Sl -U -- If~+-II-+---I--I ~ ~~~TrrrHrHrl~4+++++~~~ ~ ~~rT~++++~44~~H-I---+-~L

~ 1~~-I-J 1~ LLJ 1 r-t-t-Hl--+--I---t-+---+-Jshy

l~~TT~~~~~~~~~~

0 ~ wt-t--++-++--+-+-+-t-l-l--i--JH--l-I-L

t2 -+--I---I-J--I

~amprrrr+++~~---+-I-I-~

0 --+---+

r Assignment 17 Electric Field Problems

1) Why is an electric field a vector quantity _ _ l J -L

-- os ~ 0 ~~M amprO ClItt~CIt

2) In what direction do electric field lines go

o~ ~~ ~S~S 12 ~ ~ 3) Draw the electric field of the charge configuration below

4) What is the magnitude of the electric field strength at a point in a field where a pith ball with a 14

positive charge of 16 x 10- C experiences a ~1~0~N~f~or~c~e~_-=-~~

E ~ 10 ~~_ ~ T~l5 ~ 01 ~1t1 - 0 ~CS~ t shy

5) What is the magnitude of the electric field strength at a point in a field where an electron experiences a 10 N force ~ 0- ntt6

E Fe o~ (15 x C~ ~l~ shy - -InOC 0 ~tl~

~ Assignment 18 Electric Potential Energy amp Potential Difference

Electric Potential Energy

1 What do like charges like to do Repel each other or attract each other fI~

2 Given your answer to question 1 would it require more work to move the charges closer together or further apart mDtt weryen- -D mo~c cO~

3 Given your answer to question number 2 label which set of charges has the highest amount of energy and the lowest amount of energy

8 88 Ou) e h~h ~e

The above scenario is very similar to a spring being compressed

4 What do opposite charges like to do Repel each other or attract each other d fOLt~

5 Given your answer to questionAumber 4 would it require more work to move the charges closer together or further apart ~~~ 4 ~+


()B middot G (3It)w ~a ~~ PE

The above scenario is very similar to a spring that is being stretched

Potential Difference (aka Voltage)

Definition The potential difference between two points A and B is the difference in electric potential energy per unit of charge that is brought from A to B (or B to A)

Since it is energy per charge we can derive the formula here

Equation for Potential Difference

E

I Check your reference table Find the equation for Potential Differe1ce Was your =on co~Qt

GO~ j- _ ~ middot it

Using your equation on the previous page determine the 81 units for potential difference Fill in the middle box only Hint Think about the units for energy and the units for charge

Units for Potential

Difference

T-C

Remember that potential difference has these units

where q is the charge being brought from A to B v=W W is the work required to move it

q V is the potential difference between points A an B

80 in other words if the potential difference between two points is 9 V = 9 JIC (like the terminals of a 9 V battery) this means that the battery supplies 9 J of energy for every Coulombs worth of charge that moves across the terminals If an electrical device has a potential difference of 9 V across it this means it uses 9 J of energy for every Coulombs worth of charge that runs through it

1 Find the work done on an electron as it passes throu~h a potential difference of 1 Volt

4 W V~ (I~11 tomiddot C) tll( tomiddot l tt

a Check the front of your reference table for this S~Cifi~7)unt What does this amount represent eectn 0 T e

This is a special name given to this specific amount of energy This unit for energy is sometimes used instead of Joules to make it easier to compare amounts We can convert between these two since we know 1 eV =160 x1 019 Joules

2 30 J of work are required to bring a 1 C charge from point A to point B a Calculate the potential difference between these pOints

l~~ J( 3D 1 b How much work is required to bring a 25 C c~ to B (or B to A)

)l~ (p Xl5 C) lJ c Express your last answer in electron-volts using the conversion you discovered above

TS~ II eoI ~ ~ cr II 1 -~cr~~ l shy

3 How much work is needed to move a charge of 0003 C through a potential difference of 50 V Express your answer in both Joules and electron-volts

v~ 4= (501000~t) OIS l O S l ~ eJ (--~-----O~--~lI

-lC o-~l - shy4 Two oppositely charged plates are often used in physics to produce a uniform electric field In

the picture below an electron is brought from plate A to point B

f~cA I +1 el1l-lgt

(- - - -I a Does the potential energy of the electron increase or decrease as this happens

b If it takes 20x1 0-15 J to bring the electron from A to B what is the potential difference between the plates

HI ()r4-~ 2~ _ ~12 StJlV]

c ~ 0 c -)---shyc If the electron is released where will it go

d How much kinetic energy will it have when it gets back to the other plate Hint Conservation of Energy

- rN~1O-C~ lgte Kc t IL)t 0middot r_ --t1 ~ rI

Nedtd ~ Gh ~t ~tk - tD ~f~ No A8 ~

Assignment 19 Potential Difference Problems KE~

1) Which of the following is equivalent to a Volt 4 ~ ~ I a) CJ (9JC c) NC c d) CN

2) The work done in moving a 2 C charge in an electric field from p~t A to point B is 300 J What Is the potential difference between these paints 4 vJ ~B b) 0003 V c) 6~07- 1t d) 40 V

3) An electron placed in an electric field experiences an upwrd ~~~5wtt is the direction of the electric field at that location ~cl Wt op ~~ 00It~ ~oco shy

a) upward (0[ downw~ c) sideways d) forward

4) If 60 joules of work is required to move 50 coulombs of charge b~-fm two points in an electric field what is the potential difference between these oints - oQ

St a) 50 V b) 60 V d) 300 V

5) What happens to the force between two electrons if the distance be~tweliIIIoIiii-m is halved __-La a) halved b) doubled c) quartered d) quadruple ( bull J bull I

iJ 6) In the diagram to the right which direction would you have to move the

c) 12 V

small charge in order to increase its electrical potential energy

A D

7) Describe the energy conversion when a charged object is released in an electric field (For example the pie plates released in the electric field of the Van de Graaf generator)

~etJnC- ~4tI~ol eIU~~ S cR~

-h~~~(~~~

8) It takes 200 J of work to move a 2 C charge from point A to point B What is the potential difference between these two points

j cl 2OC)1 ~ It ~

9) How much work is needed to move a charge of 0003 C through a potential difference of 50 V Express your answer in both Joules and electron-volts

10) A uniform electric field exerts a constant force of 0003 N on a 20 x 10-6 C charge as it moves a distance of 05 m from point A to point B What is the potential difference between point A and point B

Fe OCO~ N

c)~ d~( O ~ N

I

KEy Assignment 20 Electric Fields amp Potential Difference Problems

1) How much electrical energy is required to move a 400-microcoulomb charge through a potential difference of 360 volts

~ It 0--lt bull-I 1 amp

Q -11-- _-- C) w ~

V~~ W~4~ (1~lboCe) 1Ii1(10)111

2) What is the magnitude of the electric field intensity at a point where a proton experiences an electrostatic force of magnitude 230 x 1025 newton

C~ E to (1 ~ (f-) bull fa Ll -- ~I ~IC0middot l I-_ 0middotlt) bull lI ~ J ~ 1~ () N

3) Which graph best represents the relationship between the strength of an electric field and distance from a point charge

~e Distance Distance Distance (2) (3) (4)~

4) A positive test charge is placed between an electron e and a proton p as shown in the diagram below A

r TEST CHARGE

8 ~)

When the test charge is released it will move toward regD ltplusmnf Bet) (1)A (2)8 (3)C c

5 A moving electron is deflected by two oppositely charged parallel plates as shown in the diagram below shytJC ~ The electric field between the plates is directed from e A -B

ELECTRON (-)-D I (1) A to B ~ to 0 (2)DtoC ~toA

6) Identical charges A B and C are located between two oppositely charged parallel plates as shown in the diagram below

The magnitude of the force exerted on the charges by the electric field rit i++ + + + ++++ ++1 between the plates is ~ ~ ~ W

E - h fir 0 --- ~ A~ (1) the same on A and C but less on B -L ~~ Be

ast on A and greatest on C ~ C (poundgt 3) e same for A B and C ~ 1___________ I

greatest on A and least on C -shy~ 7) The diagram below shows a pOint P located midway between two oppositely charged parallel

plates

1+ + + + +1 If an electron is introduced at point P the electron will p

(1) accelerate toward the negatively charged plate [ - -I (2) travel at constant speed toward the negatively charged plate

ravel at constant speed toward the positively charged plate (4) ccelerate toward the positively charged plate e ~~ --0

8) In an electric field 090 joule of work is required to bring 045 coulomb of charge from point A to

point B What is the electric potential difference between points A and B I = 0410-r ~2~ (2) 50 V (3) 041 V (4) 050 V ( Omiddot4S~ -7 1-

9) An object with a net charge of 480 x 10-6 ctt~mb experiences an electrostatic force having a magnitude of 600 x 10-2 newton when placed near a negativeIY(~jarged metal sphere What is the electric field strength at this location I -l E ~ ~~O-N ~88 x 10-8 NC directed away from the sphere -1 _-

88 x 10-8 NC directed toward the sphere lamp)t0 - shy3) 25 x 104 NC directed toward the sphere

125 x 104 NC directed away from the sphere 1 sao ~Ie ~ 10) The diagram below represents a source of potential difference connected to two large parallel

metal plates separated by a distance of 40 x 10-3 meter

Which statement best describes the electric field strength between the plates

(1) It is a maximum at point C (2) It is a maximum at point B ~ is zero at point BOiS the same at points A B and C

r 11) How much work is required to move a single electron through ce of 100 volts

~l ~~f 00 (ubullbullt()~~r t()~ t

12) How much work is done in moving 50 coulombs of charge against a potential difference of 12

volts t= In~s e) (1) 12 J (3) 30 J (4) 24 JB

13 The diagram below shows the arrangement of three charged hollow metal spheres A B and C The arrows indicate the direction of the electric forces acting between the spheres At least two of the spheres are positively charged

Which sphere if any could be negatively charged

(2) sphere B (3) sphere C (4) no sphere

14) In the diagram below proton p neutron n and electron e are located as shown between two oppositely charged plates

1++++++++++++++++pound++1

reg

I - ----I

The magnitude of acceleration will be greatest for the

Qlectron because it has the smallest mass ~roton because it is farthest from the negative plate

(3) neutron because it has the greatest mass (4) neutron because it is neutral

r

Assignment 21 Regents Static Electricity Problems

1) If the distance separating an electron and a proton is halved the magnitude of the electrostatic force between these charged particles will be (1) unchanged (2) doubled (3) quartered

2) Two similar metal spheres A and B have charges of +20 gtlt 10-6 coulomb and + 10gtlt 10-6 coulomb respectively as shown in the diagram below

+20 x 10-6 C +10 X 10-6 C

The magnitude of the electrostatic force on A due to B is 24 newtons What is the magnitude of the electrostatic force on ~ (1) 12 N ~ (3) 48 N (4) 96 N

3) Metal sphere A has a charge of -2 units and an identical metal sphere B has a charge of -4 units If the spheres are brought into contact with each other and then separated the charge on sphere B will be (1) 0 units (2) -2 units (4) +4 units (f-3 uni~

4) What is the net electrical charge on a magnesium ion that is formed when a neutral magnesium atom loses two electrons (1) -32 gtlt 10-19 C (2) -16 gtlt 10-19 C

5) If 10 joule of work is required to move 10 coulomb of charge between two points in an electric field the p$al difference between the two points iscrn 10 x 10deg (3) 63gtlt 1018 V (2) 90gtlt 10 V (4) 16gtlt 10-19 V

6) The diagram below represents two electrically charged identical-sized metal spheres A and B If the spheres are brought into contact which sphere will have a net gain of electrons

o 0 +10 X 10- C

(3) both A and B (4) neither A nor B

7) A subatomic farticle could have a charge of (1) 50 x 10-2 C (3) 32 x 10-19 C (2) 80 x 10-20 C X

Base your answers to questions 9 and 10 on the information and diagram below

Two small metallic spheres A and B are separated by a distance of 40 X 10-1 meter as shown The charge on each sphere is 1 0 x 10-6 coulomb Point P is located near the spheres

pbull +10 x 10-6 C +10 x 10-6 C

o 0 ---- 40 x 10-1 m 1

9 What is the magnitude of the electrostatic force between the two charged spheres amp+U ~ Hr2 b (3) 22 x 10 N(6 X10-2 (4) 56 x 104 N

10) Which arrow best represents the direction of the resultant electric field at point P due to the charges on spheres A and B

1 ( 4 ) ( 1 ) (2 )

Unit 6 Student Packet KEY

Assignment 1 Where Do Charges Come From

A Simple Atomic Model


Assignment 5 Whatrsquos a Charge


Assignment 8-21 Key

Atomic Models of Neutral Material and Charged Material View a computer simulation of atomic models with your teacher

In the charge model (your tapes) you did not consider the motion of particles The atomic model of a neutral material takes this into consideration Think about the following questions while looking at the modelsimulation of neutral tapes

1) How can you tell if a material is neutral from its atomic representation

It has equal numbers of positive and negative so that they cancel each other out

2) Which part of the atomic model is more dynamic (is more readily able to move) the positive center or the electron

Only the electrons are able to move

Your teacher will now click the button Charge the Tape to simulate the charged tapes (peeling off the tapes from each other) The top shows the interaction between charged tapes using the charge model The bottom illustrates the atomic model as if you can zoom in to see the particles

3) Record your observations of the zoomed in atomic model of the charged tapes

The left piece of tape has less electrons and is positive The right piece of tape has more electrons and is negative

4)

a) From an atomic perspective how do materials become charged through friction

Electrons leave one material (making it positive) and go to another (making it negative)

b) As shown in the atomic model two oppositely charged tapes attract each other Since each

tape contain both + and - particles why dont they repel each other due to the repulsion between + amp + particles and repulsion between minus amp minus particles Force between overall charges must be greater than that between individual charges
















ndash ndash ndash












Explanation
























charges) Charge










+3 e























Coulombs



change (still 0 C)


-6 x 106 e



(1) 80 x 10-20 C (2) 32 x 1019 C (3) 32 x 10-19 C (4) 12 x 1019 C

+2 e x 16 x 10-19 C = +32 x 10-19 C 1 e



(1) 32 x 10-19 C (2) 45 x 10-19 C (3) 96 x 10-19 C (4) 80 x 10-19 C



(1) 480 x 1019 C (2) 160 x 10-19 C (3) 480 x 10-19 C (4) 300 x 100 C








+13 ndash 10 + 27 ndash 44 + 19 = + 5 e


+5 e x 16 x 10-19 C = +8 x 10-19 C 1 e



Example



















- 5ot()e ~ pIC)f( ~





1






~ nO ne~ c~ ~ cooJT



















~ A - ~ (se e)(t--~~ -I oq









( A BIIIi

q = 1100) x 10 C



Abull B I



A Bbull l( P

bull




C 0

A B A B ~ ~~ -




~tt ~tyU 1Mtf~At




~u- 1 fte~~d


~~ w~ o_~S ~ ~(~ ~cL












So think about it


o 0 I~ ~I

r = 1 meter

So



b) By what factor










w-_ bullbulll ~0


=

















decreased

0 N




l ~6 r 0) qo N


50 x 10- C 5 N


ttO N


~S N





Explain your choice




re ~ F be~ adf

r













6



C ~ LIr to~4L Q G)I=- l aM ~ 1= +)







6 C What distance





-I tA ~ ~ ~t a ~~ C) M ~ J ) Q ~ f 1 kca ~~~


~ (i y Omiddotsl


1= - ~ =) ~ r=S~

=




answers with units)

Distance =1m

F gttJtOmiddot M

Distance = 2m

O~F ~ O

Distance =3m

F Ct 0 N

Distance =4m

F 3 ~ O~ N

Distance = 5m

f~ ~ ON

Distance =6m

F IS~OC~

Distance = 7m

f ~ 1 ~ 0 ~

Distance =8m

F ~ ~ loN

Distance =9m

F laquo1- ~ tOea N

Distance = 10m

F S~ ~ O~


rr ~d

bull

Z wft

bull-tftJ ~~


1lltW2 UlI~

~

~

~

C~

~~

1- 5 bull

~

-

~

I

g ffim ++ ~

~ g1 llJ

shy

~ i


I

-

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mEI J t m~ ~ I

~ U

II t=R=H IiIIr j

EBi~ ~

BEI11III ~ n

-000II ~ jiiiI ~

rIshy --bull J

IIIIIIII

m ffim II IJ

bull 100

I ~

bull 0








Sphere ~ Sphere 8












INSULATING STANDS













J-tf _- I - -) -~




A Test charge

~ ~ 02 rn Olm






Sphere Charge

R 0 T 4shyU




r-------~ Glass ROd


1M ~~e ~~cl ~ o~~ ~ Ju~ fOcS


-1ht -St~ -S -k ~t

r













r

a)

b)

d)

r





radius








-t

- C ~ ~ct l - fl f

shy y

I r









ceo-~~h~ ~

~ ~ CIO +-+-+-I-+shy

~ 1--+-+Ishy2 l-a

~ t-+-+I-I-+-+--I--+-+--I----l-----Ishy

f~ cOl++t+--+-+--H---+-I-I-I---+-+--t-Ishy




l~~TT~~~~~~~~~~

0 ~ wt-t--++-++--+-+-+-t-l-l--i--JH--l-I-L

t2 -+--I---I-J--I

~amprrrr+++~~---+-I-I-~

0 --+---+








E ~ 10 ~~_ ~ T~l5 ~ 01 ~1t1 - 0 ~CS~ t shy








8 88 Ou) e h~h ~e











E




Units for Potential

Difference

T-C














v~ 4= (501000~t) OIS l O S l ~ eJ (--~-----O~--~lI



f~cA I +1 el1l-lgt



HI ()r4-~ 2~ _ ~12 StJlV]











St a) 50 V b) 60 V d) 300 V



c) 12 V


A D



-h~~~(~~~


j cl 2OC)1 ~ It ~



Fe OCO~ N

c)~ d~( O ~ N

I




Q -11-- _-- C) w ~

V~~ W~4~ (1~lboCe) 1Ii1(10)111






r TEST CHARGE

8 ~)









plates















volts t= In~s e) (1) 12 J (3) 30 J (4) 24 JB





1++++++++++++++++pound++1

reg

I - ----I




r




+20 x 10-6 C +10 X 10-6 C






o 0 +10 X 10- C





pbull +10 x 10-6 C +10 x 10-6 C

o 0 ---- 40 x 10-1 m 1



1 ( 4 ) ( 1 ) (2 )







Assignment 8-21 Key
















ndash ndash ndash












Explanation
























charges) Charge










+3 e























Coulombs



change (still 0 C)


-6 x 106 e



(1) 80 x 10-20 C (2) 32 x 1019 C (3) 32 x 10-19 C (4) 12 x 1019 C

+2 e x 16 x 10-19 C = +32 x 10-19 C 1 e



(1) 32 x 10-19 C (2) 45 x 10-19 C (3) 96 x 10-19 C (4) 80 x 10-19 C



(1) 480 x 1019 C (2) 160 x 10-19 C (3) 480 x 10-19 C (4) 300 x 100 C








+13 ndash 10 + 27 ndash 44 + 19 = + 5 e


+5 e x 16 x 10-19 C = +8 x 10-19 C 1 e



Example



















- 5ot()e ~ pIC)f( ~





1






~ nO ne~ c~ ~ cooJT



















~ A - ~ (se e)(t--~~ -I oq









( A BIIIi

q = 1100) x 10 C



Abull B I



A Bbull l( P

bull




C 0

A B A B ~ ~~ -




~tt ~tyU 1Mtf~At




~u- 1 fte~~d


~~ w~ o_~S ~ ~(~ ~cL












So think about it


o 0 I~ ~I

r = 1 meter

So



b) By what factor










w-_ bullbulll ~0


=

















decreased

0 N




l ~6 r 0) qo N


50 x 10- C 5 N


ttO N


~S N





Explain your choice




re ~ F be~ adf

r













6



C ~ LIr to~4L Q G)I=- l aM ~ 1= +)







6 C What distance





-I tA ~ ~ ~t a ~~ C) M ~ J ) Q ~ f 1 kca ~~~


~ (i y Omiddotsl


1= - ~ =) ~ r=S~

=




answers with units)

Distance =1m

F gttJtOmiddot M

Distance = 2m

O~F ~ O

Distance =3m

F Ct 0 N

Distance =4m

F 3 ~ O~ N

Distance = 5m

f~ ~ ON

Distance =6m

F IS~OC~

Distance = 7m

f ~ 1 ~ 0 ~

Distance =8m

F ~ ~ loN

Distance =9m

F laquo1- ~ tOea N

Distance = 10m

F S~ ~ O~


rr ~d

bull

Z wft

bull-tftJ ~~


1lltW2 UlI~

~

~

~

C~

~~

1- 5 bull

~

-

~

I

g ffim ++ ~

~ g1 llJ

shy

~ i


I

-

Ii~

mEI J t m~ ~ I

~ U

II t=R=H IiIIr j

EBi~ ~

BEI11III ~ n

-000II ~ jiiiI ~

rIshy --bull J

IIIIIIII

m ffim II IJ

bull 100

I ~

bull 0








Sphere ~ Sphere 8












INSULATING STANDS













J-tf _- I - -) -~




A Test charge

~ ~ 02 rn Olm






Sphere Charge

R 0 T 4shyU




r-------~ Glass ROd


1M ~~e ~~cl ~ o~~ ~ Ju~ fOcS


-1ht -St~ -S -k ~t

r













r

a)

b)

d)

r





radius








-t

- C ~ ~ct l - fl f

shy y

I r









ceo-~~h~ ~

~ ~ CIO +-+-+-I-+shy

~ 1--+-+Ishy2 l-a

~ t-+-+I-I-+-+--I--+-+--I----l-----Ishy

f~ cOl++t+--+-+--H---+-I-I-I---+-+--t-Ishy




l~~TT~~~~~~~~~~

0 ~ wt-t--++-++--+-+-+-t-l-l--i--JH--l-I-L

t2 -+--I---I-J--I

~amprrrr+++~~---+-I-I-~

0 --+---+








E ~ 10 ~~_ ~ T~l5 ~ 01 ~1t1 - 0 ~CS~ t shy








8 88 Ou) e h~h ~e











E




Units for Potential

Difference

T-C














v~ 4= (501000~t) OIS l O S l ~ eJ (--~-----O~--~lI



f~cA I +1 el1l-lgt



HI ()r4-~ 2~ _ ~12 StJlV]











St a) 50 V b) 60 V d) 300 V



c) 12 V


A D



-h~~~(~~~


j cl 2OC)1 ~ It ~



Fe OCO~ N

c)~ d~( O ~ N

I




Q -11-- _-- C) w ~

V~~ W~4~ (1~lboCe) 1Ii1(10)111






r TEST CHARGE

8 ~)









plates















volts t= In~s e) (1) 12 J (3) 30 J (4) 24 JB





1++++++++++++++++pound++1

reg

I - ----I




r




+20 x 10-6 C +10 X 10-6 C






o 0 +10 X 10- C





pbull +10 x 10-6 C +10 x 10-6 C

o 0 ---- 40 x 10-1 m 1



1 ( 4 ) ( 1 ) (2 )







Assignment 8-21 Key











Explanation
























charges) Charge










+3 e























Coulombs



change (still 0 C)


-6 x 106 e



(1) 80 x 10-20 C (2) 32 x 1019 C (3) 32 x 10-19 C (4) 12 x 1019 C

+2 e x 16 x 10-19 C = +32 x 10-19 C 1 e



(1) 32 x 10-19 C (2) 45 x 10-19 C (3) 96 x 10-19 C (4) 80 x 10-19 C



(1) 480 x 1019 C (2) 160 x 10-19 C (3) 480 x 10-19 C (4) 300 x 100 C








+13 ndash 10 + 27 ndash 44 + 19 = + 5 e


+5 e x 16 x 10-19 C = +8 x 10-19 C 1 e



Example



















- 5ot()e ~ pIC)f( ~





1






~ nO ne~ c~ ~ cooJT



















~ A - ~ (se e)(t--~~ -I oq









( A BIIIi

q = 1100) x 10 C



Abull B I



A Bbull l( P

bull




C 0

A B A B ~ ~~ -




~tt ~tyU 1Mtf~At




~u- 1 fte~~d


~~ w~ o_~S ~ ~(~ ~cL












So think about it


o 0 I~ ~I

r = 1 meter

So



b) By what factor










w-_ bullbulll ~0


=

















decreased

0 N




l ~6 r 0) qo N


50 x 10- C 5 N


ttO N


~S N





Explain your choice




re ~ F be~ adf

r













6



C ~ LIr to~4L Q G)I=- l aM ~ 1= +)







6 C What distance





-I tA ~ ~ ~t a ~~ C) M ~ J ) Q ~ f 1 kca ~~~


~ (i y Omiddotsl


1= - ~ =) ~ r=S~

=




answers with units)

Distance =1m

F gttJtOmiddot M

Distance = 2m

O~F ~ O

Distance =3m

F Ct 0 N

Distance =4m

F 3 ~ O~ N

Distance = 5m

f~ ~ ON

Distance =6m

F IS~OC~

Distance = 7m

f ~ 1 ~ 0 ~

Distance =8m

F ~ ~ loN

Distance =9m

F laquo1- ~ tOea N

Distance = 10m

F S~ ~ O~


rr ~d

bull

Z wft

bull-tftJ ~~


1lltW2 UlI~

~

~

~

C~

~~

1- 5 bull

~

-

~

I

g ffim ++ ~

~ g1 llJ

shy

~ i


I

-

Ii~

mEI J t m~ ~ I

~ U

II t=R=H IiIIr j

EBi~ ~

BEI11III ~ n

-000II ~ jiiiI ~

rIshy --bull J

IIIIIIII

m ffim II IJ

bull 100

I ~

bull 0








Sphere ~ Sphere 8












INSULATING STANDS













J-tf _- I - -) -~




A Test charge

~ ~ 02 rn Olm






Sphere Charge

R 0 T 4shyU




r-------~ Glass ROd


1M ~~e ~~cl ~ o~~ ~ Ju~ fOcS


-1ht -St~ -S -k ~t

r













r

a)

b)

d)

r





radius








-t

- C ~ ~ct l - fl f

shy y

I r









ceo-~~h~ ~

~ ~ CIO +-+-+-I-+shy

~ 1--+-+Ishy2 l-a

~ t-+-+I-I-+-+--I--+-+--I----l-----Ishy

f~ cOl++t+--+-+--H---+-I-I-I---+-+--t-Ishy




l~~TT~~~~~~~~~~

0 ~ wt-t--++-++--+-+-+-t-l-l--i--JH--l-I-L

t2 -+--I---I-J--I

~amprrrr+++~~---+-I-I-~

0 --+---+








E ~ 10 ~~_ ~ T~l5 ~ 01 ~1t1 - 0 ~CS~ t shy








8 88 Ou) e h~h ~e











E




Units for Potential

Difference

T-C














v~ 4= (501000~t) OIS l O S l ~ eJ (--~-----O~--~lI



f~cA I +1 el1l-lgt



HI ()r4-~ 2~ _ ~12 StJlV]











St a) 50 V b) 60 V d) 300 V



c) 12 V


A D



-h~~~(~~~


j cl 2OC)1 ~ It ~



Fe OCO~ N

c)~ d~( O ~ N

I




Q -11-- _-- C) w ~

V~~ W~4~ (1~lboCe) 1Ii1(10)111






r TEST CHARGE

8 ~)









plates















volts t= In~s e) (1) 12 J (3) 30 J (4) 24 JB





1++++++++++++++++pound++1

reg

I - ----I




r




+20 x 10-6 C +10 X 10-6 C






o 0 +10 X 10- C





pbull +10 x 10-6 C +10 x 10-6 C

o 0 ---- 40 x 10-1 m 1



1 ( 4 ) ( 1 ) (2 )







Assignment 8-21 Key




















charges) Charge










+3 e























Coulombs



change (still 0 C)


-6 x 106 e



(1) 80 x 10-20 C (2) 32 x 1019 C (3) 32 x 10-19 C (4) 12 x 1019 C

+2 e x 16 x 10-19 C = +32 x 10-19 C 1 e



(1) 32 x 10-19 C (2) 45 x 10-19 C (3) 96 x 10-19 C (4) 80 x 10-19 C



(1) 480 x 1019 C (2) 160 x 10-19 C (3) 480 x 10-19 C (4) 300 x 100 C








+13 ndash 10 + 27 ndash 44 + 19 = + 5 e


+5 e x 16 x 10-19 C = +8 x 10-19 C 1 e



Example



















- 5ot()e ~ pIC)f( ~





1






~ nO ne~ c~ ~ cooJT



















~ A - ~ (se e)(t--~~ -I oq









( A BIIIi

q = 1100) x 10 C



Abull B I



A Bbull l( P

bull




C 0

A B A B ~ ~~ -




~tt ~tyU 1Mtf~At




~u- 1 fte~~d


~~ w~ o_~S ~ ~(~ ~cL












So think about it


o 0 I~ ~I

r = 1 meter

So



b) By what factor










w-_ bullbulll ~0


=

















decreased

0 N




l ~6 r 0) qo N


50 x 10- C 5 N


ttO N


~S N





Explain your choice




re ~ F be~ adf

r













6



C ~ LIr to~4L Q G)I=- l aM ~ 1= +)







6 C What distance





-I tA ~ ~ ~t a ~~ C) M ~ J ) Q ~ f 1 kca ~~~


~ (i y Omiddotsl


1= - ~ =) ~ r=S~

=




answers with units)

Distance =1m

F gttJtOmiddot M

Distance = 2m

O~F ~ O

Distance =3m

F Ct 0 N

Distance =4m

F 3 ~ O~ N

Distance = 5m

f~ ~ ON

Distance =6m

F IS~OC~

Distance = 7m

f ~ 1 ~ 0 ~

Distance =8m

F ~ ~ loN

Distance =9m

F laquo1- ~ tOea N

Distance = 10m

F S~ ~ O~


rr ~d

bull

Z wft

bull-tftJ ~~


1lltW2 UlI~

~

~

~

C~

~~

1- 5 bull

~

-

~

I

g ffim ++ ~

~ g1 llJ

shy

~ i


I

-

Ii~

mEI J t m~ ~ I

~ U

II t=R=H IiIIr j

EBi~ ~

BEI11III ~ n

-000II ~ jiiiI ~

rIshy --bull J

IIIIIIII

m ffim II IJ

bull 100

I ~

bull 0








Sphere ~ Sphere 8












INSULATING STANDS













J-tf _- I - -) -~




A Test charge

~ ~ 02 rn Olm






Sphere Charge

R 0 T 4shyU




r-------~ Glass ROd


1M ~~e ~~cl ~ o~~ ~ Ju~ fOcS


-1ht -St~ -S -k ~t

r













r

a)

b)

d)

r





radius








-t

- C ~ ~ct l - fl f

shy y

I r









ceo-~~h~ ~

~ ~ CIO +-+-+-I-+shy

~ 1--+-+Ishy2 l-a

~ t-+-+I-I-+-+--I--+-+--I----l-----Ishy

f~ cOl++t+--+-+--H---+-I-I-I---+-+--t-Ishy




l~~TT~~~~~~~~~~

0 ~ wt-t--++-++--+-+-+-t-l-l--i--JH--l-I-L

t2 -+--I---I-J--I

~amprrrr+++~~---+-I-I-~

0 --+---+








E ~ 10 ~~_ ~ T~l5 ~ 01 ~1t1 - 0 ~CS~ t shy








8 88 Ou) e h~h ~e











E




Units for Potential

Difference

T-C














v~ 4= (501000~t) OIS l O S l ~ eJ (--~-----O~--~lI



f~cA I +1 el1l-lgt



HI ()r4-~ 2~ _ ~12 StJlV]











St a) 50 V b) 60 V d) 300 V



c) 12 V


A D



-h~~~(~~~


j cl 2OC)1 ~ It ~



Fe OCO~ N

c)~ d~( O ~ N

I




Q -11-- _-- C) w ~

V~~ W~4~ (1~lboCe) 1Ii1(10)111






r TEST CHARGE

8 ~)









plates















volts t= In~s e) (1) 12 J (3) 30 J (4) 24 JB





1++++++++++++++++pound++1

reg

I - ----I




r




+20 x 10-6 C +10 X 10-6 C






o 0 +10 X 10- C





pbull +10 x 10-6 C +10 x 10-6 C

o 0 ---- 40 x 10-1 m 1



1 ( 4 ) ( 1 ) (2 )







Assignment 8-21 Key

















Coulombs



change (still 0 C)


-6 x 106 e



(1) 80 x 10-20 C (2) 32 x 1019 C (3) 32 x 10-19 C (4) 12 x 1019 C

+2 e x 16 x 10-19 C = +32 x 10-19 C 1 e



(1) 32 x 10-19 C (2) 45 x 10-19 C (3) 96 x 10-19 C (4) 80 x 10-19 C



(1) 480 x 1019 C (2) 160 x 10-19 C (3) 480 x 10-19 C (4) 300 x 100 C








+13 ndash 10 + 27 ndash 44 + 19 = + 5 e


+5 e x 16 x 10-19 C = +8 x 10-19 C 1 e



Example



















- 5ot()e ~ pIC)f( ~





1






~ nO ne~ c~ ~ cooJT



















~ A - ~ (se e)(t--~~ -I oq









( A BIIIi

q = 1100) x 10 C



Abull B I



A Bbull l( P

bull




C 0

A B A B ~ ~~ -




~tt ~tyU 1Mtf~At




~u- 1 fte~~d


~~ w~ o_~S ~ ~(~ ~cL












So think about it


o 0 I~ ~I

r = 1 meter

So



b) By what factor










w-_ bullbulll ~0


=

















decreased

0 N




l ~6 r 0) qo N


50 x 10- C 5 N


ttO N


~S N





Explain your choice




re ~ F be~ adf

r













6



C ~ LIr to~4L Q G)I=- l aM ~ 1= +)







6 C What distance





-I tA ~ ~ ~t a ~~ C) M ~ J ) Q ~ f 1 kca ~~~


~ (i y Omiddotsl


1= - ~ =) ~ r=S~

=




answers with units)

Distance =1m

F gttJtOmiddot M

Distance = 2m

O~F ~ O

Distance =3m

F Ct 0 N

Distance =4m

F 3 ~ O~ N

Distance = 5m

f~ ~ ON

Distance =6m

F IS~OC~

Distance = 7m

f ~ 1 ~ 0 ~

Distance =8m

F ~ ~ loN

Distance =9m

F laquo1- ~ tOea N

Distance = 10m

F S~ ~ O~


rr ~d

bull

Z wft

bull-tftJ ~~


1lltW2 UlI~

~

~

~

C~

~~

1- 5 bull

~

-

~

I

g ffim ++ ~

~ g1 llJ

shy

~ i


I

-

Ii~

mEI J t m~ ~ I

~ U

II t=R=H IiIIr j

EBi~ ~

BEI11III ~ n

-000II ~ jiiiI ~

rIshy --bull J

IIIIIIII

m ffim II IJ

bull 100

I ~

bull 0








Sphere ~ Sphere 8












INSULATING STANDS













J-tf _- I - -) -~




A Test charge

~ ~ 02 rn Olm






Sphere Charge

R 0 T 4shyU




r-------~ Glass ROd


1M ~~e ~~cl ~ o~~ ~ Ju~ fOcS


-1ht -St~ -S -k ~t

r













r

a)

b)

d)

r





radius








-t

- C ~ ~ct l - fl f

shy y

I r









ceo-~~h~ ~

~ ~ CIO +-+-+-I-+shy

~ 1--+-+Ishy2 l-a

~ t-+-+I-I-+-+--I--+-+--I----l-----Ishy

f~ cOl++t+--+-+--H---+-I-I-I---+-+--t-Ishy




l~~TT~~~~~~~~~~

0 ~ wt-t--++-++--+-+-+-t-l-l--i--JH--l-I-L

t2 -+--I---I-J--I

~amprrrr+++~~---+-I-I-~

0 --+---+








E ~ 10 ~~_ ~ T~l5 ~ 01 ~1t1 - 0 ~CS~ t shy








8 88 Ou) e h~h ~e











E




Units for Potential

Difference

T-C














v~ 4= (501000~t) OIS l O S l ~ eJ (--~-----O~--~lI



f~cA I +1 el1l-lgt



HI ()r4-~ 2~ _ ~12 StJlV]











St a) 50 V b) 60 V d) 300 V



c) 12 V


A D



-h~~~(~~~


j cl 2OC)1 ~ It ~



Fe OCO~ N

c)~ d~( O ~ N

I




Q -11-- _-- C) w ~

V~~ W~4~ (1~lboCe) 1Ii1(10)111






r TEST CHARGE

8 ~)









plates















volts t= In~s e) (1) 12 J (3) 30 J (4) 24 JB





1++++++++++++++++pound++1

reg

I - ----I




r




+20 x 10-6 C +10 X 10-6 C






o 0 +10 X 10- C





pbull +10 x 10-6 C +10 x 10-6 C

o 0 ---- 40 x 10-1 m 1



1 ( 4 ) ( 1 ) (2 )







Assignment 8-21 Key


-6 x 106 e



(1) 80 x 10-20 C (2) 32 x 1019 C (3) 32 x 10-19 C (4) 12 x 1019 C

+2 e x 16 x 10-19 C = +32 x 10-19 C 1 e



(1) 32 x 10-19 C (2) 45 x 10-19 C (3) 96 x 10-19 C (4) 80 x 10-19 C



(1) 480 x 1019 C (2) 160 x 10-19 C (3) 480 x 10-19 C (4) 300 x 100 C








+13 ndash 10 + 27 ndash 44 + 19 = + 5 e


+5 e x 16 x 10-19 C = +8 x 10-19 C 1 e



Example



















- 5ot()e ~ pIC)f( ~





1






~ nO ne~ c~ ~ cooJT



















~ A - ~ (se e)(t--~~ -I oq









( A BIIIi

q = 1100) x 10 C



Abull B I



A Bbull l( P

bull




C 0

A B A B ~ ~~ -




~tt ~tyU 1Mtf~At




~u- 1 fte~~d


~~ w~ o_~S ~ ~(~ ~cL












So think about it


o 0 I~ ~I

r = 1 meter

So



b) By what factor










w-_ bullbulll ~0


=

















decreased

0 N




l ~6 r 0) qo N


50 x 10- C 5 N


ttO N


~S N





Explain your choice




re ~ F be~ adf

r













6



C ~ LIr to~4L Q G)I=- l aM ~ 1= +)







6 C What distance





-I tA ~ ~ ~t a ~~ C) M ~ J ) Q ~ f 1 kca ~~~


~ (i y Omiddotsl


1= - ~ =) ~ r=S~

=




answers with units)

Distance =1m

F gttJtOmiddot M

Distance = 2m

O~F ~ O

Distance =3m

F Ct 0 N

Distance =4m

F 3 ~ O~ N

Distance = 5m

f~ ~ ON

Distance =6m

F IS~OC~

Distance = 7m

f ~ 1 ~ 0 ~

Distance =8m

F ~ ~ loN

Distance =9m

F laquo1- ~ tOea N

Distance = 10m

F S~ ~ O~


rr ~d

bull

Z wft

bull-tftJ ~~


1lltW2 UlI~

~

~

~

C~

~~

1- 5 bull

~

-

~

I

g ffim ++ ~

~ g1 llJ

shy

~ i


I

-

Ii~

mEI J t m~ ~ I

~ U

II t=R=H IiIIr j

EBi~ ~

BEI11III ~ n

-000II ~ jiiiI ~

rIshy --bull J

IIIIIIII

m ffim II IJ

bull 100

I ~

bull 0








Sphere ~ Sphere 8












INSULATING STANDS













J-tf _- I - -) -~




A Test charge

~ ~ 02 rn Olm






Sphere Charge

R 0 T 4shyU




r-------~ Glass ROd


1M ~~e ~~cl ~ o~~ ~ Ju~ fOcS


-1ht -St~ -S -k ~t

r













r

a)

b)

d)

r





radius








-t

- C ~ ~ct l - fl f

shy y

I r









ceo-~~h~ ~

~ ~ CIO +-+-+-I-+shy

~ 1--+-+Ishy2 l-a

~ t-+-+I-I-+-+--I--+-+--I----l-----Ishy

f~ cOl++t+--+-+--H---+-I-I-I---+-+--t-Ishy




l~~TT~~~~~~~~~~

0 ~ wt-t--++-++--+-+-+-t-l-l--i--JH--l-I-L

t2 -+--I---I-J--I

~amprrrr+++~~---+-I-I-~

0 --+---+








E ~ 10 ~~_ ~ T~l5 ~ 01 ~1t1 - 0 ~CS~ t shy








8 88 Ou) e h~h ~e











E




Units for Potential

Difference

T-C














v~ 4= (501000~t) OIS l O S l ~ eJ (--~-----O~--~lI



f~cA I +1 el1l-lgt



HI ()r4-~ 2~ _ ~12 StJlV]











St a) 50 V b) 60 V d) 300 V



c) 12 V


A D



-h~~~(~~~


j cl 2OC)1 ~ It ~



Fe OCO~ N

c)~ d~( O ~ N

I




Q -11-- _-- C) w ~

V~~ W~4~ (1~lboCe) 1Ii1(10)111






r TEST CHARGE

8 ~)









plates















volts t= In~s e) (1) 12 J (3) 30 J (4) 24 JB





1++++++++++++++++pound++1

reg

I - ----I




r




+20 x 10-6 C +10 X 10-6 C






o 0 +10 X 10- C





pbull +10 x 10-6 C +10 x 10-6 C

o 0 ---- 40 x 10-1 m 1



1 ( 4 ) ( 1 ) (2 )







Assignment 8-21 Key



Example



















- 5ot()e ~ pIC)f( ~





1






~ nO ne~ c~ ~ cooJT



















~ A - ~ (se e)(t--~~ -I oq









( A BIIIi

q = 1100) x 10 C



Abull B I



A Bbull l( P

bull




C 0

A B A B ~ ~~ -




~tt ~tyU 1Mtf~At




~u- 1 fte~~d


~~ w~ o_~S ~ ~(~ ~cL












So think about it


o 0 I~ ~I

r = 1 meter

So



b) By what factor










w-_ bullbulll ~0


=

















decreased

0 N




l ~6 r 0) qo N


50 x 10- C 5 N


ttO N


~S N





Explain your choice




re ~ F be~ adf

r













6



C ~ LIr to~4L Q G)I=- l aM ~ 1= +)







6 C What distance





-I tA ~ ~ ~t a ~~ C) M ~ J ) Q ~ f 1 kca ~~~


~ (i y Omiddotsl


1= - ~ =) ~ r=S~

=




answers with units)

Distance =1m

F gttJtOmiddot M

Distance = 2m

O~F ~ O

Distance =3m

F Ct 0 N

Distance =4m

F 3 ~ O~ N

Distance = 5m

f~ ~ ON

Distance =6m

F IS~OC~

Distance = 7m

f ~ 1 ~ 0 ~

Distance =8m

F ~ ~ loN

Distance =9m

F laquo1- ~ tOea N

Distance = 10m

F S~ ~ O~


rr ~d

bull

Z wft

bull-tftJ ~~


1lltW2 UlI~

~

~

~

C~

~~

1- 5 bull

~

-

~

I

g ffim ++ ~

~ g1 llJ

shy

~ i


I

-

Ii~

mEI J t m~ ~ I

~ U

II t=R=H IiIIr j

EBi~ ~

BEI11III ~ n

-000II ~ jiiiI ~

rIshy --bull J

IIIIIIII

m ffim II IJ

bull 100

I ~

bull 0








Sphere ~ Sphere 8












INSULATING STANDS













J-tf _- I - -) -~




A Test charge

~ ~ 02 rn Olm






Sphere Charge

R 0 T 4shyU




r-------~ Glass ROd


1M ~~e ~~cl ~ o~~ ~ Ju~ fOcS


-1ht -St~ -S -k ~t

r













r

a)

b)

d)

r





radius








-t

- C ~ ~ct l - fl f

shy y

I r









ceo-~~h~ ~

~ ~ CIO +-+-+-I-+shy

~ 1--+-+Ishy2 l-a

~ t-+-+I-I-+-+--I--+-+--I----l-----Ishy

f~ cOl++t+--+-+--H---+-I-I-I---+-+--t-Ishy




l~~TT~~~~~~~~~~

0 ~ wt-t--++-++--+-+-+-t-l-l--i--JH--l-I-L

t2 -+--I---I-J--I

~amprrrr+++~~---+-I-I-~

0 --+---+








E ~ 10 ~~_ ~ T~l5 ~ 01 ~1t1 - 0 ~CS~ t shy








8 88 Ou) e h~h ~e











E




Units for Potential

Difference

T-C














v~ 4= (501000~t) OIS l O S l ~ eJ (--~-----O~--~lI



f~cA I +1 el1l-lgt



HI ()r4-~ 2~ _ ~12 StJlV]











St a) 50 V b) 60 V d) 300 V



c) 12 V


A D



-h~~~(~~~


j cl 2OC)1 ~ It ~



Fe OCO~ N

c)~ d~( O ~ N

I




Q -11-- _-- C) w ~

V~~ W~4~ (1~lboCe) 1Ii1(10)111






r TEST CHARGE

8 ~)









plates















volts t= In~s e) (1) 12 J (3) 30 J (4) 24 JB





1++++++++++++++++pound++1

reg

I - ----I




r




+20 x 10-6 C +10 X 10-6 C






o 0 +10 X 10- C





pbull +10 x 10-6 C +10 x 10-6 C

o 0 ---- 40 x 10-1 m 1



1 ( 4 ) ( 1 ) (2 )







Assignment 8-21 Key











- 5ot()e ~ pIC)f( ~





1






~ nO ne~ c~ ~ cooJT



















~ A - ~ (se e)(t--~~ -I oq









( A BIIIi

q = 1100) x 10 C



Abull B I



A Bbull l( P

bull




C 0

A B A B ~ ~~ -




~tt ~tyU 1Mtf~At




~u- 1 fte~~d


~~ w~ o_~S ~ ~(~ ~cL












So think about it


o 0 I~ ~I

r = 1 meter

So



b) By what factor










w-_ bullbulll ~0


=

















decreased

0 N




l ~6 r 0) qo N


50 x 10- C 5 N


ttO N


~S N





Explain your choice




re ~ F be~ adf

r













6



C ~ LIr to~4L Q G)I=- l aM ~ 1= +)







6 C What distance





-I tA ~ ~ ~t a ~~ C) M ~ J ) Q ~ f 1 kca ~~~


~ (i y Omiddotsl


1= - ~ =) ~ r=S~

=




answers with units)

Distance =1m

F gttJtOmiddot M

Distance = 2m

O~F ~ O

Distance =3m

F Ct 0 N

Distance =4m

F 3 ~ O~ N

Distance = 5m

f~ ~ ON

Distance =6m

F IS~OC~

Distance = 7m

f ~ 1 ~ 0 ~

Distance =8m

F ~ ~ loN

Distance =9m

F laquo1- ~ tOea N

Distance = 10m

F S~ ~ O~


rr ~d

bull

Z wft

bull-tftJ ~~


1lltW2 UlI~

~

~

~

C~

~~

1- 5 bull

~

-

~

I

g ffim ++ ~

~ g1 llJ

shy

~ i


I

-

Ii~

mEI J t m~ ~ I

~ U

II t=R=H IiIIr j

EBi~ ~

BEI11III ~ n

-000II ~ jiiiI ~

rIshy --bull J

IIIIIIII

m ffim II IJ

bull 100

I ~

bull 0








Sphere ~ Sphere 8












INSULATING STANDS













J-tf _- I - -) -~




A Test charge

~ ~ 02 rn Olm






Sphere Charge

R 0 T 4shyU




r-------~ Glass ROd


1M ~~e ~~cl ~ o~~ ~ Ju~ fOcS


-1ht -St~ -S -k ~t

r













r

a)

b)

d)

r





radius








-t

- C ~ ~ct l - fl f

shy y

I r









ceo-~~h~ ~

~ ~ CIO +-+-+-I-+shy

~ 1--+-+Ishy2 l-a

~ t-+-+I-I-+-+--I--+-+--I----l-----Ishy

f~ cOl++t+--+-+--H---+-I-I-I---+-+--t-Ishy




l~~TT~~~~~~~~~~

0 ~ wt-t--++-++--+-+-+-t-l-l--i--JH--l-I-L

t2 -+--I---I-J--I

~amprrrr+++~~---+-I-I-~

0 --+---+








E ~ 10 ~~_ ~ T~l5 ~ 01 ~1t1 - 0 ~CS~ t shy








8 88 Ou) e h~h ~e











E




Units for Potential

Difference

T-C














v~ 4= (501000~t) OIS l O S l ~ eJ (--~-----O~--~lI



f~cA I +1 el1l-lgt



HI ()r4-~ 2~ _ ~12 StJlV]











St a) 50 V b) 60 V d) 300 V



c) 12 V


A D



-h~~~(~~~


j cl 2OC)1 ~ It ~



Fe OCO~ N

c)~ d~( O ~ N

I




Q -11-- _-- C) w ~

V~~ W~4~ (1~lboCe) 1Ii1(10)111






r TEST CHARGE

8 ~)









plates















volts t= In~s e) (1) 12 J (3) 30 J (4) 24 JB





1++++++++++++++++pound++1

reg

I - ----I




r




+20 x 10-6 C +10 X 10-6 C






o 0 +10 X 10- C





pbull +10 x 10-6 C +10 x 10-6 C

o 0 ---- 40 x 10-1 m 1



1 ( 4 ) ( 1 ) (2 )







Assignment 8-21 Key






~ nO ne~ c~ ~ cooJT



















~ A - ~ (se e)(t--~~ -I oq









( A BIIIi

q = 1100) x 10 C



Abull B I



A Bbull l( P

bull




C 0

A B A B ~ ~~ -




~tt ~tyU 1Mtf~At




~u- 1 fte~~d


~~ w~ o_~S ~ ~(~ ~cL












So think about it


o 0 I~ ~I

r = 1 meter

So



b) By what factor










w-_ bullbulll ~0


=

















decreased

0 N




l ~6 r 0) qo N


50 x 10- C 5 N


ttO N


~S N





Explain your choice




re ~ F be~ adf

r













6



C ~ LIr to~4L Q G)I=- l aM ~ 1= +)







6 C What distance





-I tA ~ ~ ~t a ~~ C) M ~ J ) Q ~ f 1 kca ~~~


~ (i y Omiddotsl


1= - ~ =) ~ r=S~

=




answers with units)

Distance =1m

F gttJtOmiddot M

Distance = 2m

O~F ~ O

Distance =3m

F Ct 0 N

Distance =4m

F 3 ~ O~ N

Distance = 5m

f~ ~ ON

Distance =6m

F IS~OC~

Distance = 7m

f ~ 1 ~ 0 ~

Distance =8m

F ~ ~ loN

Distance =9m

F laquo1- ~ tOea N

Distance = 10m

F S~ ~ O~


rr ~d

bull

Z wft

bull-tftJ ~~


1lltW2 UlI~

~

~

~

C~

~~

1- 5 bull

~

-

~

I

g ffim ++ ~

~ g1 llJ

shy

~ i


I

-

Ii~

mEI J t m~ ~ I

~ U

II t=R=H IiIIr j

EBi~ ~

BEI11III ~ n

-000II ~ jiiiI ~

rIshy --bull J

IIIIIIII

m ffim II IJ

bull 100

I ~

bull 0








Sphere ~ Sphere 8












INSULATING STANDS













J-tf _- I - -) -~




A Test charge

~ ~ 02 rn Olm






Sphere Charge

R 0 T 4shyU




r-------~ Glass ROd


1M ~~e ~~cl ~ o~~ ~ Ju~ fOcS


-1ht -St~ -S -k ~t

r













r

a)

b)

d)

r





radius








-t

- C ~ ~ct l - fl f

shy y

I r









ceo-~~h~ ~

~ ~ CIO +-+-+-I-+shy

~ 1--+-+Ishy2 l-a

~ t-+-+I-I-+-+--I--+-+--I----l-----Ishy

f~ cOl++t+--+-+--H---+-I-I-I---+-+--t-Ishy




l~~TT~~~~~~~~~~

0 ~ wt-t--++-++--+-+-+-t-l-l--i--JH--l-I-L

t2 -+--I---I-J--I

~amprrrr+++~~---+-I-I-~

0 --+---+








E ~ 10 ~~_ ~ T~l5 ~ 01 ~1t1 - 0 ~CS~ t shy








8 88 Ou) e h~h ~e











E




Units for Potential

Difference

T-C














v~ 4= (501000~t) OIS l O S l ~ eJ (--~-----O~--~lI



f~cA I +1 el1l-lgt



HI ()r4-~ 2~ _ ~12 StJlV]











St a) 50 V b) 60 V d) 300 V



c) 12 V


A D



-h~~~(~~~


j cl 2OC)1 ~ It ~



Fe OCO~ N

c)~ d~( O ~ N

I




Q -11-- _-- C) w ~

V~~ W~4~ (1~lboCe) 1Ii1(10)111






r TEST CHARGE

8 ~)









plates















volts t= In~s e) (1) 12 J (3) 30 J (4) 24 JB





1++++++++++++++++pound++1

reg

I - ----I




r




+20 x 10-6 C +10 X 10-6 C






o 0 +10 X 10- C





pbull +10 x 10-6 C +10 x 10-6 C

o 0 ---- 40 x 10-1 m 1



1 ( 4 ) ( 1 ) (2 )







Assignment 8-21 Key













~ A - ~ (se e)(t--~~ -I oq









( A BIIIi

q = 1100) x 10 C



Abull B I



A Bbull l( P

bull




C 0

A B A B ~ ~~ -




~tt ~tyU 1Mtf~At




~u- 1 fte~~d


~~ w~ o_~S ~ ~(~ ~cL












So think about it


o 0 I~ ~I

r = 1 meter

So



b) By what factor










w-_ bullbulll ~0


=

















decreased

0 N




l ~6 r 0) qo N


50 x 10- C 5 N


ttO N


~S N





Explain your choice




re ~ F be~ adf

r













6



C ~ LIr to~4L Q G)I=- l aM ~ 1= +)







6 C What distance





-I tA ~ ~ ~t a ~~ C) M ~ J ) Q ~ f 1 kca ~~~


~ (i y Omiddotsl


1= - ~ =) ~ r=S~

=




answers with units)

Distance =1m

F gttJtOmiddot M

Distance = 2m

O~F ~ O

Distance =3m

F Ct 0 N

Distance =4m

F 3 ~ O~ N

Distance = 5m

f~ ~ ON

Distance =6m

F IS~OC~

Distance = 7m

f ~ 1 ~ 0 ~

Distance =8m

F ~ ~ loN

Distance =9m

F laquo1- ~ tOea N

Distance = 10m

F S~ ~ O~


rr ~d

bull

Z wft

bull-tftJ ~~


1lltW2 UlI~

~

~

~

C~

~~

1- 5 bull

~

-

~

I

g ffim ++ ~

~ g1 llJ

shy

~ i


I

-

Ii~

mEI J t m~ ~ I

~ U

II t=R=H IiIIr j

EBi~ ~

BEI11III ~ n

-000II ~ jiiiI ~

rIshy --bull J

IIIIIIII

m ffim II IJ

bull 100

I ~

bull 0








Sphere ~ Sphere 8












INSULATING STANDS













J-tf _- I - -) -~




A Test charge

~ ~ 02 rn Olm






Sphere Charge

R 0 T 4shyU




r-------~ Glass ROd


1M ~~e ~~cl ~ o~~ ~ Ju~ fOcS


-1ht -St~ -S -k ~t

r













r

a)

b)

d)

r





radius








-t

- C ~ ~ct l - fl f

shy y

I r









ceo-~~h~ ~

~ ~ CIO +-+-+-I-+shy

~ 1--+-+Ishy2 l-a

~ t-+-+I-I-+-+--I--+-+--I----l-----Ishy

f~ cOl++t+--+-+--H---+-I-I-I---+-+--t-Ishy




l~~TT~~~~~~~~~~

0 ~ wt-t--++-++--+-+-+-t-l-l--i--JH--l-I-L

t2 -+--I---I-J--I

~amprrrr+++~~---+-I-I-~

0 --+---+








E ~ 10 ~~_ ~ T~l5 ~ 01 ~1t1 - 0 ~CS~ t shy








8 88 Ou) e h~h ~e











E




Units for Potential

Difference

T-C














v~ 4= (501000~t) OIS l O S l ~ eJ (--~-----O~--~lI



f~cA I +1 el1l-lgt



HI ()r4-~ 2~ _ ~12 StJlV]











St a) 50 V b) 60 V d) 300 V



c) 12 V


A D



-h~~~(~~~


j cl 2OC)1 ~ It ~



Fe OCO~ N

c)~ d~( O ~ N

I




Q -11-- _-- C) w ~

V~~ W~4~ (1~lboCe) 1Ii1(10)111






r TEST CHARGE

8 ~)









plates















volts t= In~s e) (1) 12 J (3) 30 J (4) 24 JB





1++++++++++++++++pound++1

reg

I - ----I




r




+20 x 10-6 C +10 X 10-6 C






o 0 +10 X 10- C





pbull +10 x 10-6 C +10 x 10-6 C

o 0 ---- 40 x 10-1 m 1



1 ( 4 ) ( 1 ) (2 )







Assignment 8-21 Key






( A BIIIi

q = 1100) x 10 C



Abull B I



A Bbull l( P

bull




C 0

A B A B ~ ~~ -




~tt ~tyU 1Mtf~At




~u- 1 fte~~d


~~ w~ o_~S ~ ~(~ ~cL












So think about it


o 0 I~ ~I

r = 1 meter

So



b) By what factor










w-_ bullbulll ~0


=

















decreased

0 N




l ~6 r 0) qo N


50 x 10- C 5 N


ttO N


~S N





Explain your choice




re ~ F be~ adf

r













6



C ~ LIr to~4L Q G)I=- l aM ~ 1= +)







6 C What distance





-I tA ~ ~ ~t a ~~ C) M ~ J ) Q ~ f 1 kca ~~~


~ (i y Omiddotsl


1= - ~ =) ~ r=S~

=




answers with units)

Distance =1m

F gttJtOmiddot M

Distance = 2m

O~F ~ O

Distance =3m

F Ct 0 N

Distance =4m

F 3 ~ O~ N

Distance = 5m

f~ ~ ON

Distance =6m

F IS~OC~

Distance = 7m

f ~ 1 ~ 0 ~

Distance =8m

F ~ ~ loN

Distance =9m

F laquo1- ~ tOea N

Distance = 10m

F S~ ~ O~


rr ~d

bull

Z wft

bull-tftJ ~~


1lltW2 UlI~

~

~

~

C~

~~

1- 5 bull

~

-

~

I

g ffim ++ ~

~ g1 llJ

shy

~ i


I

-

Ii~

mEI J t m~ ~ I

~ U

II t=R=H IiIIr j

EBi~ ~

BEI11III ~ n

-000II ~ jiiiI ~

rIshy --bull J

IIIIIIII

m ffim II IJ

bull 100

I ~

bull 0








Sphere ~ Sphere 8












INSULATING STANDS













J-tf _- I - -) -~




A Test charge

~ ~ 02 rn Olm






Sphere Charge

R 0 T 4shyU




r-------~ Glass ROd


1M ~~e ~~cl ~ o~~ ~ Ju~ fOcS


-1ht -St~ -S -k ~t

r













r

a)

b)

d)

r





radius








-t

- C ~ ~ct l - fl f

shy y

I r









ceo-~~h~ ~

~ ~ CIO +-+-+-I-+shy

~ 1--+-+Ishy2 l-a

~ t-+-+I-I-+-+--I--+-+--I----l-----Ishy

f~ cOl++t+--+-+--H---+-I-I-I---+-+--t-Ishy




l~~TT~~~~~~~~~~

0 ~ wt-t--++-++--+-+-+-t-l-l--i--JH--l-I-L

t2 -+--I---I-J--I

~amprrrr+++~~---+-I-I-~

0 --+---+








E ~ 10 ~~_ ~ T~l5 ~ 01 ~1t1 - 0 ~CS~ t shy








8 88 Ou) e h~h ~e











E




Units for Potential

Difference

T-C














v~ 4= (501000~t) OIS l O S l ~ eJ (--~-----O~--~lI



f~cA I +1 el1l-lgt



HI ()r4-~ 2~ _ ~12 StJlV]











St a) 50 V b) 60 V d) 300 V



c) 12 V


A D



-h~~~(~~~


j cl 2OC)1 ~ It ~



Fe OCO~ N

c)~ d~( O ~ N

I




Q -11-- _-- C) w ~

V~~ W~4~ (1~lboCe) 1Ii1(10)111






r TEST CHARGE

8 ~)









plates















volts t= In~s e) (1) 12 J (3) 30 J (4) 24 JB





1++++++++++++++++pound++1

reg

I - ----I




r




+20 x 10-6 C +10 X 10-6 C






o 0 +10 X 10- C





pbull +10 x 10-6 C +10 x 10-6 C

o 0 ---- 40 x 10-1 m 1



1 ( 4 ) ( 1 ) (2 )







Assignment 8-21 Key

bull




C 0

A B A B ~ ~~ -




~tt ~tyU 1Mtf~At




~u- 1 fte~~d


~~ w~ o_~S ~ ~(~ ~cL












So think about it


o 0 I~ ~I

r = 1 meter

So



b) By what factor










w-_ bullbulll ~0


=

















decreased

0 N




l ~6 r 0) qo N


50 x 10- C 5 N


ttO N


~S N





Explain your choice




re ~ F be~ adf

r













6



C ~ LIr to~4L Q G)I=- l aM ~ 1= +)







6 C What distance





-I tA ~ ~ ~t a ~~ C) M ~ J ) Q ~ f 1 kca ~~~


~ (i y Omiddotsl


1= - ~ =) ~ r=S~

=




answers with units)

Distance =1m

F gttJtOmiddot M

Distance = 2m

O~F ~ O

Distance =3m

F Ct 0 N

Distance =4m

F 3 ~ O~ N

Distance = 5m

f~ ~ ON

Distance =6m

F IS~OC~

Distance = 7m

f ~ 1 ~ 0 ~

Distance =8m

F ~ ~ loN

Distance =9m

F laquo1- ~ tOea N

Distance = 10m

F S~ ~ O~


rr ~d

bull

Z wft

bull-tftJ ~~


1lltW2 UlI~

~

~

~

C~

~~

1- 5 bull

~

-

~

I

g ffim ++ ~

~ g1 llJ

shy

~ i


I

-

Ii~

mEI J t m~ ~ I

~ U

II t=R=H IiIIr j

EBi~ ~

BEI11III ~ n

-000II ~ jiiiI ~

rIshy --bull J

IIIIIIII

m ffim II IJ

bull 100

I ~

bull 0








Sphere ~ Sphere 8












INSULATING STANDS













J-tf _- I - -) -~




A Test charge

~ ~ 02 rn Olm






Sphere Charge

R 0 T 4shyU




r-------~ Glass ROd


1M ~~e ~~cl ~ o~~ ~ Ju~ fOcS


-1ht -St~ -S -k ~t

r













r

a)

b)

d)

r





radius








-t

- C ~ ~ct l - fl f

shy y

I r









ceo-~~h~ ~

~ ~ CIO +-+-+-I-+shy

~ 1--+-+Ishy2 l-a

~ t-+-+I-I-+-+--I--+-+--I----l-----Ishy

f~ cOl++t+--+-+--H---+-I-I-I---+-+--t-Ishy




l~~TT~~~~~~~~~~

0 ~ wt-t--++-++--+-+-+-t-l-l--i--JH--l-I-L

t2 -+--I---I-J--I

~amprrrr+++~~---+-I-I-~

0 --+---+








E ~ 10 ~~_ ~ T~l5 ~ 01 ~1t1 - 0 ~CS~ t shy








8 88 Ou) e h~h ~e











E




Units for Potential

Difference

T-C














v~ 4= (501000~t) OIS l O S l ~ eJ (--~-----O~--~lI



f~cA I +1 el1l-lgt



HI ()r4-~ 2~ _ ~12 StJlV]











St a) 50 V b) 60 V d) 300 V



c) 12 V


A D



-h~~~(~~~


j cl 2OC)1 ~ It ~



Fe OCO~ N

c)~ d~( O ~ N

I




Q -11-- _-- C) w ~

V~~ W~4~ (1~lboCe) 1Ii1(10)111






r TEST CHARGE

8 ~)









plates















volts t= In~s e) (1) 12 J (3) 30 J (4) 24 JB





1++++++++++++++++pound++1

reg

I - ----I




r




+20 x 10-6 C +10 X 10-6 C






o 0 +10 X 10- C





pbull +10 x 10-6 C +10 x 10-6 C

o 0 ---- 40 x 10-1 m 1



1 ( 4 ) ( 1 ) (2 )







Assignment 8-21 Key











So think about it


o 0 I~ ~I

r = 1 meter

So



b) By what factor










w-_ bullbulll ~0


=

















decreased

0 N




l ~6 r 0) qo N


50 x 10- C 5 N


ttO N


~S N





Explain your choice




re ~ F be~ adf

r













6



C ~ LIr to~4L Q G)I=- l aM ~ 1= +)







6 C What distance





-I tA ~ ~ ~t a ~~ C) M ~ J ) Q ~ f 1 kca ~~~


~ (i y Omiddotsl


1= - ~ =) ~ r=S~

=




answers with units)

Distance =1m

F gttJtOmiddot M

Distance = 2m

O~F ~ O

Distance =3m

F Ct 0 N

Distance =4m

F 3 ~ O~ N

Distance = 5m

f~ ~ ON

Distance =6m

F IS~OC~

Distance = 7m

f ~ 1 ~ 0 ~

Distance =8m

F ~ ~ loN

Distance =9m

F laquo1- ~ tOea N

Distance = 10m

F S~ ~ O~


rr ~d

bull

Z wft

bull-tftJ ~~


1lltW2 UlI~

~

~

~

C~

~~

1- 5 bull

~

-

~

I

g ffim ++ ~

~ g1 llJ

shy

~ i


I

-

Ii~

mEI J t m~ ~ I

~ U

II t=R=H IiIIr j

EBi~ ~

BEI11III ~ n

-000II ~ jiiiI ~

rIshy --bull J

IIIIIIII

m ffim II IJ

bull 100

I ~

bull 0








Sphere ~ Sphere 8












INSULATING STANDS













J-tf _- I - -) -~




A Test charge

~ ~ 02 rn Olm






Sphere Charge

R 0 T 4shyU




r-------~ Glass ROd


1M ~~e ~~cl ~ o~~ ~ Ju~ fOcS


-1ht -St~ -S -k ~t

r













r

a)

b)

d)

r





radius








-t

- C ~ ~ct l - fl f

shy y

I r









ceo-~~h~ ~

~ ~ CIO +-+-+-I-+shy

~ 1--+-+Ishy2 l-a

~ t-+-+I-I-+-+--I--+-+--I----l-----Ishy

f~ cOl++t+--+-+--H---+-I-I-I---+-+--t-Ishy




l~~TT~~~~~~~~~~

0 ~ wt-t--++-++--+-+-+-t-l-l--i--JH--l-I-L

t2 -+--I---I-J--I

~amprrrr+++~~---+-I-I-~

0 --+---+








E ~ 10 ~~_ ~ T~l5 ~ 01 ~1t1 - 0 ~CS~ t shy








8 88 Ou) e h~h ~e











E




Units for Potential

Difference

T-C














v~ 4= (501000~t) OIS l O S l ~ eJ (--~-----O~--~lI



f~cA I +1 el1l-lgt



HI ()r4-~ 2~ _ ~12 StJlV]











St a) 50 V b) 60 V d) 300 V



c) 12 V


A D



-h~~~(~~~


j cl 2OC)1 ~ It ~



Fe OCO~ N

c)~ d~( O ~ N

I




Q -11-- _-- C) w ~

V~~ W~4~ (1~lboCe) 1Ii1(10)111






r TEST CHARGE

8 ~)









plates















volts t= In~s e) (1) 12 J (3) 30 J (4) 24 JB





1++++++++++++++++pound++1

reg

I - ----I




r




+20 x 10-6 C +10 X 10-6 C






o 0 +10 X 10- C





pbull +10 x 10-6 C +10 x 10-6 C

o 0 ---- 40 x 10-1 m 1



1 ( 4 ) ( 1 ) (2 )







Assignment 8-21 Key

=

















decreased

0 N




l ~6 r 0) qo N


50 x 10- C 5 N


ttO N


~S N





Explain your choice




re ~ F be~ adf

r













6



C ~ LIr to~4L Q G)I=- l aM ~ 1= +)







6 C What distance





-I tA ~ ~ ~t a ~~ C) M ~ J ) Q ~ f 1 kca ~~~


~ (i y Omiddotsl


1= - ~ =) ~ r=S~

=




answers with units)

Distance =1m

F gttJtOmiddot M

Distance = 2m

O~F ~ O

Distance =3m

F Ct 0 N

Distance =4m

F 3 ~ O~ N

Distance = 5m

f~ ~ ON

Distance =6m

F IS~OC~

Distance = 7m

f ~ 1 ~ 0 ~

Distance =8m

F ~ ~ loN

Distance =9m

F laquo1- ~ tOea N

Distance = 10m

F S~ ~ O~


rr ~d

bull

Z wft

bull-tftJ ~~


1lltW2 UlI~

~

~

~

C~

~~

1- 5 bull

~

-

~

I

g ffim ++ ~

~ g1 llJ

shy

~ i


I

-

Ii~

mEI J t m~ ~ I

~ U

II t=R=H IiIIr j

EBi~ ~

BEI11III ~ n

-000II ~ jiiiI ~

rIshy --bull J

IIIIIIII

m ffim II IJ

bull 100

I ~

bull 0








Sphere ~ Sphere 8












INSULATING STANDS













J-tf _- I - -) -~




A Test charge

~ ~ 02 rn Olm






Sphere Charge

R 0 T 4shyU




r-------~ Glass ROd


1M ~~e ~~cl ~ o~~ ~ Ju~ fOcS


-1ht -St~ -S -k ~t

r













r

a)

b)

d)

r





radius








-t

- C ~ ~ct l - fl f

shy y

I r









ceo-~~h~ ~

~ ~ CIO +-+-+-I-+shy

~ 1--+-+Ishy2 l-a

~ t-+-+I-I-+-+--I--+-+--I----l-----Ishy

f~ cOl++t+--+-+--H---+-I-I-I---+-+--t-Ishy




l~~TT~~~~~~~~~~

0 ~ wt-t--++-++--+-+-+-t-l-l--i--JH--l-I-L

t2 -+--I---I-J--I

~amprrrr+++~~---+-I-I-~

0 --+---+








E ~ 10 ~~_ ~ T~l5 ~ 01 ~1t1 - 0 ~CS~ t shy








8 88 Ou) e h~h ~e











E




Units for Potential

Difference

T-C














v~ 4= (501000~t) OIS l O S l ~ eJ (--~-----O~--~lI



f~cA I +1 el1l-lgt



HI ()r4-~ 2~ _ ~12 StJlV]











St a) 50 V b) 60 V d) 300 V



c) 12 V


A D



-h~~~(~~~


j cl 2OC)1 ~ It ~



Fe OCO~ N

c)~ d~( O ~ N

I




Q -11-- _-- C) w ~

V~~ W~4~ (1~lboCe) 1Ii1(10)111






r TEST CHARGE

8 ~)









plates















volts t= In~s e) (1) 12 J (3) 30 J (4) 24 JB





1++++++++++++++++pound++1

reg

I - ----I




r




+20 x 10-6 C +10 X 10-6 C






o 0 +10 X 10- C





pbull +10 x 10-6 C +10 x 10-6 C

o 0 ---- 40 x 10-1 m 1



1 ( 4 ) ( 1 ) (2 )







Assignment 8-21 Key




l ~6 r 0) qo N


50 x 10- C 5 N


ttO N


~S N





Explain your choice




re ~ F be~ adf

r













6



C ~ LIr to~4L Q G)I=- l aM ~ 1= +)







6 C What distance





-I tA ~ ~ ~t a ~~ C) M ~ J ) Q ~ f 1 kca ~~~


~ (i y Omiddotsl


1= - ~ =) ~ r=S~

=




answers with units)

Distance =1m

F gttJtOmiddot M

Distance = 2m

O~F ~ O

Distance =3m

F Ct 0 N

Distance =4m

F 3 ~ O~ N

Distance = 5m

f~ ~ ON

Distance =6m

F IS~OC~

Distance = 7m

f ~ 1 ~ 0 ~

Distance =8m

F ~ ~ loN

Distance =9m

F laquo1- ~ tOea N

Distance = 10m

F S~ ~ O~


rr ~d

bull

Z wft

bull-tftJ ~~


1lltW2 UlI~

~

~

~

C~

~~

1- 5 bull

~

-

~

I

g ffim ++ ~

~ g1 llJ

shy

~ i


I

-

Ii~

mEI J t m~ ~ I

~ U

II t=R=H IiIIr j

EBi~ ~

BEI11III ~ n

-000II ~ jiiiI ~

rIshy --bull J

IIIIIIII

m ffim II IJ

bull 100

I ~

bull 0








Sphere ~ Sphere 8












INSULATING STANDS













J-tf _- I - -) -~




A Test charge

~ ~ 02 rn Olm






Sphere Charge

R 0 T 4shyU




r-------~ Glass ROd


1M ~~e ~~cl ~ o~~ ~ Ju~ fOcS


-1ht -St~ -S -k ~t

r













r

a)

b)

d)

r





radius








-t

- C ~ ~ct l - fl f

shy y

I r









ceo-~~h~ ~

~ ~ CIO +-+-+-I-+shy

~ 1--+-+Ishy2 l-a

~ t-+-+I-I-+-+--I--+-+--I----l-----Ishy

f~ cOl++t+--+-+--H---+-I-I-I---+-+--t-Ishy




l~~TT~~~~~~~~~~

0 ~ wt-t--++-++--+-+-+-t-l-l--i--JH--l-I-L

t2 -+--I---I-J--I

~amprrrr+++~~---+-I-I-~

0 --+---+








E ~ 10 ~~_ ~ T~l5 ~ 01 ~1t1 - 0 ~CS~ t shy








8 88 Ou) e h~h ~e











E




Units for Potential

Difference

T-C














v~ 4= (501000~t) OIS l O S l ~ eJ (--~-----O~--~lI



f~cA I +1 el1l-lgt



HI ()r4-~ 2~ _ ~12 StJlV]











St a) 50 V b) 60 V d) 300 V



c) 12 V


A D



-h~~~(~~~


j cl 2OC)1 ~ It ~



Fe OCO~ N

c)~ d~( O ~ N

I




Q -11-- _-- C) w ~

V~~ W~4~ (1~lboCe) 1Ii1(10)111






r TEST CHARGE

8 ~)









plates















volts t= In~s e) (1) 12 J (3) 30 J (4) 24 JB





1++++++++++++++++pound++1

reg

I - ----I




r




+20 x 10-6 C +10 X 10-6 C






o 0 +10 X 10- C





pbull +10 x 10-6 C +10 x 10-6 C

o 0 ---- 40 x 10-1 m 1



1 ( 4 ) ( 1 ) (2 )







Assignment 8-21 Key

r













6



C ~ LIr to~4L Q G)I=- l aM ~ 1= +)







6 C What distance





-I tA ~ ~ ~t a ~~ C) M ~ J ) Q ~ f 1 kca ~~~


~ (i y Omiddotsl


1= - ~ =) ~ r=S~

=




answers with units)

Distance =1m

F gttJtOmiddot M

Distance = 2m

O~F ~ O

Distance =3m

F Ct 0 N

Distance =4m

F 3 ~ O~ N

Distance = 5m

f~ ~ ON

Distance =6m

F IS~OC~

Distance = 7m

f ~ 1 ~ 0 ~

Distance =8m

F ~ ~ loN

Distance =9m

F laquo1- ~ tOea N

Distance = 10m

F S~ ~ O~


rr ~d

bull

Z wft

bull-tftJ ~~


1lltW2 UlI~

~

~

~

C~

~~

1- 5 bull

~

-

~

I

g ffim ++ ~

~ g1 llJ

shy

~ i


I

-

Ii~

mEI J t m~ ~ I

~ U

II t=R=H IiIIr j

EBi~ ~

BEI11III ~ n

-000II ~ jiiiI ~

rIshy --bull J

IIIIIIII

m ffim II IJ

bull 100

I ~

bull 0








Sphere ~ Sphere 8












INSULATING STANDS













J-tf _- I - -) -~




A Test charge

~ ~ 02 rn Olm






Sphere Charge

R 0 T 4shyU




r-------~ Glass ROd


1M ~~e ~~cl ~ o~~ ~ Ju~ fOcS


-1ht -St~ -S -k ~t

r













r

a)

b)

d)

r





radius








-t

- C ~ ~ct l - fl f

shy y

I r









ceo-~~h~ ~

~ ~ CIO +-+-+-I-+shy

~ 1--+-+Ishy2 l-a

~ t-+-+I-I-+-+--I--+-+--I----l-----Ishy

f~ cOl++t+--+-+--H---+-I-I-I---+-+--t-Ishy




l~~TT~~~~~~~~~~

0 ~ wt-t--++-++--+-+-+-t-l-l--i--JH--l-I-L

t2 -+--I---I-J--I

~amprrrr+++~~---+-I-I-~

0 --+---+








E ~ 10 ~~_ ~ T~l5 ~ 01 ~1t1 - 0 ~CS~ t shy








8 88 Ou) e h~h ~e











E




Units for Potential

Difference

T-C














v~ 4= (501000~t) OIS l O S l ~ eJ (--~-----O~--~lI



f~cA I +1 el1l-lgt



HI ()r4-~ 2~ _ ~12 StJlV]











St a) 50 V b) 60 V d) 300 V



c) 12 V


A D



-h~~~(~~~


j cl 2OC)1 ~ It ~



Fe OCO~ N

c)~ d~( O ~ N

I




Q -11-- _-- C) w ~

V~~ W~4~ (1~lboCe) 1Ii1(10)111






r TEST CHARGE

8 ~)









plates















volts t= In~s e) (1) 12 J (3) 30 J (4) 24 JB





1++++++++++++++++pound++1

reg

I - ----I




r




+20 x 10-6 C +10 X 10-6 C






o 0 +10 X 10- C





pbull +10 x 10-6 C +10 x 10-6 C

o 0 ---- 40 x 10-1 m 1



1 ( 4 ) ( 1 ) (2 )







Assignment 8-21 Key



6 C What distance





-I tA ~ ~ ~t a ~~ C) M ~ J ) Q ~ f 1 kca ~~~


~ (i y Omiddotsl


1= - ~ =) ~ r=S~

=




answers with units)

Distance =1m

F gttJtOmiddot M

Distance = 2m

O~F ~ O

Distance =3m

F Ct 0 N

Distance =4m

F 3 ~ O~ N

Distance = 5m

f~ ~ ON

Distance =6m

F IS~OC~

Distance = 7m

f ~ 1 ~ 0 ~

Distance =8m

F ~ ~ loN

Distance =9m

F laquo1- ~ tOea N

Distance = 10m

F S~ ~ O~


rr ~d

bull

Z wft

bull-tftJ ~~


1lltW2 UlI~

~

~

~

C~

~~

1- 5 bull

~

-

~

I

g ffim ++ ~

~ g1 llJ

shy

~ i


I

-

Ii~

mEI J t m~ ~ I

~ U

II t=R=H IiIIr j

EBi~ ~

BEI11III ~ n

-000II ~ jiiiI ~

rIshy --bull J

IIIIIIII

m ffim II IJ

bull 100

I ~

bull 0








Sphere ~ Sphere 8












INSULATING STANDS













J-tf _- I - -) -~




A Test charge

~ ~ 02 rn Olm






Sphere Charge

R 0 T 4shyU




r-------~ Glass ROd


1M ~~e ~~cl ~ o~~ ~ Ju~ fOcS


-1ht -St~ -S -k ~t

r













r

a)

b)

d)

r





radius








-t

- C ~ ~ct l - fl f

shy y

I r









ceo-~~h~ ~

~ ~ CIO +-+-+-I-+shy

~ 1--+-+Ishy2 l-a

~ t-+-+I-I-+-+--I--+-+--I----l-----Ishy

f~ cOl++t+--+-+--H---+-I-I-I---+-+--t-Ishy




l~~TT~~~~~~~~~~

0 ~ wt-t--++-++--+-+-+-t-l-l--i--JH--l-I-L

t2 -+--I---I-J--I

~amprrrr+++~~---+-I-I-~

0 --+---+








E ~ 10 ~~_ ~ T~l5 ~ 01 ~1t1 - 0 ~CS~ t shy








8 88 Ou) e h~h ~e











E




Units for Potential

Difference

T-C














v~ 4= (501000~t) OIS l O S l ~ eJ (--~-----O~--~lI



f~cA I +1 el1l-lgt



HI ()r4-~ 2~ _ ~12 StJlV]











St a) 50 V b) 60 V d) 300 V



c) 12 V


A D



-h~~~(~~~


j cl 2OC)1 ~ It ~



Fe OCO~ N

c)~ d~( O ~ N

I




Q -11-- _-- C) w ~

V~~ W~4~ (1~lboCe) 1Ii1(10)111






r TEST CHARGE

8 ~)









plates















volts t= In~s e) (1) 12 J (3) 30 J (4) 24 JB





1++++++++++++++++pound++1

reg

I - ----I




r




+20 x 10-6 C +10 X 10-6 C






o 0 +10 X 10- C





pbull +10 x 10-6 C +10 x 10-6 C

o 0 ---- 40 x 10-1 m 1



1 ( 4 ) ( 1 ) (2 )







Assignment 8-21 Key




answers with units)

Distance =1m

F gttJtOmiddot M

Distance = 2m

O~F ~ O

Distance =3m

F Ct 0 N

Distance =4m

F 3 ~ O~ N

Distance = 5m

f~ ~ ON

Distance =6m

F IS~OC~

Distance = 7m

f ~ 1 ~ 0 ~

Distance =8m

F ~ ~ loN

Distance =9m

F laquo1- ~ tOea N

Distance = 10m

F S~ ~ O~


rr ~d

bull

Z wft

bull-tftJ ~~


1lltW2 UlI~

~

~

~

C~

~~

1- 5 bull

~

-

~

I

g ffim ++ ~

~ g1 llJ

shy

~ i


I

-

Ii~

mEI J t m~ ~ I

~ U

II t=R=H IiIIr j

EBi~ ~

BEI11III ~ n

-000II ~ jiiiI ~

rIshy --bull J

IIIIIIII

m ffim II IJ

bull 100

I ~

bull 0








Sphere ~ Sphere 8












INSULATING STANDS













J-tf _- I - -) -~




A Test charge

~ ~ 02 rn Olm






Sphere Charge

R 0 T 4shyU




r-------~ Glass ROd


1M ~~e ~~cl ~ o~~ ~ Ju~ fOcS


-1ht -St~ -S -k ~t

r













r

a)

b)

d)

r





radius








-t

- C ~ ~ct l - fl f

shy y

I r









ceo-~~h~ ~

~ ~ CIO +-+-+-I-+shy

~ 1--+-+Ishy2 l-a

~ t-+-+I-I-+-+--I--+-+--I----l-----Ishy

f~ cOl++t+--+-+--H---+-I-I-I---+-+--t-Ishy




l~~TT~~~~~~~~~~

0 ~ wt-t--++-++--+-+-+-t-l-l--i--JH--l-I-L

t2 -+--I---I-J--I

~amprrrr+++~~---+-I-I-~

0 --+---+








E ~ 10 ~~_ ~ T~l5 ~ 01 ~1t1 - 0 ~CS~ t shy








8 88 Ou) e h~h ~e











E




Units for Potential

Difference

T-C














v~ 4= (501000~t) OIS l O S l ~ eJ (--~-----O~--~lI



f~cA I +1 el1l-lgt



HI ()r4-~ 2~ _ ~12 StJlV]











St a) 50 V b) 60 V d) 300 V



c) 12 V


A D



-h~~~(~~~


j cl 2OC)1 ~ It ~



Fe OCO~ N

c)~ d~( O ~ N

I




Q -11-- _-- C) w ~

V~~ W~4~ (1~lboCe) 1Ii1(10)111






r TEST CHARGE

8 ~)









plates















volts t= In~s e) (1) 12 J (3) 30 J (4) 24 JB





1++++++++++++++++pound++1

reg

I - ----I




r




+20 x 10-6 C +10 X 10-6 C






o 0 +10 X 10- C





pbull +10 x 10-6 C +10 x 10-6 C

o 0 ---- 40 x 10-1 m 1



1 ( 4 ) ( 1 ) (2 )







Assignment 8-21 Key


rr ~d

bull

Z wft

bull-tftJ ~~


1lltW2 UlI~

~

~

~

C~

~~

1- 5 bull

~

-

~

I

g ffim ++ ~

~ g1 llJ

shy

~ i


I

-

Ii~

mEI J t m~ ~ I

~ U

II t=R=H IiIIr j

EBi~ ~

BEI11III ~ n

-000II ~ jiiiI ~

rIshy --bull J

IIIIIIII

m ffim II IJ

bull 100

I ~

bull 0








Sphere ~ Sphere 8












INSULATING STANDS













J-tf _- I - -) -~




A Test charge

~ ~ 02 rn Olm






Sphere Charge

R 0 T 4shyU




r-------~ Glass ROd


1M ~~e ~~cl ~ o~~ ~ Ju~ fOcS


-1ht -St~ -S -k ~t

r













r

a)

b)

d)

r





radius








-t

- C ~ ~ct l - fl f

shy y

I r









ceo-~~h~ ~

~ ~ CIO +-+-+-I-+shy

~ 1--+-+Ishy2 l-a

~ t-+-+I-I-+-+--I--+-+--I----l-----Ishy

f~ cOl++t+--+-+--H---+-I-I-I---+-+--t-Ishy




l~~TT~~~~~~~~~~

0 ~ wt-t--++-++--+-+-+-t-l-l--i--JH--l-I-L

t2 -+--I---I-J--I

~amprrrr+++~~---+-I-I-~

0 --+---+








E ~ 10 ~~_ ~ T~l5 ~ 01 ~1t1 - 0 ~CS~ t shy








8 88 Ou) e h~h ~e











E




Units for Potential

Difference

T-C














v~ 4= (501000~t) OIS l O S l ~ eJ (--~-----O~--~lI



f~cA I +1 el1l-lgt



HI ()r4-~ 2~ _ ~12 StJlV]











St a) 50 V b) 60 V d) 300 V



c) 12 V


A D



-h~~~(~~~


j cl 2OC)1 ~ It ~



Fe OCO~ N

c)~ d~( O ~ N

I




Q -11-- _-- C) w ~

V~~ W~4~ (1~lboCe) 1Ii1(10)111






r TEST CHARGE

8 ~)









plates















volts t= In~s e) (1) 12 J (3) 30 J (4) 24 JB





1++++++++++++++++pound++1

reg

I - ----I




r




+20 x 10-6 C +10 X 10-6 C






o 0 +10 X 10- C





pbull +10 x 10-6 C +10 x 10-6 C

o 0 ---- 40 x 10-1 m 1



1 ( 4 ) ( 1 ) (2 )







Assignment 8-21 Key








Sphere ~ Sphere 8












INSULATING STANDS













J-tf _- I - -) -~




A Test charge

~ ~ 02 rn Olm






Sphere Charge

R 0 T 4shyU




r-------~ Glass ROd


1M ~~e ~~cl ~ o~~ ~ Ju~ fOcS


-1ht -St~ -S -k ~t

r













r

a)

b)

d)

r





radius








-t

- C ~ ~ct l - fl f

shy y

I r









ceo-~~h~ ~

~ ~ CIO +-+-+-I-+shy

~ 1--+-+Ishy2 l-a

~ t-+-+I-I-+-+--I--+-+--I----l-----Ishy

f~ cOl++t+--+-+--H---+-I-I-I---+-+--t-Ishy




l~~TT~~~~~~~~~~

0 ~ wt-t--++-++--+-+-+-t-l-l--i--JH--l-I-L

t2 -+--I---I-J--I

~amprrrr+++~~---+-I-I-~

0 --+---+








E ~ 10 ~~_ ~ T~l5 ~ 01 ~1t1 - 0 ~CS~ t shy








8 88 Ou) e h~h ~e











E




Units for Potential

Difference

T-C














v~ 4= (501000~t) OIS l O S l ~ eJ (--~-----O~--~lI



f~cA I +1 el1l-lgt



HI ()r4-~ 2~ _ ~12 StJlV]











St a) 50 V b) 60 V d) 300 V



c) 12 V


A D



-h~~~(~~~


j cl 2OC)1 ~ It ~



Fe OCO~ N

c)~ d~( O ~ N

I




Q -11-- _-- C) w ~

V~~ W~4~ (1~lboCe) 1Ii1(10)111






r TEST CHARGE

8 ~)









plates















volts t= In~s e) (1) 12 J (3) 30 J (4) 24 JB





1++++++++++++++++pound++1

reg

I - ----I




r




+20 x 10-6 C +10 X 10-6 C






o 0 +10 X 10- C





pbull +10 x 10-6 C +10 x 10-6 C

o 0 ---- 40 x 10-1 m 1



1 ( 4 ) ( 1 ) (2 )







Assignment 8-21 Key






INSULATING STANDS













J-tf _- I - -) -~




A Test charge

~ ~ 02 rn Olm






Sphere Charge

R 0 T 4shyU




r-------~ Glass ROd


1M ~~e ~~cl ~ o~~ ~ Ju~ fOcS


-1ht -St~ -S -k ~t

r













r

a)

b)

d)

r





radius








-t

- C ~ ~ct l - fl f

shy y

I r









ceo-~~h~ ~

~ ~ CIO +-+-+-I-+shy

~ 1--+-+Ishy2 l-a

~ t-+-+I-I-+-+--I--+-+--I----l-----Ishy

f~ cOl++t+--+-+--H---+-I-I-I---+-+--t-Ishy




l~~TT~~~~~~~~~~

0 ~ wt-t--++-++--+-+-+-t-l-l--i--JH--l-I-L

t2 -+--I---I-J--I

~amprrrr+++~~---+-I-I-~

0 --+---+








E ~ 10 ~~_ ~ T~l5 ~ 01 ~1t1 - 0 ~CS~ t shy








8 88 Ou) e h~h ~e











E




Units for Potential

Difference

T-C














v~ 4= (501000~t) OIS l O S l ~ eJ (--~-----O~--~lI



f~cA I +1 el1l-lgt



HI ()r4-~ 2~ _ ~12 StJlV]











St a) 50 V b) 60 V d) 300 V



c) 12 V


A D



-h~~~(~~~


j cl 2OC)1 ~ It ~



Fe OCO~ N

c)~ d~( O ~ N

I




Q -11-- _-- C) w ~

V~~ W~4~ (1~lboCe) 1Ii1(10)111






r TEST CHARGE

8 ~)









plates















volts t= In~s e) (1) 12 J (3) 30 J (4) 24 JB





1++++++++++++++++pound++1

reg

I - ----I




r




+20 x 10-6 C +10 X 10-6 C






o 0 +10 X 10- C





pbull +10 x 10-6 C +10 x 10-6 C

o 0 ---- 40 x 10-1 m 1



1 ( 4 ) ( 1 ) (2 )







Assignment 8-21 Key




A Test charge

~ ~ 02 rn Olm






Sphere Charge

R 0 T 4shyU




r-------~ Glass ROd


1M ~~e ~~cl ~ o~~ ~ Ju~ fOcS


-1ht -St~ -S -k ~t

r













r

a)

b)

d)

r





radius








-t

- C ~ ~ct l - fl f

shy y

I r









ceo-~~h~ ~

~ ~ CIO +-+-+-I-+shy

~ 1--+-+Ishy2 l-a

~ t-+-+I-I-+-+--I--+-+--I----l-----Ishy

f~ cOl++t+--+-+--H---+-I-I-I---+-+--t-Ishy




l~~TT~~~~~~~~~~

0 ~ wt-t--++-++--+-+-+-t-l-l--i--JH--l-I-L

t2 -+--I---I-J--I

~amprrrr+++~~---+-I-I-~

0 --+---+








E ~ 10 ~~_ ~ T~l5 ~ 01 ~1t1 - 0 ~CS~ t shy








8 88 Ou) e h~h ~e











E




Units for Potential

Difference

T-C














v~ 4= (501000~t) OIS l O S l ~ eJ (--~-----O~--~lI



f~cA I +1 el1l-lgt



HI ()r4-~ 2~ _ ~12 StJlV]











St a) 50 V b) 60 V d) 300 V



c) 12 V


A D



-h~~~(~~~


j cl 2OC)1 ~ It ~



Fe OCO~ N

c)~ d~( O ~ N

I




Q -11-- _-- C) w ~

V~~ W~4~ (1~lboCe) 1Ii1(10)111






r TEST CHARGE

8 ~)









plates















volts t= In~s e) (1) 12 J (3) 30 J (4) 24 JB





1++++++++++++++++pound++1

reg

I - ----I




r




+20 x 10-6 C +10 X 10-6 C






o 0 +10 X 10- C





pbull +10 x 10-6 C +10 x 10-6 C

o 0 ---- 40 x 10-1 m 1



1 ( 4 ) ( 1 ) (2 )







Assignment 8-21 Key




r-------~ Glass ROd


1M ~~e ~~cl ~ o~~ ~ Ju~ fOcS


-1ht -St~ -S -k ~t

r













r

a)

b)

d)

r





radius








-t

- C ~ ~ct l - fl f

shy y

I r









ceo-~~h~ ~

~ ~ CIO +-+-+-I-+shy

~ 1--+-+Ishy2 l-a

~ t-+-+I-I-+-+--I--+-+--I----l-----Ishy

f~ cOl++t+--+-+--H---+-I-I-I---+-+--t-Ishy




l~~TT~~~~~~~~~~

0 ~ wt-t--++-++--+-+-+-t-l-l--i--JH--l-I-L

t2 -+--I---I-J--I

~amprrrr+++~~---+-I-I-~

0 --+---+








E ~ 10 ~~_ ~ T~l5 ~ 01 ~1t1 - 0 ~CS~ t shy








8 88 Ou) e h~h ~e











E




Units for Potential

Difference

T-C














v~ 4= (501000~t) OIS l O S l ~ eJ (--~-----O~--~lI



f~cA I +1 el1l-lgt



HI ()r4-~ 2~ _ ~12 StJlV]











St a) 50 V b) 60 V d) 300 V



c) 12 V


A D



-h~~~(~~~


j cl 2OC)1 ~ It ~



Fe OCO~ N

c)~ d~( O ~ N

I




Q -11-- _-- C) w ~

V~~ W~4~ (1~lboCe) 1Ii1(10)111






r TEST CHARGE

8 ~)









plates















volts t= In~s e) (1) 12 J (3) 30 J (4) 24 JB





1++++++++++++++++pound++1

reg

I - ----I




r




+20 x 10-6 C +10 X 10-6 C






o 0 +10 X 10- C





pbull +10 x 10-6 C +10 x 10-6 C

o 0 ---- 40 x 10-1 m 1



1 ( 4 ) ( 1 ) (2 )







Assignment 8-21 Key

r













r

a)

b)

d)

r





radius








-t

- C ~ ~ct l - fl f

shy y

I r









ceo-~~h~ ~

~ ~ CIO +-+-+-I-+shy

~ 1--+-+Ishy2 l-a

~ t-+-+I-I-+-+--I--+-+--I----l-----Ishy

f~ cOl++t+--+-+--H---+-I-I-I---+-+--t-Ishy




l~~TT~~~~~~~~~~

0 ~ wt-t--++-++--+-+-+-t-l-l--i--JH--l-I-L

t2 -+--I---I-J--I

~amprrrr+++~~---+-I-I-~

0 --+---+








E ~ 10 ~~_ ~ T~l5 ~ 01 ~1t1 - 0 ~CS~ t shy








8 88 Ou) e h~h ~e











E




Units for Potential

Difference

T-C














v~ 4= (501000~t) OIS l O S l ~ eJ (--~-----O~--~lI



f~cA I +1 el1l-lgt



HI ()r4-~ 2~ _ ~12 StJlV]











St a) 50 V b) 60 V d) 300 V



c) 12 V


A D



-h~~~(~~~


j cl 2OC)1 ~ It ~



Fe OCO~ N

c)~ d~( O ~ N

I




Q -11-- _-- C) w ~

V~~ W~4~ (1~lboCe) 1Ii1(10)111






r TEST CHARGE

8 ~)









plates















volts t= In~s e) (1) 12 J (3) 30 J (4) 24 JB





1++++++++++++++++pound++1

reg

I - ----I




r




+20 x 10-6 C +10 X 10-6 C






o 0 +10 X 10- C





pbull +10 x 10-6 C +10 x 10-6 C

o 0 ---- 40 x 10-1 m 1



1 ( 4 ) ( 1 ) (2 )







Assignment 8-21 Key

r

a)

b)

d)

r





radius








-t

- C ~ ~ct l - fl f

shy y

I r









ceo-~~h~ ~

~ ~ CIO +-+-+-I-+shy

~ 1--+-+Ishy2 l-a

~ t-+-+I-I-+-+--I--+-+--I----l-----Ishy

f~ cOl++t+--+-+--H---+-I-I-I---+-+--t-Ishy




l~~TT~~~~~~~~~~

0 ~ wt-t--++-++--+-+-+-t-l-l--i--JH--l-I-L

t2 -+--I---I-J--I

~amprrrr+++~~---+-I-I-~

0 --+---+








E ~ 10 ~~_ ~ T~l5 ~ 01 ~1t1 - 0 ~CS~ t shy








8 88 Ou) e h~h ~e











E




Units for Potential

Difference

T-C














v~ 4= (501000~t) OIS l O S l ~ eJ (--~-----O~--~lI



f~cA I +1 el1l-lgt



HI ()r4-~ 2~ _ ~12 StJlV]











St a) 50 V b) 60 V d) 300 V



c) 12 V


A D



-h~~~(~~~


j cl 2OC)1 ~ It ~



Fe OCO~ N

c)~ d~( O ~ N

I




Q -11-- _-- C) w ~

V~~ W~4~ (1~lboCe) 1Ii1(10)111






r TEST CHARGE

8 ~)









plates















volts t= In~s e) (1) 12 J (3) 30 J (4) 24 JB





1++++++++++++++++pound++1

reg

I - ----I




r




+20 x 10-6 C +10 X 10-6 C






o 0 +10 X 10- C





pbull +10 x 10-6 C +10 x 10-6 C

o 0 ---- 40 x 10-1 m 1



1 ( 4 ) ( 1 ) (2 )







Assignment 8-21 Key

r





radius








-t

- C ~ ~ct l - fl f

shy y

I r









ceo-~~h~ ~

~ ~ CIO +-+-+-I-+shy

~ 1--+-+Ishy2 l-a

~ t-+-+I-I-+-+--I--+-+--I----l-----Ishy

f~ cOl++t+--+-+--H---+-I-I-I---+-+--t-Ishy




l~~TT~~~~~~~~~~

0 ~ wt-t--++-++--+-+-+-t-l-l--i--JH--l-I-L

t2 -+--I---I-J--I

~amprrrr+++~~---+-I-I-~

0 --+---+








E ~ 10 ~~_ ~ T~l5 ~ 01 ~1t1 - 0 ~CS~ t shy








8 88 Ou) e h~h ~e











E




Units for Potential

Difference

T-C














v~ 4= (501000~t) OIS l O S l ~ eJ (--~-----O~--~lI



f~cA I +1 el1l-lgt



HI ()r4-~ 2~ _ ~12 StJlV]











St a) 50 V b) 60 V d) 300 V



c) 12 V


A D



-h~~~(~~~


j cl 2OC)1 ~ It ~



Fe OCO~ N

c)~ d~( O ~ N

I




Q -11-- _-- C) w ~

V~~ W~4~ (1~lboCe) 1Ii1(10)111






r TEST CHARGE

8 ~)









plates















volts t= In~s e) (1) 12 J (3) 30 J (4) 24 JB





1++++++++++++++++pound++1

reg

I - ----I




r




+20 x 10-6 C +10 X 10-6 C






o 0 +10 X 10- C





pbull +10 x 10-6 C +10 x 10-6 C

o 0 ---- 40 x 10-1 m 1



1 ( 4 ) ( 1 ) (2 )







Assignment 8-21 Key

I r









ceo-~~h~ ~

~ ~ CIO +-+-+-I-+shy

~ 1--+-+Ishy2 l-a

~ t-+-+I-I-+-+--I--+-+--I----l-----Ishy

f~ cOl++t+--+-+--H---+-I-I-I---+-+--t-Ishy




l~~TT~~~~~~~~~~

0 ~ wt-t--++-++--+-+-+-t-l-l--i--JH--l-I-L

t2 -+--I---I-J--I

~amprrrr+++~~---+-I-I-~

0 --+---+








E ~ 10 ~~_ ~ T~l5 ~ 01 ~1t1 - 0 ~CS~ t shy








8 88 Ou) e h~h ~e











E




Units for Potential

Difference

T-C














v~ 4= (501000~t) OIS l O S l ~ eJ (--~-----O~--~lI



f~cA I +1 el1l-lgt



HI ()r4-~ 2~ _ ~12 StJlV]











St a) 50 V b) 60 V d) 300 V



c) 12 V


A D



-h~~~(~~~


j cl 2OC)1 ~ It ~



Fe OCO~ N

c)~ d~( O ~ N

I




Q -11-- _-- C) w ~

V~~ W~4~ (1~lboCe) 1Ii1(10)111






r TEST CHARGE

8 ~)









plates















volts t= In~s e) (1) 12 J (3) 30 J (4) 24 JB





1++++++++++++++++pound++1

reg

I - ----I




r




+20 x 10-6 C +10 X 10-6 C






o 0 +10 X 10- C





pbull +10 x 10-6 C +10 x 10-6 C

o 0 ---- 40 x 10-1 m 1



1 ( 4 ) ( 1 ) (2 )







Assignment 8-21 Key

~ ~ CIO +-+-+-I-+shy

~ 1--+-+Ishy2 l-a

~ t-+-+I-I-+-+--I--+-+--I----l-----Ishy

f~ cOl++t+--+-+--H---+-I-I-I---+-+--t-Ishy




l~~TT~~~~~~~~~~

0 ~ wt-t--++-++--+-+-+-t-l-l--i--JH--l-I-L

t2 -+--I---I-J--I

~amprrrr+++~~---+-I-I-~

0 --+---+








E ~ 10 ~~_ ~ T~l5 ~ 01 ~1t1 - 0 ~CS~ t shy








8 88 Ou) e h~h ~e











E




Units for Potential

Difference

T-C














v~ 4= (501000~t) OIS l O S l ~ eJ (--~-----O~--~lI



f~cA I +1 el1l-lgt



HI ()r4-~ 2~ _ ~12 StJlV]











St a) 50 V b) 60 V d) 300 V



c) 12 V


A D



-h~~~(~~~


j cl 2OC)1 ~ It ~



Fe OCO~ N

c)~ d~( O ~ N

I




Q -11-- _-- C) w ~

V~~ W~4~ (1~lboCe) 1Ii1(10)111






r TEST CHARGE

8 ~)









plates















volts t= In~s e) (1) 12 J (3) 30 J (4) 24 JB





1++++++++++++++++pound++1

reg

I - ----I




r




+20 x 10-6 C +10 X 10-6 C






o 0 +10 X 10- C





pbull +10 x 10-6 C +10 x 10-6 C

o 0 ---- 40 x 10-1 m 1



1 ( 4 ) ( 1 ) (2 )







Assignment 8-21 Key








E ~ 10 ~~_ ~ T~l5 ~ 01 ~1t1 - 0 ~CS~ t shy








8 88 Ou) e h~h ~e











E




Units for Potential

Difference

T-C














v~ 4= (501000~t) OIS l O S l ~ eJ (--~-----O~--~lI



f~cA I +1 el1l-lgt



HI ()r4-~ 2~ _ ~12 StJlV]











St a) 50 V b) 60 V d) 300 V



c) 12 V


A D



-h~~~(~~~


j cl 2OC)1 ~ It ~



Fe OCO~ N

c)~ d~( O ~ N

I




Q -11-- _-- C) w ~

V~~ W~4~ (1~lboCe) 1Ii1(10)111






r TEST CHARGE

8 ~)









plates















volts t= In~s e) (1) 12 J (3) 30 J (4) 24 JB





1++++++++++++++++pound++1

reg

I - ----I




r




+20 x 10-6 C +10 X 10-6 C






o 0 +10 X 10- C





pbull +10 x 10-6 C +10 x 10-6 C

o 0 ---- 40 x 10-1 m 1



1 ( 4 ) ( 1 ) (2 )







Assignment 8-21 Key






8 88 Ou) e h~h ~e











E




Units for Potential

Difference

T-C














v~ 4= (501000~t) OIS l O S l ~ eJ (--~-----O~--~lI



f~cA I +1 el1l-lgt



HI ()r4-~ 2~ _ ~12 StJlV]











St a) 50 V b) 60 V d) 300 V



c) 12 V


A D



-h~~~(~~~


j cl 2OC)1 ~ It ~



Fe OCO~ N

c)~ d~( O ~ N

I




Q -11-- _-- C) w ~

V~~ W~4~ (1~lboCe) 1Ii1(10)111






r TEST CHARGE

8 ~)









plates















volts t= In~s e) (1) 12 J (3) 30 J (4) 24 JB





1++++++++++++++++pound++1

reg

I - ----I




r




+20 x 10-6 C +10 X 10-6 C






o 0 +10 X 10- C





pbull +10 x 10-6 C +10 x 10-6 C

o 0 ---- 40 x 10-1 m 1



1 ( 4 ) ( 1 ) (2 )







Assignment 8-21 Key


Units for Potential

Difference

T-C














v~ 4= (501000~t) OIS l O S l ~ eJ (--~-----O~--~lI



f~cA I +1 el1l-lgt



HI ()r4-~ 2~ _ ~12 StJlV]











St a) 50 V b) 60 V d) 300 V



c) 12 V


A D



-h~~~(~~~


j cl 2OC)1 ~ It ~



Fe OCO~ N

c)~ d~( O ~ N

I




Q -11-- _-- C) w ~

V~~ W~4~ (1~lboCe) 1Ii1(10)111






r TEST CHARGE

8 ~)









plates















volts t= In~s e) (1) 12 J (3) 30 J (4) 24 JB





1++++++++++++++++pound++1

reg

I - ----I




r




+20 x 10-6 C +10 X 10-6 C






o 0 +10 X 10- C





pbull +10 x 10-6 C +10 x 10-6 C

o 0 ---- 40 x 10-1 m 1



1 ( 4 ) ( 1 ) (2 )







Assignment 8-21 Key


v~ 4= (501000~t) OIS l O S l ~ eJ (--~-----O~--~lI



f~cA I +1 el1l-lgt



HI ()r4-~ 2~ _ ~12 StJlV]











St a) 50 V b) 60 V d) 300 V



c) 12 V


A D



-h~~~(~~~


j cl 2OC)1 ~ It ~



Fe OCO~ N

c)~ d~( O ~ N

I




Q -11-- _-- C) w ~

V~~ W~4~ (1~lboCe) 1Ii1(10)111






r TEST CHARGE

8 ~)









plates















volts t= In~s e) (1) 12 J (3) 30 J (4) 24 JB





1++++++++++++++++pound++1

reg

I - ----I




r




+20 x 10-6 C +10 X 10-6 C






o 0 +10 X 10- C





pbull +10 x 10-6 C +10 x 10-6 C

o 0 ---- 40 x 10-1 m 1



1 ( 4 ) ( 1 ) (2 )







Assignment 8-21 Key







St a) 50 V b) 60 V d) 300 V



c) 12 V


A D



-h~~~(~~~


j cl 2OC)1 ~ It ~



Fe OCO~ N

c)~ d~( O ~ N

I




Q -11-- _-- C) w ~

V~~ W~4~ (1~lboCe) 1Ii1(10)111






r TEST CHARGE

8 ~)









plates















volts t= In~s e) (1) 12 J (3) 30 J (4) 24 JB





1++++++++++++++++pound++1

reg

I - ----I




r




+20 x 10-6 C +10 X 10-6 C






o 0 +10 X 10- C





pbull +10 x 10-6 C +10 x 10-6 C

o 0 ---- 40 x 10-1 m 1



1 ( 4 ) ( 1 ) (2 )







Assignment 8-21 Key



Fe OCO~ N

c)~ d~( O ~ N

I




Q -11-- _-- C) w ~

V~~ W~4~ (1~lboCe) 1Ii1(10)111






r TEST CHARGE

8 ~)









plates















volts t= In~s e) (1) 12 J (3) 30 J (4) 24 JB





1++++++++++++++++pound++1

reg

I - ----I




r




+20 x 10-6 C +10 X 10-6 C






o 0 +10 X 10- C





pbull +10 x 10-6 C +10 x 10-6 C

o 0 ---- 40 x 10-1 m 1



1 ( 4 ) ( 1 ) (2 )







Assignment 8-21 Key




Q -11-- _-- C) w ~

V~~ W~4~ (1~lboCe) 1Ii1(10)111






r TEST CHARGE

8 ~)









plates















volts t= In~s e) (1) 12 J (3) 30 J (4) 24 JB





1++++++++++++++++pound++1

reg

I - ----I




r




+20 x 10-6 C +10 X 10-6 C






o 0 +10 X 10- C





pbull +10 x 10-6 C +10 x 10-6 C

o 0 ---- 40 x 10-1 m 1



1 ( 4 ) ( 1 ) (2 )







Assignment 8-21 Key






plates















volts t= In~s e) (1) 12 J (3) 30 J (4) 24 JB





1++++++++++++++++pound++1

reg

I - ----I




r




+20 x 10-6 C +10 X 10-6 C






o 0 +10 X 10- C





pbull +10 x 10-6 C +10 x 10-6 C

o 0 ---- 40 x 10-1 m 1



1 ( 4 ) ( 1 ) (2 )







Assignment 8-21 Key




volts t= In~s e) (1) 12 J (3) 30 J (4) 24 JB





1++++++++++++++++pound++1

reg

I - ----I




r




+20 x 10-6 C +10 X 10-6 C






o 0 +10 X 10- C





pbull +10 x 10-6 C +10 x 10-6 C

o 0 ---- 40 x 10-1 m 1



1 ( 4 ) ( 1 ) (2 )







Assignment 8-21 Key

r




+20 x 10-6 C +10 X 10-6 C






o 0 +10 X 10- C





pbull +10 x 10-6 C +10 x 10-6 C

o 0 ---- 40 x 10-1 m 1



1 ( 4 ) ( 1 ) (2 )







Assignment 8-21 Key




pbull +10 x 10-6 C +10 x 10-6 C

o 0 ---- 40 x 10-1 m 1



1 ( 4 ) ( 1 ) (2 )







Assignment 8-21 Key

Assignment #1: Where Do Charges Come From...c.) Polarization involves the use of a charged object to...

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Transcript of Assignment #1: Where Do Charges Come From...c.) Polarization involves the use of a charged object to...