<-- (This is the Word 1997-2004 version of the following outline. I would suggest using it instead.)

Otherwise...

ATTENTION: It has come to my attention that the wikispaces page refuses to display Greek letters, which are ESSENTIAL to this unit. My Word document on my computer displays them, but wikispaces must not know how to. So physics fans, I have a conversion chart for you:

p-->pi

eo--> epsilon naught (the permittivity of free space constant)

x--> Actually dot product. Don't ask why.

f-->closed integral sign.

Also, no superscripts and subscripts are shown. Please try to use your imagination. It should still be pretty straightforward. And every time there's a closed bracket symbol after a word in all caps, there's a footnote at the very bottom which provides definitions.

Outline:

Allow me to put this unit in very basic terms. If I were to sum up this unit in four words, it would be the following: Electricity is a force. If I had one more word, that statement would be: Electricity is a measurable force. And if I had three MORE words (scary, isn’t it?), it would be Electricity is a measureable force in many ways. Has your brain melted yet?

Okay, so we’re doing fine as of now. Now let’s begin to get into the real math and science behind this unit. You see, all objects possess CHARGE[[#_ftn1|[1]]] in the object itself. Just pick any object. Go ahead. I dare you. What’s that you say? You can’t think of anything. Hmph. Allow me to demonstrate… Imagine a plate. You know, like the ones you eat dinner on. Paper? Plastic? Porcelain? Doesn’t matter what it’s made out of. Just any plate, like the following:

You see? Now, let me tell you this awesome Physics fact: This plate has charge. But wait, there’s more! This plate has an even amount of charge distributed throughout. And just pick any material you can find. Machinery, turntables, vegetables, closets, even you and me, all have charge. Now, what makes electricity interesting are the kinds of charges. To be frank: there are two. They are: Positive + and Negative –

These are just names. It’s like naming charges “Kelsey” charges and “Gavin” charges. (Thank you, Mr. Wadness for that lovely example). Now, here’s another key concept to learn. This one is what makes electricity work. Opposite charges attract, like charges repel. Okay folks bear with me. A positive and a negative charge with attract each other, while a pair of positives and a pair of negatives will repel each other. That’s how it works in electricity. And so, every object has positive and negative charges. That’s a fact. But there’s one more state of charge to keep in mind. If an object has the same number of positive and negative charges scattered evenly throughout, that object is part of the third charge state, which is: Neutral +/- A neutral object doesn’t appear to feel an electric force, because the charges when distributed throughout aren’t strong enough to produce a measurable force. Remember the plate from before? Now let me illustrate what it looks like with charge in it:

The red pluses and the blue minuses each represent positive and negative charges respectfully. They’re scattered throughout, so they don’t have enough power to feel an electric attraction or repulsion, if say a positive point charge passed by it. And even if you had a strong positive charge next to the plate, it still wouldn’t move. That’s because the plate is ceramic, and ceramic is a NONCONDUCTOR[[#_ftn2|[2]]]. The plate can’t feel an attractive force, unless all the positives and negatives were rearranged to opposite sides. But what happens if the plate were made out of, say Copper? Well, Copper is a CONDUCTOR[[#_ftn3|[3]]], which makes electrons travel through the material very easily. So, if a positive test charge were right by it, the positives would shift away from the test charge (repulsion) and the negatives would shift towards it. The Copper would then be POLARIZED[[#_ftn4|[4]]] and feel electric force. Consider the following:

The copper plate on the left has no test charge near it. When a positive test charge is added near one end on the right, the negatives align to try and meet the positive charge. Now, imagine if instead of a plate, we had just two charges in open space. I’ll illustrate:

The two charges will attract each other, because they apply force on each other. It’s the same deal if the charges are the same, except that they’ll repel instead. So, big question: How can we measure this force? The answer: Coulomb already has. Charles Augustin de Coulomb studied charged objects in the late 1700’s. After making his experiments, he discovered the fundamental law to find the force on two charges. In 1783 he published his discovery of what became known as COULOMB’S LAW[[#_ftn5|[5]]] as a fundamental law to find the force between two charges, given that their charges and the distance between them is known. It looks like this: Felectric=Kq1q2 r2 Okay, so the force is stated F=kq1q2/r2. “k” is a constant. Don’t worry. It’s expressed as 8.98x109 Nm2/c2. The two q’s are the two charges involved in the equation are the two charges in question. “r” is the distance in meters between the two. (There’s a good reason why this looks like the universal gravity law). But now there’s the million-dollar question (estimated to be worth $1,024,956 adjusted for inflation) where does that force come from? So let me briefly (and I mean BRIEFLY) explain that it is caused by ELECTRIC FIELD[[#_ftn6|[6]]]. The electric field is what supplies force to all charged objects. All charged objects have one. It looks like this:

Those arrows represent potential force that will someday affect an electrically charged object. Once you know the electric force on an object, the field at that point is easy to calculate: E=F Q See? The electric field is equal to the force divided by the charge that’s causing it. The field is measured in units of Newtons per coulomb (N/C) or volts per meter (V/m). Extra credit for anybody who shows me a detailed unit conversion explaining why the two are equal by the end of next week. Do the conversion backwards and forwards N/C to V/m and V/m to N/C. They’re different.) ONE MORE IMPORTANT THING: The electric field is defined by which way a POSITIVE (+) charge would move in it. So, for the above picture, if the point charge were negative, the lines would be drawn into the charge, because that’s how a positive charge would move. So essentially, electric field provides force. So, vectorally, the electric field applies force and moves a charge. Add the resultant vectors, and you’ve calculated yourself one awesome force! (Isn’t physics just lovely?) Another equation ofr electric field (derived by adding coulomb’s law to the aforementioned one and doing calculus: E=q/4pe0r2 So we get it? Electric field can be solved for multiple ways, depending on what we know. Okay so let’s see what we’ve gone through. Charge? Check. Coulomb’s law? Check. Electric Field? Check-a-roo. But suppose we want to calculate the electric field for something that’s not as simple as a point charge? Hm? Who do we call for that? WAIT! I know who we call! Mr. Carl Friedrich Gauss! Recognize the name? Of course you do. He’s the discoverer of a little function we call GAUSS’ LAW[[#_ftn7|[7]]]. Gauss stated that the electric flux that passed through any surface was proportional to the charge inside it. Did I lose you? Nuts. Let me explain. When we drew the electric field earlier, remember how all those lines surrounded the charge? Well those lines are actually called “electric flux”, and they have a value. In Gauss’ law, the flux is defined by the amount of charge that’s enclosed in a certain area. That law can be expressed mathematically: fE×dA=Qenclosed/e0 (I know that’s not actually the sign for “closed integral” and is instead the Greek lowercase letter “Iota”. Just keep it on the DL and maybe Mr. Wadness won’t notice.) Okay, so this is Gauss’ law. It’s all pretty straightforward, except for the dA, which is just the surface area of what you’re measuring. (The area that the flux passes through. We’ll learn why this is important later). So say for a sphere it would be 4pr2. For a cylinder it would be 2prl (we’ll learn why we use half the formula for a cylinder later). So, let’s say we have a metal conducting sphere. Like this picture:

The sphere has a positive charge of, let’s say +Q. Because it’s a sphere, we want to plug in the correct surface area formula, which is 4pr2. The yellow circle on the inside is r1. As you can see, because in a conductor, the charge is located on the edge, r1 encloses no charge. Q is zero, and consequently, E is zero. But the pink circle, which is r2 fully encloses the sphere, contains the entire charge for the sphere. Therefore Qenc=+Q. dA, is 4pr2 as mentioned earlier, and everything else stays the same. So our equation is (drumroll please) E(4pr2)=+Q/e0 Or E=+Q/4pr2e0 It’s so simple you can do it at home! (If you haven’t already you most likely failed physics for the third quarter.) Now, in a cylinder things get tricky. If a cylinder has charge, we want to use dA, right? So we plug in the surface are formula for a cylinder, right? That formula is 2prl+2(pr2). But it’s tricky. Get this: we only want to include the parts of the formula that the flux goes through. Okay, so in a cylinder the charge resides in the middle, producing only flux that goes through the CURVED part of the cylinder. No flux goes through the ends! Therefore the calculation for the surface area of the ends, which is 2(pr2), can go. We are left with 2prl for the dA of a cylinder. The rest is algebra. For a flat sheet of charge, dA would be just length times width. For a thick sheet of charge dA would be 2(length times width), because the flux only goes through two sides. Get it? I hope so. What happens when a nonconductor is used for Gauss’ law? Keep in mind, the charges don’t flow to the end like they would in a conductor. Given the charge density, and the amount of area enclosed, you should be able to find the amount of charge enclosed. Just multiply the two values together, and you’ll get the total charge enclosed by whatever radius is made. Is that it? I think that’s it. Wow, I am long-winded. So congratulations, you now know about electric force and fields! Wait, you didn’t just skip to the last paragraph, did you? DID YOU?! Wow, that’s cheap man, that’s real cheap. Do you read the last page of a mystery book first, too? Wow, I hope you have a crushing feeling of guilt now. I mean, think about all the wonderful PHYSICS[[#_ftn8|[8]]] you could have learned by reading through. You could have learned all about Coulomb’s law, charges, the electric field, electric forces, and Gaussian surfaces, but no, Mr. Hotshot has to skip to the end I order to feign accomplishment to add excitement to his drab wretched life! Well, listen here Mr. Hotshot, I have better things to do than waste my time with inconsiderate people who have no regard for what person has to offer intellectually, especially when its as valuable as this! God, I hope you’re crushed with shame now. For those of you who did read through, good job. It’s not easy to get at first, but you did really well. Let’s go on to see what other physics marvels await us!

[[#_ftnref|[1]]] CHARGE: A property that exists in all normal matter that allows it to experience an electrical force.

[[#_ftnref|[2]]] NONCONDUCTOR: Any material that does not allow the free passage of charge.

[[#_ftnref|[3]]] CONDUCTOTOR: Any material that allows for the passage of electrons (charge). Opposite of nonconductor.

[[#_ftnref|[4]]] POLARIZATION: Formation of opposite sections of charge in an object.

[[#_ftnref|[5]]] COULOMB’S LAW: Formula for finding force between two charges.

[[#_ftnref|[6]]] ELECTRIC FIELD: An invisible field that surrounds charged objects and provides electric force to other charges nearby.

[[#_ftnref|[7]]] GAUSS’ LAW: For determining the electric field of an object that isn’t as simple as a point-charge.

1. The positive charge has a charge of 5 C. At this instant, the forces are .2 m apart. The positive charge feels a force of -2.25x1012 N. The charge on the negative charge is most nearly a. .5 C b. 1 C c. 2 C d. 4 C e. 5 C 2. The acceleration for each charge if it has a mass, m can be expressed by which setup? a. Kq1/r2 b. (Kq1/r2)/m c. Kq1q2/r2 d. Kq1q2m/r2 e. (Kq1q2/r2)/m

3. A solid conducting sphere is placed inside a conducting spherical shell. The solid sphere is given a charge +Q. The shell is given a charge –Q. Where would the strongest electric field be located? a. R=0 b. 0<R<R1 c. R=R1 d. R1<R<R2 e. R>R2 4. The electric field of a conducting sphere of charge Q can be determined by which of the following setups? a. Q/2prle0 b. Q/4prle0 c. Q/4pr2e0 d. 4Q/3pre0 e. 4Q/3prle0

5. Two positive and two negative charges are aligned in a square formation, with rides length r. Which is most likely going to happen to the charge on the bottom right hand corner? (Assume that the diagonal distances are too far to produce a measureable force. a. It will accelerate diagonally upwards and to the left. b. It will accelerate towards the charge to the right. c. It will accelerate towards the charge directly upwards. d. It will not accelerate, because the vectors of the electric fields cancel out. e. This answer cannot be determined without knowing the values of the charges.

INVESTIGATION: Here's how you can use charges to see how electrical force works. Rub a plastic rod with a piece of cloth repeatedly. This moves the electrons around in the rod, and leaves it with a negative charge. Place this rod near a small, hanging conductor. Preferably something like a small ball of aluminum foil, no bigger than a marble. Bring the rod close to it, but don't touch the two together. With your finger, tap the ball. What happens? Does the ball move? Does it appear to have any attraction to the rod? INTENDED RESULTS: The ball should move. You see, when the rod (let's say it has negative charge) moves close to the ball, the positive and negative charges in the ball align to try and meet each other. Because there's no contact, they shouldn't,, but the positive charges will align along the edge closest to the rod, and the negative charges will move as far away as possible. That's why, when the ball is tapped it moves with the rod's motion. By tapping the ball, you give the like negative charges room to get further away from negative charges, which is what they will do. The ball remains positively charged, and attracted by the rod's negative charge. The ball should swing in the direction the rod is in.

Open Response Question:

In the above diagram, a wire (which may be thought of as a cylinder) has actually two wires inside it. The first one is hollow and rests above a layer of insulation. The second one is solid and rests below the insulation layer. The inner wire carries a current of +I0 while the outer wire carries a charge of –I0. Both wires are conductors. R1=.02 m, R2=.05 m, R3=.04 m and the wire is 10 m long.

(a) What is the electric field at R1 for five seconds if the current going through there is .01 A? (b) Suppose the outer wire has a current density of 4 C/m2. If the inner wire has a charge of .5 C pass through it in one instant, what is the electric field when r=.045 m? (c) Given that the amount of charge passing through each wire is equal but opposite, what must the electric field be at R2? (d) Based on your last answer, what is the electric field at any point outside the wire? (e) The current is then turned off and the wire is allowed to settle. The wire this then given a charge of 4 C. What happens to the neutral outer wire? Please draw a diagram of what happens. What can be determined about the electric field between R1 and R3, between R3 and R2 and outside R2?

Thank you so much for using this outline! This has been a ZecRabbit Project. I hope you enjoyed it.

## Otherwise...

## ATTENTION: It has come to my attention that the wikispaces page refuses to display Greek letters, which are ESSENTIAL to this unit. My Word document on my computer displays them, but wikispaces must not know how to. So physics fans, I have a conversion chart for you:

## p-->pi

## eo--> epsilon naught (the permittivity of free space constant)

## x--> Actually dot product. Don't ask why.

## f-->closed integral sign.

## Also, no superscripts and subscripts are shown. Please try to use your imagination. It should still be pretty straightforward. And every time there's a closed bracket symbol after a word in all caps, there's a footnote at the very bottom which provides definitions.

Outline:Allow me to put this unit in very basic terms. If I were to sum up this unit in four words, it would be the following:

Electricity is a force.If I had one more word, that statement would be:Electricity is a measurable force.And if I had three MORE words (scary, isn’t it?), it would beElectricity is a measureable force in many ways.Has your brain melted yet?Okay, so we’re doing fine as of now. Now let’s begin to get into the real math and science behind this unit. You see, all objects possess CHARGE[[#_ftn1|[1]]] in the object itself. Just pick any object. Go ahead. I dare you. What’s that you say? You can’t think of anything. Hmph. Allow me to demonstrate… Imagine a plate. You know, like the ones you eat dinner on. Paper? Plastic? Porcelain? Doesn’t matter what it’s made out of. Just any plate, like the following:

You see? Now, let me tell you this awesome Physics fact:

This plate has charge.But wait, there’s more!This plate has an even amount of charge distributed throughout.And just pick any material you can find. Machinery, turntables, vegetables, closets, even you and me, all have charge. Now, what makes electricity interesting are the kinds of charges. To be frank: there are two. They are:Positive + and Negative –

These are just names. It’s like naming charges “Kelsey” charges and “Gavin” charges. (Thank you, Mr. Wadness for that lovely example).

Now, here’s another key concept to learn. This one is what makes electricity work.

Opposite charges attract, like charges repel.Okay folks bear with me. A positive and a negative charge with attract each other, while a pair of positives and a pair of negatives will repel each other. That’s how it works in electricity. And so, every object has positive and negative charges. That’s a fact. But there’s one more state of charge to keep in mind. If an object has the same number of positive and negative charges scattered evenly throughout, that object is part of the third charge state, which is:Neutral +/-

A neutral object doesn’t appear to feel an electric force, because the charges when distributed throughout aren’t strong enough to produce a measurable force. Remember the plate from before? Now let me illustrate what it looks like with charge in it:

The red pluses and the blue minuses each represent positive and negative charges respectfully. They’re scattered throughout, so they don’t have enough power to feel an electric attraction or repulsion, if say a positive point charge passed by it. And even if you had a strong positive charge next to the plate, it still wouldn’t move. That’s because the plate is ceramic, and ceramic is a NONCONDUCTOR[[#_ftn2|[2]]]. The plate can’t feel an attractive force, unless all the positives and negatives were rearranged to opposite sides.

But what happens if the plate were made out of, say Copper? Well, Copper is a CONDUCTOR[[#_ftn3|[3]]], which makes electrons travel through the material very easily. So, if a positive test charge were right by it, the positives would shift away from the test charge (repulsion) and the negatives would shift towards it. The Copper would then be POLARIZED[[#_ftn4|[4]]] and feel electric force. Consider the following:

The copper plate on the left has no test charge near it. When a positive test charge is added near one end on the right, the negatives align to try and meet the positive charge. Now, imagine if instead of a plate, we had just two charges in open space. I’ll illustrate:

The two charges will attract each other, because they apply force on each other. It’s the same deal if the charges are the same, except that they’ll repel instead. So, big question: How can we measure this force?

The answer: Coulomb already has. Charles Augustin de Coulomb studied charged objects in the late 1700’s. After making his experiments, he discovered the fundamental law to find the force on two charges. In 1783 he published his discovery of what became known as COULOMB’S LAW[[#_ftn5|[5]]] as a fundamental law to find the force between two charges, given that their charges and the distance between them is known. It looks like this:

Felectric=Kq1q2r2Okay, so the force is stated F=kq1q2/r2. “k” is a constant. Don’t worry. It’s expressed as 8.98x109 Nm2/c2. The two q’s are the two charges involved in the equation are the two charges in question. “r” is the distance in meters between the two. (There’s a good reason why this looks like the universal gravity law).

But now there’s the million-dollar question (estimated to be worth $1,024,956 adjusted for inflation) where does that force come from? So let me briefly (and I mean BRIEFLY) explain that it is caused by ELECTRIC FIELD[[#_ftn6|[6]]]. The electric field is what supplies force to all charged objects. All charged objects have one. It looks like this:

Those arrows represent potential force that will someday affect an electrically charged object. Once you know the electric force on an object, the field at that point is easy to calculate:

E=FQSee? The electric field is equal to the force divided by the charge that’s causing it. The field is measured in units of Newtons per coulomb (N/C) or volts per meter (V/m). Extra credit for anybody who shows me a detailed unit conversion explaining why the two are equal by the end of next week. Do the conversion backwards and forwards N/C to V/m and V/m to N/C. They’re different.)

ONE MORE IMPORTANT THING: The electric field is defined by which way a POSITIVE (+) charge would move in it. So, for the above picture, if the point charge were negative, the lines would be drawn into the charge, because that’s how a positive charge would move.

So essentially, electric field provides force. So, vectorally, the electric field applies force and moves a charge. Add the resultant vectors, and you’ve calculated yourself one awesome force!

(Isn’t physics just lovely?)

Another equation ofr electric field (derived by adding coulomb’s law to the aforementioned one and doing calculus:

E=q/4pe0r2

So we get it? Electric field can be solved for multiple ways, depending on what we know.

Okay so let’s see what we’ve gone through. Charge? Check. Coulomb’s law? Check. Electric Field? Check-a-roo. But suppose we want to calculate the electric field for something that’s not as simple as a point charge? Hm? Who do we call for that? WAIT! I know who we call! Mr. Carl Friedrich Gauss! Recognize the name? Of course you do. He’s the discoverer of a little function we call GAUSS’ LAW[[#_ftn7|[7]]]. Gauss stated that the

electric flux that passed through any surface was proportional to the charge inside it.Did I lose you? Nuts. Let me explain. When we drew the electric field earlier, remember how all those lines surrounded the charge? Well those lines are actually called “electric flux”, and they have a value. In Gauss’ law, the flux is defined by the amount of charge that’s enclosed in a certain area. That law can be expressed mathematically:

fE×dA=Qenclosed/e0(I know that’s not actually the sign for “closed integral” and is instead the Greek lowercase letter “Iota”. Just keep it on the DL and maybe Mr. Wadness won’t notice.) Okay, so this is Gauss’ law. It’s all pretty straightforward, except for the dA, which is just the surface area of what you’re measuring. (The area that the flux passes through. We’ll learn why this is important later). So say for a sphere it would be 4pr2. For a cylinder it would be 2prl (we’ll learn why we use half the formula for a cylinder later).

So, let’s say we have a metal conducting sphere. Like this picture:

The sphere has a positive charge of, let’s say +Q. Because it’s a sphere, we want to plug in the correct surface area formula, which is 4pr2. The yellow circle on the inside is r1. As you can see, because in a conductor, the charge is located on the edge, r1 encloses no charge. Q is zero, and consequently, E is zero. But the pink circle, which is r2 fully encloses the sphere, contains the entire charge for the sphere. Therefore Qenc=+Q. dA, is 4pr2 as mentioned earlier, and everything else stays the same. So our equation is (drumroll please)

E(4pr2)=+Q/e0

Or

E=+Q/4pr2e0

It’s so simple you can do it at home! (If you haven’t already you most likely failed physics for the third quarter.)

Now, in a cylinder things get tricky. If a cylinder has charge, we want to use dA, right? So we plug in the surface are formula for a cylinder, right? That formula is 2prl+2(pr2). But it’s tricky. Get this: we only want to include the parts of the formula that the flux goes through. Okay, so in a cylinder the charge resides in the middle, producing only flux that goes through the CURVED part of the cylinder. No flux goes through the ends! Therefore the calculation for the surface area of the ends, which is 2(pr2), can go. We are left with 2prl for the dA of a cylinder. The rest is algebra. For a flat sheet of charge, dA would be just length times width. For a thick sheet of charge dA would be 2(length times width), because the flux only goes through two sides. Get it? I hope so.

What happens when a nonconductor is used for Gauss’ law? Keep in mind, the charges don’t flow to the end like they would in a conductor. Given the charge density, and the amount of area enclosed, you should be able to find the amount of charge enclosed. Just multiply the two values together, and you’ll get the total charge enclosed by whatever radius is made.

Is that it? I think that’s it. Wow, I am long-winded. So congratulations, you now know about electric force and fields! Wait, you didn’t just skip to the last paragraph, did you? DID YOU?! Wow, that’s cheap man, that’s real cheap. Do you read the last page of a mystery book first, too? Wow, I hope you have a crushing feeling of guilt now. I mean, think about all the wonderful PHYSICS[[#_ftn8|[8]]] you could have learned by reading through. You could have learned all about Coulomb’s law, charges, the electric field, electric forces, and Gaussian surfaces, but no, Mr. Hotshot has to skip to the end I order to feign accomplishment to add excitement to his drab wretched life! Well, listen here Mr. Hotshot, I have better things to do than waste my time with inconsiderate people who have no regard for what person has to offer intellectually, especially when its as valuable as this! God, I hope you’re crushed with shame now. For those of you who did read through, good job. It’s not easy to get at first, but you did really well. Let’s go on to see what other physics marvels await us!

[[#_ftnref|[1]]] CHARGE: A property that exists in all normal matter that allows it to experience an electrical force.

[[#_ftnref|[2]]] NONCONDUCTOR: Any material that does not allow the free passage of charge.

[[#_ftnref|[3]]] CONDUCTOTOR: Any material that allows for the passage of electrons (charge). Opposite of nonconductor.

[[#_ftnref|[4]]] POLARIZATION: Formation of opposite sections of charge in an object.

[[#_ftnref|[5]]] COULOMB’S LAW: Formula for finding force between two charges.

[[#_ftnref|[6]]] ELECTRIC FIELD: An invisible field that surrounds charged objects and provides electric force to other charges nearby.

[[#_ftnref|[7]]] GAUSS’ LAW: For determining the electric field of an object that isn’t as simple as a point-charge.

[[#_ftnref|[8]]] PHYSICS: Sheer bliss, believe me.

_

Five multiple choice questions:

Questions 1 and 2 refer to the following diagram:

1. The positive charge has a charge of 5 C. At this instant, the forces are .2 m apart. The positive charge feels a force of -2.25x1012 N. The charge on the negative charge is most nearly

a. .5 C

b. 1 C

c. 2 C

d. 4 C

e. 5 C

2. The acceleration for each charge if it has a mass, m can be expressed by which setup?

a. Kq1/r2

b. (Kq1/r2)/m

c. Kq1q2/r2

d. Kq1q2m/r2

e. (Kq1q2/r2)/m

3. A solid conducting sphere is placed inside a conducting spherical shell. The solid sphere is given a charge +Q. The shell is given a charge –Q. Where would the strongest electric field be located?

a. R=0

b. 0<R<R1

c. R=R1

d. R1<R<R2

e. R>R2

4. The electric field of a conducting sphere of charge Q can be determined by which of the following setups?

a. Q/2prle0

b. Q/4prle0

c. Q/4pr2e0

d. 4Q/3pre0

e. 4Q/3prle0

5. Two positive and two negative charges are aligned in a square formation, with rides length r. Which is most likely going to happen to the charge on the bottom right hand corner? (Assume that the diagonal distances are too far to produce a measureable force.

a. It will accelerate diagonally upwards and to the left.

b. It will accelerate towards the charge to the right.

c. It will accelerate towards the charge directly upwards.

d. It will not accelerate, because the vectors of the electric fields cancel out.

e. This answer cannot be determined without knowing the values of the charges.

INVESTIGATION: Here's how you can use charges to see how electrical force works. Rub a plastic rod with a piece of cloth repeatedly. This moves the electrons around in the rod, and leaves it with a negative charge. Place this rod near a small, hanging conductor. Preferably something like a small ball of aluminum foil, no bigger than a marble. Bring the rod close to it, but don't touch the two together. With your finger, tap the ball. What happens? Does the ball move? Does it appear to have any attraction to the rod?

INTENDED RESULTS: The ball should move. You see, when the rod (let's say it has negative charge) moves close to the ball, the positive and negative charges in the ball align to try and meet each other. Because there's no contact, they shouldn't,, but the positive charges will align along the edge closest to the rod, and the negative charges will move as far away as possible. That's why, when the ball is tapped it moves with the rod's motion. By tapping the ball, you give the like negative charges room to get further away from negative charges, which is what they will do. The ball remains positively charged, and attracted by the rod's negative charge. The ball should swing in the direction the rod is in.

Open Response Question:

In the above diagram, a wire (which may be thought of as a cylinder) has actually two wires inside it. The first one is hollow and rests above a layer of insulation. The second one is solid and rests below the insulation layer. The inner wire carries a current of +I0 while the outer wire carries a charge of –I0. Both wires are conductors. R1=.02 m, R2=.05 m, R3=.04 m and the wire is 10 m long.

(a) What is the electric field at R1 for five seconds if the current going through there is .01 A?

(b) Suppose the outer wire has a current density of 4 C/m2. If the inner wire has a charge of .5 C pass through it in one instant, what is the electric field when r=.045 m?

(c) Given that the amount of charge passing through each wire is equal but opposite, what must the electric field be at R2?

(d) Based on your last answer, what is the electric field at any point outside the wire?

(e) The current is then turned off and the wire is allowed to settle. The wire this then given a charge of 4 C. What happens to the neutral outer wire? Please draw a diagram of what happens. What can be determined about the electric field between R1 and R3, between R3 and R2 and outside R2?

## Thank you so much for using this outline! This has been a ZecRabbit Project. I hope you enjoyed it.