Electric Charge and Atomic Structure Return to Table of Contents - - PDF document

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AP Physics C - E & M

Electric Charge & Field

2015-12-29 www.njctl.org

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Electric Charge and Field

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· Electric Charge and Atomic Structure · Conductors and Insulators · Coulomb's Law · Electric Field of Point Charges · Electric Field of Continuous Charge Distributions · Electric Dipole

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Electric Charge and Atomic Structure

Slide 4 / 95 Electric Charge

When you take two non metallic objects, such as a plastic ruler and animal fur and rub them together, you get an interesting

• effect. Before they are rubbed, the plastic ruler is held over bits
• f paper and nothing happens.

After the rubbing, the plastic ruler is held over the bits of paper and they are accelerated towards the ruler. There must be a force that was created by the rubbing of the ruler and fur together. In ancient Greece, people noticed that when thread was spun

• ver an amber spindle, the thread was attracted to the spindle.

The Greek word for amber was "elektron," hence this force was called electric.

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Further experimentation showed that dissimilar materials would attract each other after rubbing, while similar materials would repel each other. These effects would not happen before the contact, and would fade after a period of time. This led to the proposal that something was being exchanged between the materials - later named "charge." Because there was an attractive and a repulsive force, it was further proposed that the charge came in two types.

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Electric Charge

In the 18th century, Benjamin Franklin named the two types of charge when he observed that when a rubber rod was rubbed by animal fur, the objects attracted each other. Benjamin named the charge on the rod, negative, and the charge on the fur, positive. If he named them oppositely, we wouldn't have to deal with the difference between actual current flow and the defined "conventional current." More on this in the Current and Circuits unit. Like charges repel. Opposite charges attract.

Slide 7 / 95 Conservation of Charge

No new charge was created when dissimilar objects are rubbed against each other - it is just separated. The positive charge acquired by one object is exactly equal in magnitude and opposite in sign to the charge lost by the other object. Electric charge is conserved - the total amount of electric charge in a closed system is neither created nor destroyed - just like energy, linear momentum and angular momentum.

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1 A plastic rod is rubbed with a piece of wool. During the process the plastic rod acquires a positive charge and the wool: A acquires an equal positive charge. B acquires an equal negative charge. C acquires a smaller positive charge. D acquires a smaller negative charge. E remains neutral.

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1 A plastic rod is rubbed with a piece of wool. During the process the plastic rod acquires a positive charge and the wool: A acquires an equal positive charge. B acquires an equal negative charge. C acquires a smaller positive charge. D acquires a smaller negative charge. E remains neutral.

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Answer B

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2 Two pith spheres covered with conducting paint are hanging from two insulating threads. When the spheres are brought close to each other, they attract each other. What type of charge is on the spheres? After they touch, will they separate or cling together? Discuss all possibilities.

Students type their answers here

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2 Two pith spheres covered with conducting paint are hanging from two insulating threads. When the spheres are brought close to each other, they attract each other. What type of charge is on the spheres? After they touch, will they separate or cling together? Discuss all possibilities.

Students type their answers here

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Answer Since the spheres attract each other, they have opposite charges. If the spheres have equal amounts of charge, they will neutralize after touching and hang from the threads

• vertically. If one sphere has a larger

amount of charge, they will share the charge after touching (same charge

• n each) and repel each other.

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Atomic Structure

When we rub two objects together, why does one attain a positive charge and the other one a negative charge? The answer lies in the structure of the atom. As you already know, an atom is comprised of electrons, protons, and neutrons. What does it look like?

Slide 11 / 95 Atomic Structure

Here's a look at a Helium atom. The nucleus is buried deep within the atom and is 1,000,000 times smaller than the atom. The two protons and two neutrons are shown in red and purple - the width of the nucleus is 1x10-7 nm. The diagram shows a magnified view of the nucleus. What is the significance of the dark circle surrounded by the lighter shades of gray?

http://commons.wikimedia.org/wiki/File:Helium_atom_QM_rev1.svg

Slide 12 / 95 Atomic Structure

You know that Helium has two electrons - yet they're not shown

• n this picture.

That's because we don't know exactly where those electrons

• are. We only know a probability
• f where they might be.

The darker the shade means that it is more probable that the electrons are found within that shape. For more information, refer to the Quantum Physics and Atomic Models chapter of the AP Physics 2 class.

http://commons.wikimedia.org/wiki/File:Helium_atom_QM_rev1.svg

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Atomic Structure

http://commons.wikimedia.org/wiki/File:Helium_atom_QM_rev1.svg

It has been shown that electric charges move between objects. Because they are more lightly bound to the atom than the protons and the neutrons, the electrons are the particles that move between atoms and carry this charge. The electrons are fundamental particles - physicists have not found any underlying structure - yet.

Slide 14 / 95 The Electron

We'll focus on the electron in this chapter as it is the source

• f most of the physical phenomena in the Electricity and

Magnetism course. J.J.Thomson discovered the electron in 1897, and concluded it was part of the atom - the first evidence that the atom had an underlying structure. In a series of experiments between 1909 and 1913, Robert Millikan and his graduate student, Harvey Fletcher, established the value of the charge, "e," on an electron.

Slide 15 / 95 Measurement of Charge

Millikan and Fletcher's work and subsequent experiments have established the value of "e" as 1.602 x 10-19 Coulombs. The charge on an electron is -e and the charge on a proton is +e. It has also been demonstrated that this is the smallest value of charge (with the exception of quarks which were covered in the AP Physics 2 course) and all larger charges are an integral multiple of this number. Later, the electron's mass was found to be 9.1 x 10

• 31 kg.

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3 Which of these could be the charge on an object? A 0.80 x 10-19 C B 2.00 x 10-19 C C 3.2 x 10-19 C D 4.0 x 10-19 C E All of the above

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3 Which of these could be the charge on an object? A 0.80 x 10-19 C B 2.00 x 10-19 C C 3.2 x 10-19 C D 4.0 x 10-19 C E All of the above

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Answer C

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4 Which of these could be the charge on an object? A 0.40 mC B 4.8 mC C 5.0 C D 6.5 C E All of the above

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4 Which of these could be the charge on an object? A 0.40 mC B 4.8 mC C 5.0 C D 6.5 C E All of the above

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Answer B

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Conductors and Insulators

Slide 19 / 95 Conductors

A conductor is an element or material which allows the free flow

• f charge.

If a metal sphere is given a charge Q, that charge will be uniformly distributed over the surface of the sphere. This is because the charges repel one another and the material will not hinder the charge movement so they are able to move all the way to the surface.

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Insulator

A insulator is an element or material which prevents charge from freely flowing. If a charge Q is uniformly distributed throughout the spherical insulator, the charge will remain where it is because even though the charges repel one another, the insulator prevents them from moving very far.

Slide 21 / 95 Semi-Conductor

Semi-Conductors have properties of both conductors and

• insulators. These properties can be altered by adding small

amounts of other elements. They are crucial to the operation of transistors which are then used in cell phones, computers, recording equipment, music playback devices and many other applications. Silicon and Germanium are semi-conductors.

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5 A conducting sphere is charged with a negative charge

• Q. Which statement about the charge distribution is

correct? A Charge in concentrated at the center of the sphere. B Charge is concentrated at the bottom part of the sphere. C Charge is evenly distributed throughout the entire volume of the sphere. D Charge is evenly distributed on the surface of the sphere. E More information is required.

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5 A conducting sphere is charged with a negative charge

• Q. Which statement about the charge distribution is

correct? A Charge in concentrated at the center of the sphere. B Charge is concentrated at the bottom part of the sphere. C Charge is evenly distributed throughout the entire volume of the sphere. D Charge is evenly distributed on the surface of the sphere. E More information is required.

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Answer D

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6 An insulated sphere is charged uniformly with a positive charge +Q. Which statement about the charge distribution is correct? A Charge in concentrated at the center of the sphere. B Charge is concentrated at the bottom part of the sphere. C Charge is evenly distributed throughout the entire volume of the sphere. D Charge is evenly distributed on the surface of the sphere. E More information is required.

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6 An insulated sphere is charged uniformly with a positive charge +Q. Which statement about the charge distribution is correct? A Charge in concentrated at the center of the sphere. B Charge is concentrated at the bottom part of the sphere. C Charge is evenly distributed throughout the entire volume of the sphere. D Charge is evenly distributed on the surface of the sphere. E More information is required.

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Answer C

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7 A positively charged sphere is brought near end A of the uncharged metal bar shown below. Ends A and B of the metal bar will be charged: A positive, negative B negative, positive C positive, positive D negative, negative E neutral, neutral

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7 A positively charged sphere is brought near end A of the uncharged metal bar shown below. Ends A and B of the metal bar will be charged: A positive, negative B negative, positive C positive, positive D negative, negative E neutral, neutral

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Answer B

Slide 25 (Answer) / 95 Conduction

Conduction is the process of charging an object through contact, and the charge will flow until the electric potential of both objects is equal. Electric potential was defined in AP Physics 2, and will be covered in this course a little later. The net charge on the charging rod decreases, and the net charge on the sphere increases. The total charge remains the same.

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If a negatively charged rod is brought close to a neutral metal sphere, the rod will induce a charge on the sphere's surface. The negative charges will move away from the rod, leaving a positive charge near the rod. If the sphere is then grounded, the negative charges will be forced through the wire into the earth. The ground wire is then removed and the rod is taken

• away. The sphere now has

an induced positive charge that spreads evenly over the

• surface. This order is critical.

If the ground wire is removed first, the negative charges will return to the sphere, resulting in a neutral charge.

Induction

Induction is the process of charging a metallic object without

• contact. The charge on the rod remains the same. The increase in

charge of the rod/sphere system comes from the ground.

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Gold Leaves Conductor

• +

+ + + +

• +
• The Electroscope

The electroscope measures electrical charge (both sign and magnitude). The conductor rod is insulated from the glass container. When the scope is neutral, the leaves hang down to due to their own mass. Electroscopes can be charged by conduction or induction.

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8 A neutral electroscope is touched by a positively charged rod and then the rod is removed. Then a negatively charged rod is brought close to the electroscope. What happens to the leaves of the electroscope? A Move apart, then move farther apart. B Move apart, then move closer together. C Move apart, then remain the same distance apart. D Do not move, then move apart. E More information is required.

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8 A neutral electroscope is touched by a positively charged rod and then the rod is removed. Then a negatively charged rod is brought close to the electroscope. What happens to the leaves of the electroscope? A Move apart, then move farther apart. B Move apart, then move closer together. C Move apart, then remain the same distance apart. D Do not move, then move apart. E More information is required.

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Answer B

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9 A neutral electroscope is grounded with a wire. A positively charged wire is brought close to the electroscope; the wire is cut first and the rod is removed. A negatively charged rod is then brought near. What happens to the leaves of the electroscope? A The leaves don't move, then move farther apart. B The leaves separate, then remain the same. C The leaves separate, then move closer together. D The leaves separate, then move farther apart. E More information is required.

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9 A neutral electroscope is grounded with a wire. A positively charged wire is brought close to the electroscope; the wire is cut first and the rod is removed. A negatively charged rod is then brought near. What happens to the leaves of the electroscope? A The leaves don't move, then move farther apart. B The leaves separate, then remain the same. C The leaves separate, then move closer together. D The leaves separate, then move farther apart. E More information is required.

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Answer D

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10 A physics student wants to determine the polarity of an initially charged electroscope in the lab. She has two

• ppositely charged rods and brings each one separately

near the electroscope. The positive rod causes the leaves to move farther apart and the negative rod causes the leaves to move closer togethere. What type of charge is on the electroscope? A Positive. B Negative. C Neutral. D Cannot be determined because the electroscope wasn't grounded. E Cannot be determined because the rods were not neutral.

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10 A physics student wants to determine the polarity of an initially charged electroscope in the lab. She has two

• ppositely charged rods and brings each one separately

near the electroscope. The positive rod causes the leaves to move farther apart and the negative rod causes the leaves to move closer togethere. What type of charge is on the electroscope? A Positive. B Negative. C Neutral. D Cannot be determined because the electroscope wasn't grounded. E Cannot be determined because the rods were not neutral.

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Answer A

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Coulomb's Law

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Coulomb's Law

The magnitude of the force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. The direction of the force is along a line connecting the

• charges. Like charges repel. Opposite charges attract.

The proportionality constant, ke was determined using almost the same apparatus (torsion balance) that Henry Cavendish used to determine the Gravitational constant, G.

Slide 33 / 95 Coulomb's Law

This is the vector version of Coulomb's Law, replacing the Coulomb constant κe with ε0: ke is related to the permeability of free space, ε0, which is used in Maxwell's equations, by: is a unit vector on the line connecting q1 to q2.

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11 Two charged objects with an equal charge of Q are separated by a distance r and attract each other with a certain force. If the charges on both objects are doubled and the separation is halved, the force between them will be: A 4 times greater B 2 times greater C 4 times less D 16 times greater E 16 times less

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11 Two charged objects with an equal charge of Q are separated by a distance r and attract each other with a certain force. If the charges on both objects are doubled and the separation is halved, the force between them will be: A 4 times greater B 2 times greater C 4 times less D 16 times greater E 16 times less

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Answer D

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12 Two identical conducting spheres are charged to +Q and -3Q and separated by a distance r. The attractive force between the spheres is F. The two spheres are brought in brief contact and then moved to their

• riginal positions. If the new electrostatic force between

the spheres is F', which of the following is true? A F = F' B F' = 3F C F' = F/3 D F' = 9F E F' = F/9

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12 Two identical conducting spheres are charged to +Q and -3Q and separated by a distance r. The attractive force between the spheres is F. The two spheres are brought in brief contact and then moved to their

• riginal positions. If the new electrostatic force between

the spheres is F', which of the following is true? A F = F' B F' = 3F C F' = F/3 D F' = 9F E F' = F/9

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Answer C

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13 What is the magnitude of the net force on charge C due to charges A and B? A B C D E

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13 What is the magnitude of the net force on charge C due to charges A and B? A B C D E

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Answer E

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Electric Field of Point Charges

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Slide 39 / 95 Slide 40 / 95 Electric Field

The electric field is defined by assuming a small positive test charge, q, is near a large charge, Q, and this large charge is generating a field. The test charge is small, as we don't want its field involved in the

• calculation. We then immediately divide it out:

Electric Field due to a point charge

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Electric Field

Electric Field lines radiate inward for a negative charge Electric Field lines radiate outward from a positive charge

The Electric Field at any point in space describes the force that a positive charge at that point will feel - in magnitude and direction. The Electric Field is represented by lines with a convention that the more lines per area show a stronger field.

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14 An electric charge Q is placed at the origin. If the magnitude of the electric field at point A is E, what is the electric field at point B? A 4E B 2E C E D E/2 E E/4

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14 An electric charge Q is placed at the origin. If the magnitude of the electric field at point A is E, what is the electric field at point B? A 4E B 2E C E D E/2 E E/4

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Answer E

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15 A small sphere with charge q and mass m is attached to one end of an insulating string of length L. The other end is attached to a negatively charged wall that produces a constant E field in the vicinity

• f the charged
• sphere. The string makes a constant

angle of with the

• vertical. What is the sign and

magnitude of charge q? A Positive and a magnitude of B Positive and a magnitude of C Negative and a magnitude of D Negative and a magnitude of E Negative and a magnitude of

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15 A small sphere with charge q and mass m is attached to one end of an insulating string of length L. The other end is attached to a negatively charged wall that produces a constant E field in the vicinity

• f the charged
• sphere. The string makes a constant

angle of with the

• vertical. What is the sign and

magnitude of charge q? A Positive and a magnitude of B Positive and a magnitude of C Negative and a magnitude of D Negative and a magnitude of E Negative and a magnitude of

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Answer C

Slide 44 (Answer) / 95 Superposition of Electric Fields

How is the Electric Field calculated when there are multiple charges in the same area (assume they are all large enough, or there are enough of them to contribute to the total field)?

+

• +

P E1 E3 E2

q1 q3 q2

Just like Electric Force and the physical forces in Dynamics, the net Electric Field at a given point is the vector sum of each charge's Electric Field. This principle can be extended to charge distributions with a linear, surface, or volumetric charge density.

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16 What is the direction of the net electric field at Point P? A B C D E

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16 What is the direction of the net electric field at Point P? A B C D E

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Answer C

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17 A positive charge Q

1 = 4.0 μC is placed at point -2 m. A

charge Q2 is placed at point +3 m. The net electric field at the origin is zero. What is the magnitude of charge Q

2?

A +9.0 μC B +6.0 μC C +3.0 μC D

• 6.0 μC

E -9.0 μC

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17 A positive charge Q

1 = 4.0 μC is placed at point -2 m. A

charge Q2 is placed at point +3 m. The net electric field at the origin is zero. What is the magnitude of charge Q

2?

A +9.0 μC B +6.0 μC C +3.0 μC D

• 6.0 μC

E -9.0 μC

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Answer E

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Electric Field of Continuous Charge Distributions

Slide 48 / 95 Electric Field due to Continuous Charge Distributions

Now that you've had some practice in calculating the electric field due to various configurations of individual charges, it is time to move to more realistic scenarios - where an object of some symmetric shape is considered. These are called continuous charge distributions. The continuous nature of the object requires the use of integral calculus to sum up the contribution of each little piece of charge that makes up the object. And, we are limited at this mathematical level to highly symmetric charge distributions.

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Electric Field due to Continuous Charge Distributions

Consider a shape with a uniform distribution of charge, and find the Electric field at a distance r, from the shape. due to the segment of charge Δq adding up all the small segments, and then shrinking Δq - integral calculus

Slide 50 / 95 Electric Field due to Continuous Charge Distributions

This is a vector sum, and would be very difficult to calculate for most random shapes. But there are three specific configurations, where the symmetry of the objects lend themselves to a relatively straight forward integration: · Uniformly Charged Rod · Uniform Ring of Charge · Uniform Disc of Charge

Slide 51 / 95 Electric Field due to Continuous Charge Distributions

In addition to the symmetry requirement, the charge, Q, must be distributed uniformly so the following definitions are used: Linear charge density: Surface charge density: Volume charge density: Since we'll be adding the contributions of each little piece of charge the following relationships will be used as appropriate:

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18 What conditions are required in order to find the Electric Field due to a continuous charge distribution through simple integration techniques? A I and II B I and III C II and III D I, II and III E None of the above I - Uniform Charge Density II - Symmetric charge distribution III - Positive charge distribution

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18 What conditions are required in order to find the Electric Field due to a continuous charge distribution through simple integration techniques? A I and II B I and III C II and III D I, II and III E None of the above I - Uniform Charge Density II - Symmetric charge distribution III - Positive charge distribution

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Answer B

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19 What mathematical process should be used to verify an Electric Field calculation due to a continuous charge distribution? A Take the first derivative of the solution and set it equal to zero. B Take the second derivative of the solution and set it equal to zero. C Check if the solution is the same with an oppositely charged distribution. D Take a limiting case and see if it reduces to a known Electric Field. E None of the above.

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19 What mathematical process should be used to verify an Electric Field calculation due to a continuous charge distribution? A Take the first derivative of the solution and set it equal to zero. B Take the second derivative of the solution and set it equal to zero. C Check if the solution is the same with an oppositely charged distribution. D Take a limiting case and see if it reduces to a known Electric Field. E None of the above.

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Answer D

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20 What does the Electric Field of a line of continuous positive charge look like at a distance much greater than the length of the rod? A A positively charged cylinder. B A negatively charged cylinder. C A positive sheet of charge. D A positive point charge. E A negative sheet of charge.

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20 What does the Electric Field of a line of continuous positive charge look like at a distance much greater than the length of the rod? A A positively charged cylinder. B A negatively charged cylinder. C A positive sheet of charge. D A positive point charge. E A negative sheet of charge.

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Answer D

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Uniform Ring of Charge

Consider a uniformly positively charged ring of radius a, with total charge Q. Find the electric field at point P, a distance x away from the center of the ring. First, sketch the ring and the geometry. Pick a small segment of the ring and work with the charge, dq, of that segment. Draw the electric field at point P resulting from that charge, and resolve it into its x and y components.

x a P dq r dEx dEy dE θ

Slide 59 / 95 Uniform Ring of Charge

The location of dq is arbitrary - as the next step would be to choose another dq in the clockwise (also arbitrary) direction, and find the electric field at point P due to it. Then choose another, and then another, until we have gone around the entire ring. The final step is to superimpose all these electric fields to find the total field at point P. We did that earlier with a group of a few charges - to get an infinite number, we integrate.

x a P dq r dEx dEy dE θ

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x a P r

dEx1 dEy1 dE1

θ

dq2

r

dE2 dEx2 dEy2 dq1

Before we get to an infinite amount of charge, pick a second charge dq2, directly opposite the first charge on the ring. Draw the electric field vectors.

Uniform Ring of Charge

The y components of the electric field of each charge cancel out. The x components reinforce each other. Every piece of charge, dq, will have a charge

• pposite it which will have

the same effect. This is the use of symmetry to simplify the problem.

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x a P dq r dEx dEy dE θ θ

Uniform Ring of Charge

Only the x component of the electric field due to each dq in the ring will survive. Calculate dEx due to one dq: Using trigonometry, and noting that x is constant (we're used to it being a variable), along with a and r, we get the equations to the right:

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x a P dq r dEx dEy dE θ θ

Uniform Ring of Charge

Find the total E by integrating Ex, recalling that Ey = 0, for each element of charge on the ring. This is somewhat of a "phony" integration - the only variable is dq, and when it is summed from 0 charge to the total charge Q, we get Q.

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21 What is the value of the electric field at point P, at a distance x, in the y direction due to a uniform ring of positive charge? A B C D E

x a P r θ y

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21 What is the value of the electric field at point P, at a distance x, in the y direction due to a uniform ring of positive charge? A B C D E

x a P r θ y

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Answer E

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22 What is the value of the electric field at point P, at a distance x, in the x direction due to a uniform ring of positive charge? A B C D E

x a P r θ y

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22 What is the value of the electric field at point P, at a distance x, in the x direction due to a uniform ring of positive charge? A B C D E

x a P r θ y

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Answer D

Slide 67 (Answer) / 95 Uniform Disc of Charge

Calculate the electric field at point P, along the center axis of a uniform disc,of radius R and having a uniform positive charge.

x P dq dEx

r dr R

Evaluate the disc as a series of concentric rings of charge, dq, centered on its center axis (along the x axis). Take advantage of the previous problem where the electric field of a ring of charge was

• calculated. Construct a ring of charge of radius r,

and thickness dr - the blue shaded area in the diagram.

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Slide 69 / 95 Uniform Disc of Charge

x P dq dEx

r dr R

Substitute for dq: Integrate over the total disc by summing up all the rings - integration limits are r = 0 to r = R.

Slide 70 / 95 Uniform Disc of Charge

There are two ways to integrate this expression - the longer way is to use integration by substitution. The second way is to understand how the integral really works and to do it more

• quickly. The first way will be shown first.

Let: Limits of integration change to:

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Slide 72 / 95 Uniform Disc of Charge

The electric field for a uniform disc of charge is along the x axis

• f the disc and is shown above.

Evaluate this expression very close to the disc - where R >> x, so the right hand term in the parentheses reduces to 0. We will see this expression again in the Gauss's Law unit when calculating the Electric Field near an infinite plane of charge.

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23 What is the Electric Field of a Uniform Disc of Charge at a distance, x, where x >> R, the radius of the disc? A B C D E

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23 What is the Electric Field of a Uniform Disc of Charge at a distance, x, where x >> R, the radius of the disc? A B C D E

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Answer A

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Electric Dipole

Slide 75 / 95 Electric Dipoles

An electric dipole is a combination of point charges with the same magnitude, but opposite signs, separated by a distance d. This has applications in such different fields as chemistry, and radio wave transmission, so its behavior will now be discussed.

d

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Electric Dipoles

Let's start with chemistry. Many molecules can be represented by an electric

• dipole. The most famous
• ne is water.

H O H

+

• The covalent bond which holds the water molecule together

shifts the probability of finding electrons closer to the Oxygen atom - making the top half of the molecule positive and the lower half negative.

Slide 77 / 95 Electric Dipole

H O H

+

• +
• d (along dipole axis)

The water molecule is represented as a positive charge separated from a negative charge at a distance, d - an electric dipole. The dipole nature of water makes it a universal solvent - it separates ionic compounds into their constituent atoms and keeps them apart. Na+ and Cl- separate from the salt molecule as they're attracted to either end of the water molecule dipole. The solvent property of water makes possible the chemical reactions responsible for life.

Slide 78 / 95 Electric Field of an Electric Dipole

d

The Electric Field at any point is a tangent line to the field lines. The calculation of the Electric Field at any point in space can be accomplished by superimposing the field due to each charge. To simplify the algebra, we'll take the case where it is calculated at a point along the perpendicular bisector of the dipole.

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• +

x y r

Electric Field of an Electric Dipole

P

Find the Electric Field at point P, at a perpendicular distance y, from the line connecting the positive and negative charges (dipole axis). E1 is the Electric Field due to the positive charge and E2 is the Electric Field due to the negative charge. The y components of the two charges cancel out. The x components add to each

• ther in the postive x direction.

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• +

x y r

Electric Field of an Electric Dipole

P

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• +

x y r

Electric Field of an Electric Dipole

P

As always, let's take a limiting case to see what results. At a large distance from the dipole, y >> d, the equation reduces to: The mathematics won't be done here, but this expression holds true in every direction from the dipole.

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x y r

Electric Field of an Electric Dipole

P

The Electric Field of a point charge falls

• ff as 1/r2, while the Electric field of a

dipole falls off as 1/r3, since the equal and opposite charges tend to cancel each other as r increases. Dipole antennas are used to transmit radio waves - this will be covered in the EM Induction unit. But it makes you wonder how radio waves have such an impossibly great range, if the field decreases so quickly, doesn't it? You'll have to wait for Faraday and Maxwell to understand this.

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24 What is the magnitude and direction of the electric field at point P due to the electric dipole? A to the right. B to the left. C to the right. D to the left. E up.

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x y r

P

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24 What is the magnitude and direction of the electric field at point P due to the electric dipole? A to the right. B to the left. C to the right. D to the left. E up.

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x y r

P

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Answer A

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25 The Electric Field due to an electric dipole falls off as a factor of: A 1/r B 1/r2 C 1/r3 D remains constant. E is zero at a distance equal to twice the separation

• f the charges.

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25 The Electric Field due to an electric dipole falls off as a factor of: A 1/r B 1/r2 C 1/r3 D remains constant. E is zero at a distance equal to twice the separation

• f the charges.

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Answer C

Slide 85 (Answer) / 95 Torque on Electric Dipoles

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• An electric dipole is placed within a uniform electric field.

Assume that d is small enough so that the magnitude of the electric force on each charge is the same. The net electric force is then zero - there will be no translational motion. Both forces will cause a clockwise rotation - a torque is created by the electric force about the center of the dipole, which will cause it to align with the field.

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Torque on Electric Dipoles

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• Each force creates a torque of magnitude:

The magnitude of the total torque is:

Slide 87 / 95 Torque on Electric Dipoles

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• Define the electric dipole moment magnitude as:

The magnitude of the torque is: Vector expression for torque where the dipole electric moment points along the dipole axis from the negative to the positive charge:

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• The electric dipole moment is zero when the dipole axis is

parallel to the Electric Field, and is at a maximum when they are perpendicular.

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Work done by the Electric Field

• n an Electric Dipole

In the rotational motion unit, it was presented that the change in work is equal to the the torque times the angular displacement. Substitute the electric dipole torque and use a negative sign because torque is acting in the direction of decreasing θ. Then integrate from θ1 to θ2 to find the work done by the Electric Field as the dipole rotates:

Slide 90 / 95 Potential Energy of an Electric Dipole in an Electric Field

The change in potential energy of a system is equal to the negative

• f the work done on a particle by the field, or W = U(θ1) - U(θ2).

Combine this with the above equation, and we get:

Slide 91 / 95 Potential Energy of an Electric Dipole in an Electric Field

An electric dipole placed in an external Electric Field will realign itself so it has the lowest potential energy. This occurs when θ = 00 (U = -pE). It has a maximum potential energy (U = +pE) at θ =1800. The potential energy is zero when the dipole is perpendicular to the Electric Field. Be careful - that doesn't mean the dipole won't rotate; it will rotate to achieve the lower potential energy of -pE.

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26 An electric dipole is placed in a uniform electric field that goes from left to right. At which orientation of the dipole is the torque on the dipole a minimum? A The electric dipole moment is pointed to the right. B The electric dipole moment is pointed to the left. C The electric dipole moment is pointed straight up. D The electric dipole moment is pointed straight down. E The electric dipole moment makes an angle of 450 with the electric field.

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26 An electric dipole is placed in a uniform electric field that goes from left to right. At which orientation of the dipole is the torque on the dipole a minimum? A The electric dipole moment is pointed to the right. B The electric dipole moment is pointed to the left. C The electric dipole moment is pointed straight up. D The electric dipole moment is pointed straight down. E The electric dipole moment makes an angle of 450 with the electric field.

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Answer A

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27 An electric dipole is placed in a uniform electric field that goes from left to right. At which orientation of the dipole is the potential energy of the system a maximum? A The electric dipole moment is pointed to the right. B The electric dipole moment is pointed to the left. C The electric dipole moment is pointed straight up. D The electric dipole moment is pointed straight down. E The electric dipole moment makes an angle of 450 with the electric field.

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27 An electric dipole is placed in a uniform electric field that goes from left to right. At which orientation of the dipole is the potential energy of the system a maximum? A The electric dipole moment is pointed to the right. B The electric dipole moment is pointed to the left. C The electric dipole moment is pointed straight up. D The electric dipole moment is pointed straight down. E The electric dipole moment makes an angle of 450 with the electric field.

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Answer B

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