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Transcript
AP Physics - Static Electricity Wow! The course is just zooming by. Aren’t you like totally impressed? We’ve been doing a lot of good physics in the old classic mode, but it is, now, in the words of Monte Python, time for “something completely different”. It is time for electric fields and static electricity! History: Electricity was first described (that we know of) by an ancient Greek, the philosopher Thales (640? - 546 B.C.E.) in the 580’s B.C. What Thales noted was that when he rubbed a chunk of amber with a piece of cloth, the amber would attract small bits of stuff – fibers, dust bunny bits, fluff, etc. The word for amber in Greek, elektron, gave its name to the phenomenon. Thales didn’t do much with the idea which, truthfully, isn’t all that exciting to begin with, so the thing sort of languished around for a very long time. We now make a giant time shift leap type thingee to the 17 th century. To 1600 to be specific. It was in this year that the English physician and physicist William Gilbert, having begun to play around with the attractive force of electricity found other substances that could be charged up besides amber. He divided materials up into classes. The classes were: Electrics - stuff that gains charge and can attract things when rubbed. Nonelectrics - stuff that doesn't gain a charge. Electrics were materials like glass, amber, silk, and rubber. Nonelectrics were mostly metals In the 1660's, the German physicist Otto von Guericke build the first static electric charging machine. He took a large ball of sulfur and rigged it so he could rotate it with a crank. As the ball rotated he pressed a piece of leather against it and the ball would gain an electric charge. Later he built an improved machine that used a glass globe. He used the machine to generate large sparks. At this time, electricity was thought to be a fluid. They also thought that fire was a fluid, so electricity was commonly called “electric fire”. Guericke also discovered that static electricity could both be attractive and repulsive. Around 1709 Francis Hauksbee, a British Scientist, began to experiment with Guericke’s machine. In one of his experiments, he placed a small amount of mercury inside the globe. He cranked the machine in the standard way. When he touched the surface of the globe with his hand, the globe gave off a flash of light. In 1729 Stephen Gray, an English physicist, found cork acted as a conductor. A conductor is a substance that allows charge to travel through it. This introduced the idea that there were materials that were conductors and materials that were insulators. It turns out that conductors are Gilbert’s nonelectrics and insulators are the electrics. Insulators stop the flow of charge. In 1733 Charles Du Fay, a French physicist, discovered that there were two types of electric charge. He found that the charge on a glass rod was different than the charge on a piece of amber. He 263 named the charge on a glass rod vitreous electricity. The electricity on the amber was called resinous electricity. In the 1740’s, Benjamin Franklin, the famous American printer and founding father, found that the vitreous electric charge could cancel out the resinous electric charge. He also came up with the names we use today for the differing charges, positive and negative charge. Electric Basics: Electricity is an aspect of one of the four fundamental forces in the universe, the electromagnetic force. It involves attraction and repulsion between charged particles. The source of the charge is two subatomic particles, the electron and the proton. Electrons have a negative charge and protons have a positive charge. The magnitude of the charge is the same for each particle. We say that an electron has a charge of “minus one” or – 1. The proton has charge of “plus one” or + 1. All this minus one or plus one stuff is mainly a chemistry thing. In physics we use a different unit for charge, as we shall see. Atoms are electrically neutral – they have no charge. This is because they have the same number of electrons as protons and their charges cancel each other out. If an atom gains or loses electrons, it gains a charge and becomes an ion. We say it is ionized. Ions are a really big deal in chemistry, but not much of a thing in physics. Different elements vary widely in their ability to gain or lose electrons. This is what is involved when you rub an object with a cloth to give it a charge. You blow up a rubber balloon and rub it with a bit of wool. The balloon is more attractive to electrons than is the cloth, so during the rubbing, electrons from the cloth jump onto the balloon. This gives the balloon a negative charge because it now has more electrons than protons. The cloth gains a positive charge (it has more protons than electrons so the net charge is positive). Rubber objects almost always gain a negative charge during rubbing operations. A glass rod rubbed with silk will gain a positive charge. Rubber and glass rods are often used in experiments to establish known charges for use in comparison tests. The generation of charge by friction is called triboelectrity. Isn’t it wonderful how physics has phancy names for everything? Charged objects exert forces on one another and obey the fundamental law of static electricity: Fundamental Law of Static Electricity Like charges repel; opposite charges attract. Two balloons that have been given a negative charge will repel each other. A negatively charged object will attract a positively charged object. Here’s another key concept, the principle of conservation of charge. 264 Principle of Conservation of Charge charge is not created or destroyed, merely transferred from one system to another. Back to Insulators and Conductors: Conductors are usually metals. The charge is carried through the material by the free electrons that metals have because of their metallic bonds. Insulators are non-metals; materials like plastic, rubber, ceramics, etc. These substances have their electrons tightly bound in their chemical bonds. The charge can’t go anywhere in these substances because there’s nothing to carry the charge. The electrons are not free to move, don’t you see. When a charge is placed on an insulator, the charge stays where you put it. When a charge is placed on a conductor it will immediately spread out over the entire object (actually, as we shall see, it travels to the outer surface of the conductor). Electrolytes are liquid solutions that can conduct electricity. The electrolyte contains ions that transfer charge. Charging Objects: There are two methods that can be used to charge objects; charging by conduction and charging by induction. Charging by conduction is very simple. An object is given a charge – we rub a rubber rod with a rabbit fur. The rod now has a negative charge. We also have a metal sphere attached to an insulated stand. We touch the sphere with the charged rod and some of the extra electrons on the rod will flow onto the sphere, giving it a negative charge. A common way to transfer charge by conduction is through the use of a small metal disc attached to an insulated rod. This device is called a proof plane (the Physics Kahuna has no idea why). When it is touched to a charged object it gains the same charge by conduction. This charge can then be transferred to some other object or tested. Charging by Conduction 265 Charging by induction is a bit more complicated. We start out with a charged object and an uncharged object. Charge is transferred, but there is no physical contact between the two objects. There are two ways to do this. Let’s look at the first method. We have a negatively charged rubber rod and an insulated metal sphere. You cannot charge an insulator by induction, the method works only with objects that have conductive surfaces. 1. Bring the charged object near the insulated sphere. The sphere is electrically neutral – same number of electrons as protons, duh. The negative charge on the rod will repel the free electrons on the sphere’s surface, however. They will collect on the opposite side of the sphere from the rod - they’re trying to get as far away from the negative charge on the rod as they can. Remember, in a conductor, they are free to move about. If the charged rod is moved away from the sphere, the electrons will redistribute themselves over the surface and the sphere will remain electrically neutral. With the charged rod near, the sphere is polarized – one side has a positive charge and the other side has a negative charge. Even though the net charge on the object is still zero, so the thing is still electrically neutral. 2. Keeping the charged rod near the sphere (but not touching it) touch the opposite side of the sphere with a finger. This will “ground” the sphere. The electrons, wanting to get away from the negative charge on the rod (like objects repel) will flow into your finger. This is because you are a big electron sink that the electrons can go into. The earth is an even bigger sink, so anytime electrons are given a path to travel into a big sink we say they have been “grounded”. The big sink is a “ground”. 3. Take the finger away and remove the rod. 4. The sphere, having lost a good many of its electrons, will now have a positive charge. It is positive because there are more protons than electrons. It has been charged by induction. 1 2 3 4 The second induction method involves two insulated objects. 1. Place the two objects in physical contact with one another – they must touch. In this example we have two insulated conductive spheres. These are metal spheres on insulated stands. 266 2. Bring a charged object to the side of one of the spheres. Don’t touch the sphere; just bring the object nearby. In this example we have a negatively charged rubber rod. 1 2 3 4 Electrons in the first sphere, the one near the charged rod, are repelled and will move into the second sphere. They try to get as far away from the rod as possible. As long as the rod near one of the spheres, the two sphere system will be polarized. 3. Move the sphere opposite of the one near the charged rod away from the other sphere so they no longer touch. The sphere you moved will have an excess of electrons, the ones from the other sphere, and will have a negative charge. The other sphere, having lost some of its electrons, will have a positive charge. 4. Pull the rod away. The two spheres are now charged – charged by induction! Polarizing Objects: Briefly mentioned in the explanation of charging by induction was the term “polarized”. Polarizing is important in many of the electrostatic phenomenon that we have played around with. For example, why did the rubber rod attract bits of paper? The rubber rod had a charge, but the paper bits did not, they were quite neutral, wasn’t they? Why did the charged balloon stick to a person’s clothing or the wall? Why was the 2 x 4 attracted to the charged PVC pipe? These things happen because of polarization. When you bring a charged object near an uncharged object, the uncharged object gets polarized. Think of the molecules that make up the uncharged object. The molecule has no charge. But when the charged rod comes near it, electrons are either attracted or repelled toward the rod. This motion happens because of the nature of the covalent bond, you know, the old “sharing of electrons” thing. So the electrons try to get away form a negatively charged object, or they are attracted to a positively charged object. The net effect is to cause the molecules to have a positive side and a negative side, even though overall it is electrically neutral. As long as the charged object is nearby, the molecule stays polarized. Polarized Molecules The charged balloon sticks to the wall because it polarizes the molecules in the wall and the negative charge of the balloon is attracted to the positive end of the wall’s molecules. 267 Sparks: One of the things you will have observed is that when a charge goes from one object to another, it can travel through the air in the form of a spark. Sparks can be barely noticeable or they can be so massive that it is impossible to ignore them, like a lightening bolt. The bigger the charge difference between two objects, the more the charge wants to move. If the charge is big enough, it will ionize the air molecules between the objects. The charge can then travel through the ionized air. This is what we call a spark. The Electrophorus: The electrophorus is a wonderful device that was invented by Alessandro Volta in 1775. He called it the elettrofore perpetuo. It seemed to an almost endless source of charge. The electrophorus has two parts: a base made from an insulator and a metal plate that has an insulated handle. Here’s how to use the electrophorus. 1. The first step is to charge the insulator base by rubbing it wait a cloth or fur. Then you, holding onto the insulated handle, place the metal plate onto the charged insulator base. insulated handle metal plate charged insulator base 2. The metal plate will be polarized. Free electrons will collect on the upper surface of the metal plate. 3. Touch the top surface of the metal plate with a finger while the plate sits on the base. Immediately remove your finger from the plate after the touch. You should feel a small spark. See, here’s what happened, when you touched the plate, the free electrons had a path to ground and went into your finger. The plate now has a positive charge. 4. Pick the plate up with the handle. It is charged and you can transfer the charge to some other object. The plate did not remove any charge from the base, so if you place it on the base again, you can repeat the charging operation again and again. To the 18th century physicists, it seemed like a magical source of charge. Sadly, however, the charge on the insulated plate will eventually leak off into the air, so you can’t use it forever. On a humid day, you might be lucky to charge the plate two or three times before the charge leaks off. This is because humid air ionizes much faster than does dry air. 268 Electroscopes: A useful device to measure charge is the electroscope. There are several different styles that are available. Each of them will indicate whether an object is charged. Foil on thread type: One type is made from a small, light piece of metal foil that is attached to an insulated thread. A charged object is brought near the foil. The foil will be polarized and attracted to the charged object. When it touches the charged object it will be charged by conduction. Now having the same charge, it will be repelled. The angle formed by the thread with the vertical is proportional to the charge. If the foil is charged by a rubber rod, then one knows that it has a negative charge. An unknown charge can now be brought near the foil. If the charged foil is attracted, then you know that the object has a positive charge – remember, unlike charges attract. If the foil is repelled, then you know that the object has a negative charge – like charges repel. Twin hanging leaf type: This type of electroscope is usually enclosed in a transparent box or beaker. A metal rod has a metal ball mounted on its top. We will call this the electrode. The rod goes into the box where it has two very thin metal foil leaves attached to it. The metal leaves hang downward and are separated by a small space. They are both attached to the rod. A negatively charged rod is brought to the electrode on the top. The electrode is charged by conduction. The charge travels throughout the rod and into the metal foil leaves. They each have a negative charge and will repel each other. This means that they will lift away from the vertical. The angle they make is proportional to the No charge Charged charge. You charge up an electrophorus and give it a positive charge. You bring it to an electroscope to transfer the charge. How does a positive charge transfer? As the electrophorus nears the ball on the electrode of the electroscope, the electroscope becomes polarized. Free electrons collect on the ball, attracted to the positive charge on the plate of the electrophorus. 269 The electrophorus has a pretty good charge, so usually the electrons will be so attracted to the plate as it nears that they will ionize the air and jump to the electrophorus in the form of a spark. The electrophorus is slightly less positive having collected some electrons and the electroscope is positive, having lost free electrons. Leyden Jar: Ewald George von Kleist in 1745 and Pieter van Musschenbroek in 1746 developed this wonderful device. It was the first method discovered to store a significant amount of charge for later use. The Leyden jar was an accidental discovery. Kleist apparantly thought that electricity was a liquid. It made sense that it might dissolve in water, so an attempt was made to put some charge in a beaker of water. He hooked a wire in the beaker to an electrostatic generating machine. Holding the beaker in his hand, he cranked up the machine to generate charge. It turned out that electricity did go into the beaker, but it didn’t dissolve into the water. This was discovered a bit later when an assistant touched the wire in the beaker while holding the outside of the thing and got a tremendous shock. The shock was so painful and shocking that Kleist gave up on the whole thing. Electrolyte Foil The Leyden jar was the first capacitor. The first one wasn’t a real proper Leyden jar, but it certainly acted like one. A Leyden jar is simply a jar that has a conductor on the inside lining its inner surface and a conductor on the outside surface. An electrode sticks up above the jar through the lid. The electrode is connected to the inner conductor. Here is the classic Leyden jar. It has a metal foil lining on the inside lower half and a similar coating on the outside. The electrode in the center makes electrical contact with the inside metal layer via a chain. The key thing though is that you must have two conductors separated by an insulator. Another common way to make a Leyden jar (the one we will use) is to have an electrolyte solution on the inside and a foil lining on the outside. The electrode pokes down into the electrolyte. Recall that electrolytes are solutions that conduct electricity. We will Polarized Leyden jar 270 use salt water. Charge and the Leyden Jar: When a charged object is brought near the center electrode, the inner conductor becomes polarized. In the drawing (which has a greatly exaggerated conductor thickness), a positive charge is brought near the electrode. The free electrons flow to the top ball, attracted by the positive charge. The inner conductor has a positive charge because the free electrons have made their departure. The outer conductor is also polarized. Just across the insulator is a strong positive charge, so the free electrons are attracted to the surface of the insulator. To charge the Leyden jar, the outside conductor must be grounded as the charged object is brought to the electrode ball. Usually the charge will jump to the electrode in a spark. If the outside of the jar isn’t grounded, you can’t store charge in the jar. The grounding is usually done by holding the jar in you hand. With a ground (your hand) attached to the outer conductor, electrons flow from the ground into the outer conductor. The free electrons at the top of the electrode flow into the positively charged object. You end up with a negative charge on the outer conductor and a positive charge on the inner conductor. They stay in place because they attract each other across the insulator. You can pick it up by the outer conductor and the charge stays in place. This is how the Leyden jar stores charge. Charged Leyden jar electrons flow to positive charge electrons flow from ground into outer conductor Polarized Leyden jar To ground 271 AP Physics - Coulomb's Law We’ve learned that electrons have a minus one charge and protons have a positive one charge. This plus and minus one business doesn’t work very well when we go in and try to do the old major figuring stuff out deal– it’s kind of arbitrary thing, see. So we need a really useful unit of charge. Well, we got one. It is your basic standard unit of charge; a thing called a Coulomb (C). The symbol for charge is Q however q is used as well. One Coulomb is equal to the charge of 6.25 x 10 18 electrons or protons. The charge of a single electron is - 1.60 x 10-19 C. The charge of a proton is + 1.60 x 10-19 C. The Coulomb is a large amount of charge, so it is very common to use micro Coulombs and milli Colulombs. 1 mC = 10-3 C 1 C = 10-6 C What is the charge of 1.35 x 1017 electrons? This is a simple dimensional analysis type o’problemo. 1.60 x 1019 C 1.35 x 10 e 1e 17 2.06 x 102 C 1 C 1 2.06 x 102 C 2.06 x 10 C 3 1 x 10 C or 20.6 C Coulomb’s Law: Charged objects exert forces on one another. This is very similar to what happens with gravity between two objects that have mass. Recall that Newton’s universal law of gravity can be used to calculate the force between two objects that have mass. It turns out that there is a similar law that can be used to calculate the force between two objects that have charge. This law is called Coulomb’s law. Here it is: F= 1 q1 q 2 4πε0 r 2 F is the force exerted between the two charges. q1 and q2 are the two charges. (Note, we will actually use the absolute value of the charges - we simply don’t care about whether they are 272 positive or negative.) r is the distance between the two charges and 1 is called Coulomb’s 4πεo Constant. It is similar to the universal gravitational constant. The value for Coulomb’s constant is: 1 = 9 x 109 Nm2 / C 2 4 πε 0 Coulomb’s law in most physics books is usually written in a slightly different form: k q q F e 12 2 r or k qq F e 12 2 r But we won’t use that form, because the wonderful AP Physics folks use the first one that the Physics Kahuna gave you. The force between two charged objects can be either attractive or repulsive, depending on whether the charges are like or unlike. We will also assume that the charges are concentrated into a small area – point charges. Example Problems: Two point charges are 5.0 m apart. If the charges are 0.020 C and 0.030 C, what is the force between them and is it attractive or repulsive? 1 q1q2 F 4 0 r 2 Nm 2 0.020 C 0.030 C 8.99 x 10 C 2 5.0 m 2 9 F 0.000216 x 109 N 2.2 x 105 N The force is repulsive - both charges are positive. A force of 1.6 x 10-3 N exists between 2 charges; 1.3 C and 3.5C. How far apart are they? F 1 q1q2 4 0 r 2 r 1 q1q2 4 0 F 273 Nm 2 8.99 x 10 1.3 x 106 C 2 C r 1.6 x 103 N 9 r 25.57 x 100 m 2 3.5 x 10 6 6C 5.1 m Electric Force and Gravity: Both gravity and the electric force are fundamental forces. The equations for the gravity and the electromagnetic force have the same form; they are both inverse square relationships. Where the F= 1 q1q 2 4πε o r122 F G m1m2 r2 and 1 term is the constant for Coulomb’s law and G is the constant for the law of 4 0 gravity. There are four really significant differences in the two forces: Gravity is always attractive. The electromagnetic force can be either attractive or repulsive. Gravity is much weaker. Gravity has a much greater range within which it is a significant force. The electric force can be shielded, blocked, or cancelled. You cannot do any of these things with the gravity force. The force of gravity is around 1040 times smaller than the electromagnetic force. This can be seen in a comparison of the two proportionality constants. 1 = 8.99 x 109 Nm2/C2 for the electric force 4 0 G = 6.67 x 10-11 Nm2/kg2 for the gravitational force The two constants differ by a factor of 1020! 274 Gravity extends out to great distances. The sun is 92 million miles away from the earth, yet the force of gravity is large enough to cause the earth to be locked in an orbit around the sun. Now for electricity, the force between two charges drops off very quickly with distance. This is because the magnitude of the two charges is very small – at the most, maybe a few Coulombs. But with gravity, we are dealing with enormous masses - the mass of the sun is 1.99 x 1030 kg! Because of these large masses, even though the gravity constant is very small, the force of gravity between really large masses ends up being a really big force that reaches out over distances of billions and even trillions of kilometers. Let us compare forces in a hydrogen atom. The hydrogen atom is made up of a proton and an electron. The two particles attract each other because they both have mass and they also have opposite charges. The magnitude of the electron/proton charge is 1.60 x 10-19 C. The distance between them in a hydrogen atom is around 5.3 x 10-11 m. For the mass of an electron we’ll use 9.1 x 10 –31 kg. For the mass of a proton we’ll use 1.7 x 10 –27 kg. We can now calculate the force of gravity between the two particles: Gm1m2 F r2 F 6.67 x 10 11 27 31 Nm 2 1.7 x 10 kg 9.1 x 10 kg 2 kg 2 5.3 x 1011 m 3.7 x 1047 N (Pretty small, ain’t it?) Next, we solve for the electromagnetic force. 1 q1q2 F 4 0 r 2 F 0.82 x 107 N 19 19 Nm 2 1.60 x 10 C 1.60 x 10 C 8.99 x 10 2 C 2 5.3 x 1011 m 9 8.2 x 108 N Looking at the two forces, we see that gravity is much weaker. The electromagnetic force is 2.2 x 1039 times bigger! 275 Gravity vs Electricmagnetic Force: Gravity fields and electric fields are both similar and different at the same time. Here is a handy little table to organize things: Gravity Force Attracts inverse square law surround objects cannot be shielded incredibly weaker Electricmagnetic Force attracts and repels inverse square law surround objects can be shielded enormously stronger Superposition Principle: When we have more than two charges in proximity, the forces between them get more complicated. But, please to relax, even though things seem complicated, they actually ain’t and it is fairly simple to work things out. The forces, being vectors, just have to be added up. We call this the superposition principle. Superposition Principle The resultant force on a charge is the vector sum of the forces exerted on it by other charges. Let’s look at a system of three charges. The charges are arranged as shown in the drawing. q1 is 3.00 m from q2. q2 is 4.00 m from the q3. (We immediately spot this as one of those “345” triangle deals, so we know that q1 is 5.00 m from q3). What is the net force acting on q3? q2 - 4.00 m F23 F13 + 37.0 0 q3 3.00 m 5.00 m + q1 q3 is attracted to q2 (they have opposite charges) and repulsed by q1 (they have the same charge). The two force vectors have been drawn and labeled, F23 and F13. q1 6.00 x 109 C q2 2.00 x 109 C q3 5.00 x 109 C The net force on q3 is F23 + F13. The first step is solving the problem is to find the magnitude of the 2 forces: 276 9 9 Nm 2 5.00 x 10 C 6.00 x 10 C 8.99 x 10 2 C2 5.00 m 1 q1q2 F13 4 0 r 2 9 F13 10.8 x 109 N 1.08 x 108 N 9 9 Nm 2 2.00 x 10 C 5.00 x 10 C 8.99 x 10 2 C2 4.00 m 1 q1q2 F23 4 0 r 2 9 5.62 x 109 N F23 The next step is to break the two vectors down into their horizontal and vertical components and add the two vectors in the x and y directions. This gives us the components of the resultant vector, FX and FY: F sin F Fx F13 cos F23 8 9 o Fx 1.08 x 10 N cos37.0 5.62 x 10 N Fx 8.63 x 109 N 5.62 x 109 N F F cos 3.01 x 109 N Fy F13 sin Fy 1.08 x 108 N sin 37.0o 6.50 x 109 N Now we can find the resultant force: F Fy 2 Fx 2 F 6.50 x 109 N 2 3.01 x 109 N 2 51.31 x 1018 N 2 7.16 x 109 N Now we can find the direction or the resultant force: 1 F tan Y FX x 109 N tan 9 x N 3.01 10 1 6.50 65.2o with the x axis Isn’t it great to solve these problems? Let’s do another! 277 Two 0.25 g metal spheres have identical positive charges. They hang down from light strings that are 32 cm long commonly attached to the ceiling. If the angle the strings form with the vertical is 7.0, what is the magnitude of the charges? T First let’s draw a FBD for the forces on one of the spheres: FE 32 cm There are three forces: the tension T in the string holding the sphere up, the electric force FE, and the weight of the sphere mg. mg Because the system is static, the sum of the forces must be zero. The forces in the x direction are: FE T sin 0 The forces in the y direction are: T cos mg 0 Now things become simpler, we can use this last equation to find the tension. Armed with the tension we can find the electric force. Using the electric force we can find the charge on the sphere. Let’s go ahead and find the tension. T cos mg 0 T mg cos m 1 0.025 kg 9.8 2 0.247 N o cos 7.0 s Now we find FE: FE T sin 0 FE T sin 0.247 N sin 7.0o 0.0301 N Coulomb’s law is next: F 1 q1q2 4 0 r 2 The 2 charges have the same value; we need to find the distance between the two spheres r. l l=32 cm We can see that half of r is l sin. So r is 2l sin . r 278 F 1 q1q2 4 0 r 2 But the two charges are the same, so: 1 qq F 2 4 0 r 1 q r F 1 4 0 1 q 2 4 0 r 1 2 l sin F 1 4 0 q 2 0.32 m sin 7.0o 1 0.0301 N 2 Nm 9 8.99 x 10 C2 q 0.0780 0.003348 x 109 C 2 q 1 2 2 q Fr 1 4 0 2 0.0780 0.03348 x 1010 C 2 0.0143 x 105 C 1.43 x 107 C 279 AP Physics - Electric Fields The earth is surrounded by a gravity field. Any object with mass will have a force exerted on it by the earth’s gravity (and it will exert an equal force on the earth – third law, natch). This is fairly simple to picture in one’s mind. You have the earth pulling things down with the force of gravity, but because the force of gravity is so weak, we don’t have to worry about the force of gravity between objects that are in the earth’s gravity field. This is not true for electric charges. If you have a group of them, the forces they exert on each other are significant and we can’t ignore them. It would be like having 3 or 4 planets scattered around in a volume a thousand times smaller than the solar system. They would all exert tremendously large forces on each other. Because of this we have to treat electric forces between charges way different than how we treat gravitational forces between masses. Electric Lines of Force: An electric field exists around any charged object in space. A second charged object brought into this field will experience a force according to Coulomb's law. Of course this is also true for gravity. Objects with gravity are surrounded by a gravity field. The thing is that gravity fields aren’t very useful (because gravity is pretty simple), but electric fields are. The electric field E is a vector quantity. It has both magnitude and direction. The direction of the field is the direction a small positive charge would be subject to. In the drawings below, Q represents the charge causing the field. q represents a small test charge. In this first case, Q is positive and the test charge is positive. +Q q In the example above, a positive charge Q exerts a force on a small positive test charge q. The direction of the force is away from Q. If we move the test charge to a new location, the force exerted on it will have a new direction, but it will still be away from Q. q +Q We can represent the field and the direction of the forces it will exert by drawing in lines that show how the forces would be directed. We call these electric lines of force. 280 Electric lines of force lines drawn so that a tangent to the line shows the direction of the electric force. 1. The number of lines per unit area is proportional to the strength of the field. 2. Where E (the electric field strength) is large, the lines will be close together. Where E is small, the lines will be far apart. The electric field around a single point positive charge would look like this: The direction of the arrows is the direction of the force that would be exerted on a positive test charge placed at that point. The lines show that any positive charge in the field would experience a force that wold be directed directly away from the positive charge in the center. The lines of force around a negative charge would look like this: Note: the direction is towards the source of the field. This is because a positive test charge would be attracted to the negative charge in the center. Here are electric lines of force between two positive charges: 281 Lines of force between two unlike charges: Field Intensity: The field intensity is a measure of the strength of an electric field. It is represented by the symbol E. E F q Here, E represents field intensity, and F represents the force in Newtons acting on a test charge q0, which is a test charge that is being acted upon. This charge, q0, is not the charge that has set up the field! q0 is a test charge that is in the field made by Q (which is some other charge)! The unit for the field intensity is a Newton or N/C. Coulomb Electric lines of force are very useful to figure out what sort of forces will act on charges in a given electric field. You can quickly determine the direction of a force acting on a charged particle due to an electric field. In the above example (in the drawing) there is a uniform electric field E. A proton will experience an electric force from the field, FE as shown in the drawing below. 282 FE Proton in field Electric Field E Direction of force on proton Fb b a Fa An electric field surrounds a positive charge. Two small positive test charges are placed in the field; one at point a and the other at point b. We can determine the direction of the forces acting on the test charges and also determine their relative magnitudes. The force on the particle at a will be greater than the force at b because the lines of force are closer together where a is located. Closer lines of force means a bigger force. An electric field has a field intensity of 2.0 x 104 N/C. If the force acting on a test charge is 6.2 N, what is the magnitude of the test charge? This is a simple plug and chug problem. E F q q F E 6.2 N 2.0 x 104 N C 3.1 x 104 C Once we can find forces, we can use Newton’s laws to calculate all sorts of wondrous things. Like velocity or acceleration! 283 An electron travelling at 2.3 x 105 m/s with a direction as shown in the drawing enters a uniform 280 N/C electric field. (a) Analyzing the electric lines of force, what is the direction of the force acting on the particle? (b) What is the magnitude of the force? (c) What is the acceleration acting on the electron? (d) If the electron travels a distance of 3.0 mm in the field, what is the distance it will be deflected from its original path? v E (a) The direction of the force will be down. The arrows on the lines of force show the direction of a force acting on a positive test charge. Since the electron has a negative charge, the force will be in the opposite direction. (b) We know the charge of an electron, so we can figure out the force acting on it. E F q F Eq 280 N 1.6 x 1019 C 448 x 1019 N C 4.5 x 1017 N (c) We can use the second law to find the acceleration. We’ll need the mass of the electron, but we can look that up. F ma a a 0.494 x 1014 m s2 F m 4.5 x 1017 kg m 1 2 s 9.11 x 1031 kg 4.9 x 1013 m s2 (d) The electron is moving horizontally at a constant velocity. It will be accelerated downward by an electromagnetic force and also by gravity. Looking at the acceleration from the electric field, we can see that the acceleration from gravity is way way smaller, so we can ignore gravity - it is totally insignificant. (Hey, what is a lousy 9.8 m/s2 compared with 1013 m/s2?) We need to figure out the length of time it will be accelerated. It is moving through a field – this is when it will be accelerated – a distance of 3.0 mm. We know the horizontal speed, so we can find the time to travel that distance. x vt t x v 1 0.0030 m 0.0013 x 105 s 1.3 x 108 s m 2.3 x 105 s 284 Armed with the time, we can find the distance it will be displaced. 1 y at 2 2 1 m 3.2 x 1013 2 1.3 x 108 s 2 s 2 2.1 x 103 m or 2.1 mm So the electron will be deflected downwards a distance of 2.1 mm as it travels through the field. This problem looked really horrible, even the Physics Kahuna must admit this, but it actually turned out to be quite simple. Other than the E F equation, the problem dealt with a force, an q acceleration, a constant velocity over a given distance, and a displacement caused by a force. All of which is stuff you’ve done before. Electric Fields and Objects: The electric charge on an object, such as a conductive sphere, say for example, always is on the outside of the object. It is on the outer surface. Why is this so? Well, it’s very fundamental. The free electrons repel each other. This means that they try to get as far away from one another as they possibly can. In order to do this, they collect on the outside surface and spread out. If they were on the inside, they would be closer together, so they don’t do that. Electrons repel each other to the outside Electrons end up on the outer surface 285 AP Physics – Electric Potential Energy Indulge the Physics Kahuna for a bit of a quick review of some vital previously covered material. The importance of the earlier concepts will be made clear as we proceed. W Fr Work takes place when a force acts over a distance. The unit for work is the joule (J). A joule is a newton meter. Potential energy from gravity is U g mgh . Fg The force of gravity acting on an object with mass is always directed towards the center of the earth. increase in gravitational potential energy Remember? When you lift an object you do work on it. As its height is increased, its potential energy increases. The work that you do lifting it is equal to the increase in potential energy of the thing. Imagine a skyscraper. An object that is on the 75 floor has more potential energy than does an object on the 2nd floor, true? If we move an object from the 10th floor to the 20th floor, we’ve (a) done work on it and (b) increased its potential energy. If we move an identical object from the 25th floor to the 35th floor, we’ve (a) done work on it and (b) increased its potential energy. In fact we’ve done the same amount of work on both occasions and we’ve increased the potential energy of the object by the same amount, true? U U We’ve done ten floors worth of work and increased the potential energy ten floors worth. We could make it even more general by looking at potential energy changes per kilogram. Then we wouldn’t have to worry about the objects all having the same mass. We would be dealing with a change in potential energy per kilogram. Analyzing changes in gravitational potential energy doesn’t turn out to be terribly useful in dealing with gravity, energy, and work. So why did the Physics Kahuna waste your precious time with the pointless discussion? U For the same height change, potential energy change is the same 286 Well, it isn’t useful for gravity fields, but it is terribly useful when dealing with electric fields. Looking at an electric field around a charged object we find that we also have work being done and potential energy changes being made when we move charged objects about. We have a comparable situation to that of gravity when we have two opposite charges. We have a large charge Q. Placed near this charge is a small test charge q. As we increase the distance between q and Q, we have to do work to move the charge and we increase its potential energy . If the charges are the same, then we have a different situation. It takes work to move the test charge closer to the other charge (the force is repulsive between them). As the two charges move closer together, the potential energy increases. FE + + - increase in electric potential energy - FE FE - - - + Unlike charges Force is attractive increase in electric potential energy + F E + + Like charges Force is repulsive Electric Potential Energy: Let’s go ahead and develop an equation for the potential energy of a charge in an electric field. Let’s look a simple situation with two charges. Now, deriving an equation for electrical potential energy sounds too hard. We haven’t studied anything about electrical potential energy, have we? Oh, ye, of little faith. We don’t need to study anything special, we already know all that we need to in order to figure this out. U Fr We know that potential energy is force times distance. The force acting between the electron and the proton is: F 1 q1q2 4 0 r 2 We can plug in Coulomb’s law for the force in the potential energy equation. 1 q1q2 U r 2 4 0 r 1 q1q2 4 0 r 287 UE So there it is. An equation for electrical potential energy: 1 q1q2 4 0 r An electron in a hydrogen atom is at its lowest energy state. It is at a distance of 5.29 x 10-11 m from the proton in the nucleus. What is its potential energy relative to the proton? We can look up the charges for the particle, although by now, you probably know what they are by heart. Anyway, here are the charges: q1 1.60 x 1019 C U q2 1.60 x 1019 C 1 q1q2 4 0 r Nm 2 1.60 x 1019 C 1.60 x 1019 C U 8.99 x 10 2 C 9 U 5.29 x 101 11 m 4.35 x 1018 J Work in the Electric Field: Earlier we mentioned that work was done when a particle was moved from one point to another in an electric field. This is accompanied by an increase in the charge’s potential energy (the charge that gets moved). If the two charges are unlike: Pull the charges apart – increase potential energy (just like with gravity). Push the charges together - potential energy decreases. If the two charges are the same: Pull the charges apart - decrease the potential energy. Push together - potential energy increases. The important thing here is that the work done in moving the charge is equal to the potential energy difference between the two points The critical thing is the change in displacement – the difference between the starting point and the final position of the charge. We don’t care about what the actual distance between the charges is. We know that work is: W Fr The force exerted on a test charge in an electric field is: 288 E F q F qE We now put these two equations together and we get: W Fd qE d qEd This gives us an equation for the work done in an electric field in moving a charge form one position to another: W qEd W is the work done in Joules, q is the charge of the test charge (not the charge making the field), E is the electric field strength, and d is the displacement of the charge. E In the drawing to the right, we have a uniform electric field of strength E. A charge is moved from point A to point B, which is a distance d. The work needed to do this is given by the equation we just developed. The magnitude of the work done in moving the charge is equal to the change in potential energy. Therefore we can write: A B d U W The negative sign simply means that the potential energy will increase if the charge is negative or decrease if the charge is positive. We recall the equation for work: W qEd We plug this into the equation for the change in potential energy. U qEd This is another useful equation. 289 Potential Difference: We can make things even more general. Instead of looking at the total potential energy, we can look at the potential energy per Coulomb of charge. This, the potential energy per charge, is called potential difference. Potential Difference Change in potential energy per Charge The symbol for potential difference is V or V. V Potential Difference V U q If we look at the equation for potential energy and divide both sides by the charge q, then we get the potential difference. U qEd We divide both sides by q (which is mathematically legal). U qEd q q V Ed This equation is commonly rearranged to this form: E V d This is the equation that you will have available for the AP Physics Test. Notice also that the electric field can be expressed in units of V/m as well as N/C. Here is the formal definition for potential difference between points a and b: Potential difference is the change in potential energy as a charge q is moved from a to b divided by the charge q. If V Ed , then U qEd V Ed so U qV On the AP Physics Exam, the equations for potential energy are combined together into one, looks like this: U E qV 1 q1q2 4 0 r 290 And of course, the work is equal to the change in potential energy, so W qV 1 q1q2 4 0 r The unit for potential difference is the volt (V). The volt is a joule/Coulomb. 1V1 J C A potential difference of one volt means that it takes one joule of energy to move a one Coulomb charge in the field. A potential difference of 1000 V means that it would take 1000 J to move 1 C of charge in the field. Potential difference is also called electric potential or voltage. Potential difference is one of the most important concepts in physics when you are dealing with electricity, so you want to take special care to master the thing. It is essential that you have a solid understanding about potential difference and how it works. Charge moves because of the potential difference, V. The bigger the potential difference between two points, the more the charge will want to move. Electric potential is a scalar quantity. Here are the equations that you will have to work with on the AP Physics Test: E F q U E qV 1 q1q2 4 0 r E Avg V d We’ll now make use of them to solve the odd problem or two. The potential difference between two points is 12.0 V. What amount of work is needed to move a 2.00 C charge within the field? U E qV W qV 1 q1q2 4 0 r U E qV W 2.00 x 106 C 12.0 V 24.0 x 106 J 2.40 x 105 J 291 Two charged plates are 5.00 cm apart. The electric field between them is 775 N/C. What is (a) the potential difference between the plates and (b) what work is done moving an electron from one plate to another? (a) E V d V Ed 775 N J 0.050 m 38.8 C C 38.8 V (b) We know the electron’s charge, so this is fairly simple: W qV 1.60 x 1019 C 38.8 V 62.1 x 1019 J 6.21 x 1018 J The voltage between two big charged up parallel plates is -120.0 V. The plates are 2.50 cm apart. What is the electric field between them? E V d J 1 120.0 C 0.0250 m 4800 N C 4.80 x 103 N C Notice that the Physics Kahuna has abandoned showing how the units all work out. This is because in the electric world with all of its wacky units, it is just too much trouble. Here’s what you do. You put everything into standard units: meters, Newtons, joules, Coulombs, Volts, etc. Then you just put down the proper unit that the answer is supposed to have. A proton is released from rest in a uniform electric field, E = 8.0 x 104 V/m. It is directed along the x axis. It’s displacement is 0.50 m in the direction of the field. (a) What is the change in potential difference? (b) What is the change in electrical potential energy? (c) What is its velocity after it traveled the 0.50 m? (a) We can find the potential difference since we know the electric field strength and the displacement. E V d V Ed V 8.0 x 104 0.50 m m 4.0 x 104V (b) The potential energy is easy to find. We know the charge of a proton and we know the potential difference it is exposed to. U E qV J 1.60 x 1019 C 4.0 x 104 C 6.40 x 1015 J (c) The proton’s potential energy will be converted into kinetic energy as it moves through the 0.50 m. So we can set the kinetic energy equal to the potential energy and solve for the particle’s velocity. 292 1 U mv 2 2 2U m v We can look up the mass of a proton. 2 1 15 kg m v 2 6.40 x 10 27 s 2 1.67 x 10 kg v 2 12 m 7.66 x 10 2 s 2.8 x 106 m s Potential Difference and Conductors: In a conductor electrons, for all practical purposes, are free to move about every which way but loose. This means that there is no potential difference within a conductor from one point to another. If one end of a wire has a potential difference of 12 V, the other end will also have a potential of 12 V. We physics types say that all points in a conductor are at an equipotential. When conductors are joined together, the charge ends up being evenly distributing throughout the newly formed conductive system. The new conductor system will also be at an equipotential. 293 AP Physics – Potential Difference - 5 By convention, a point in an electric circuit is said to have zero electric potential (potential difference) if is grounded (connected to earth). The other way to have zero potential difference is to have a point charge be located at an infinite distance from another charge. We’ve been looking at a test charge placed in an electric field; we learned how to determine the potential difference on the test charge, the change in potential energy of charge, and the work needed to move the charge within the field. We can look at the field itself and calculate the potential difference at a point within the field using this equation: V 1 q 4 0 r V is the potential difference at a point that is a distance r from the charge q that created the field. The Potential difference depends only on charge and the distance from the charge. Superposition Principle: When there are two or more charges in proximity to one another, the change in potential energy is the sum of the potential difference for each charge. Potential difference is a scalar, therefore: Total electric potential at some point near several point charges is the algebraic sum of the electric potentials from each of the charges This can be written in general form as: V 1 4 0 q ri i i This shows that for a number of charges, the voltage is the algebraic sum of 1 q for 4 0 r each charge. This is the equation you will be provided on the AP Physics Test. 294 q1 q3 r r 1 r 3 2 q2 Three charges are arranged as shown in the drawing to the right. To find the potential difference at the point in the center, you calculate the potential difference for q1, q2, and q3. You can do this if you know the value for the charge and the distance to the central point. Lastly, to find the total potential difference, you simply add up the three V’s that you found for the three points. r1 Let’s put some numbers into the mix. Let q1 = 1.5 C , q2 = 2.0 C, and q3 = 2.5 C. For the distances, r1 = 2.0 cm, r2 = 1.8 cm, and r3 = 2.7 cm. V 1 4 0 q3 q1 r3 r2 q2 q ri i i 1 q1 V1 4 0 r1 2 1 6 9 Nm 5 1.5 x 10 C 8.99 x 10 6.7 x 10 V 2 2 C 2.0 x 10 m Nm 2 1 5 V2 8.99 x 109 2 2.0 x 106 C 10 x 10 V 2 C 1.8 x 10 m 2 1 9 Nm 6 V3 8.99 x 10 x C 2.5 10 8.3 x 105 V 2 2 C 2.7 x 10 m Now, having found each of the potential differences for the charges, we can find the total potential difference. V V1 V2 V3 V 6.7 x 105 V 10 x 105 V 8.3 x 105 V 25 x 105 V 2.5 x 106 V An important thing to understand is that potential difference is not a vector, it’s a scalar, so you add them up algebraically. 295 Two charges are situated as shown. The distances are given in the drawing. The charges have the following values: q1 is 5.0 C and q2 is -2.0 C. (a) Find the potential difference at point P. (b) How much work is required to bring a third point charge of 4.0 C from infinity to P? P (a) Find the potential difference at P. V 1 q 4 0 r r2 for q1 2 1 9 Nm V1 8.99 x 10 5.0 x 106 C 2 C 4.0 m 3 V1 11.2 x 10 V 4.0 m q1 4 1.12 x 10 V 3.0 m - q2 We must now find r2: r2 4.0 m 2 3.0 m 2 5.0 m 2 1 9 Nm 6 4 V2 8.99 x 10 2.0 x 10 C 0.360 x 10 V 2 C 5.0 m V V1 V2 V 1.12 x 104V 0.360 x 104V (b) 0.760 x 104V 7.6 x 103 V How much work to bring a third point charge of 4.0 C from infinity to P? The work is equal to the potential energy change of the move. U qV W qV W 30.4 x 103 J J 4.0 x 106 C 7.6 x 103 C 3.0 x 102 J Velocity and Electric Fields: We can use all this impressive new knowledge about potential difference, potential energy, and work to solve impressive problems. In fact, you will be required to solve such problems to be successful on the AP Physics Test. Calculate (a) the energy of proton that is accelerated from rest through a potential difference of 120 V and (b) the speed of an electron accelerated from rest through a potential difference of 120 V. 296 Potential energy that the proton gains will be equal to the kinetic energy it will have after it has been accelerated through the field. 1 W K mv 2 2 U qV W v 2qV m J 2 1.60 x 1019 C 120 C 27 1.67 x 10 kg 2qV v m v 1 2 mv qV 2 so 2 8m 229.9 x 10 2 s 15 x 104 m s 1.5 x 103 m s Through what potential difference would an electron need to accelerate to achieve a velocity of 1.80 x 107 m/s? We employ the same reasoning as in the previous problem: the work is the same as the potential energy that goes into the system. We can then set that equal to the kinetic energy that the electron ends up with after its been accelerated through the field. 1 2 mv qV 2 U qV W mv 2 V 2q Solve for potential difference m 9.11 x 10 kg 1.80 x 107 s 19 2 1.60 x 10 C 31 2 9.22 x 102 V An electron is 3.00 cm from the center of a uniformly charged sphere of radius 2.00 cm. The sphere’s charge is 1.00 x 10 –9 C. How fast will the electron be traveling when it hits the surface of the sphere? Conservation of Energy: Ui K f ke q1q2 1 2 mv 2 r 2keq1q2 v rm 297 2 9 kg m m 2 8.99 x 10 1.60 x 1019 C 1.00 x 109 C 2 2 s C v 0.0300 m 0.0200 m 9.11 x 1031kg v 2 12 m 315.8 x 10 2 17.8 x 106 s m s 1.78 x 107 m s An electron is fired into at the midpoint of a field between two charged plates. The initial velocity of the electron is 5.6 x 106 m/s. The plates are 2.00 mm apart. The V for the plates is 100.0 V. (a) Find the magnitude of the electric field. (b) Make a drawing of the two plates and put in some arrows showing the direction of the field. (c) Determine where the electron will hit on the upper plate. (a) We need to find the magnitude of the field. We can use the equation for the electric field to find the field. We’re ignoring the minus sign as we’re only interested in the magnitude of the field. E V d 100 V 0.0020 m 5.00 x 104 D V m (b) From the drawing, we see that the field is oriented so that the force on the electron is upward. This means that the field must point downwards. Recall that the direction of the field is the direction that a positive test charge would be deflected. A negative charge would go in the opposite direction. (c) An upward force is exerted on the electron as it travels through the field. We can find the force from the field: E F q N F Eq 5.00 x 104 1.60 x 1019 C C 8.00 x 1015 N Since we know the force acting on the electron, we can use the second law to find the acceleration. F ma 8.00 x 10 a F m 15 kg m a s2 9.11 x 1031 kg m 0.878 x 1016 2 s m 8.78 x 1015 2 s 298 Note -- acceleration of gravity is insignificant compared with the acceleration form the field, so we can ignore it. Next we find the time it takes for the electron to hit the upper plate. Since it’s in the center of the plates when it enters, it will be accelerated a vertical distance of 1 mm. 1 y at 2 2 2y t a y is 1.00 mm or 10-3 m 2 1.00 x 103 m m 8.78 x 1015 2 s 0.2278 x 1018 s 2 0.477 x 109 s We know the time till the electron hits, we also know its initial horizontal velocity, and we know that it will travel horizontally at a constant speed. So knowing the time and the speed, we can find the horizontal distance it travels before the electron hits. x v xt 5.6 x 106 m 0.477 x 109 s s 2.7 x 103 m 2.7 mm Equipotential Curves: Equipotential curves can be drawn for a charge. For example if we look at a positively charged conducting sphere, the lines of force would look like this: Now if we draw a curve around the sphere that represents places that have the same potential difference, we get a sphere. It will be a circle on our two dimensional drawing. This is because the voltage only depends on the distance from the charge. The charge is, of course, evenly distributed on the sphere. The voltage drops off with distance, so the curves nearest the surface of the sphere would have the greatest potential. The equation for potential difference is simply: V 1 4 0 q ri i i 299 Which simplifies to: 1 q V 4 0 r As the distance (r) increases, the potential difference decreases. So if we draw equipotential curves that represent a specific voltage difference – say each curve represents a decrease of 5 volts, it would look like this: 300 AP Physics – More Electric Fields - 6 We have learned how to calculate the force that acts on a test charge that is near two or more charges. We can also find the potential difference. But what about the electric field? q1 q2 P - r2 r1 What is the electric field at point P between the two charges? V The voltage at P from the charges is: 1 q 4 0 r E The electric field is related to the potential difference by: V d Let’s solve this equation for the potential difference: E V d V Ed V Er V 1 q 4 0 r the distance d is simply r, so: We can plug this into the equation for potential difference above: Er 1 q 4 0 r E 1 q 4 0 r 2 We can draw the two electric field vectors: q1 E1 E2 P q2 - Let’s put some numbers to the thing. Let q1 = 22.5 C and q2 = -35.5 C. r1 = 22.0 cm and r2 = 42.5 cm. Find the electric field at P. 301 1 q1 E1 4 0 r12 E1 1 q1 4 0 r12 2 6 9 Nm 22.5 x 10 C 8.99 x 10 2 2 C 0.22 m N 4179 x 103 C Nm 2 33.5 x 106 C 8.99 x 109 2 C 0.425 m 2 N 1667 x 103 C Since the two vectors are in opposite directions, we can just add them: E E1 E2 4179 x 103 N N 1667 x 103 C C 2510 Two charges are situated near point P. The angle is 33. q1 = 0.025 C and q2 = 0.12 C. Find (a) the potential difference at P? (b) The electric field strength at point P? N C P 3.0 m (a) We can find the voltage (potential difference) at P from each of the two charges, then add them up. V 1 q 4 0 r V1 q1 2.5 m r1,2 r q2 1 q1 4 0 r1 2 1 1 q1 9 Nm 6 8.99 x 10 0.0749 x 103V V1 0.025 x 10 C 2 4 0 r1 C 3.0 m 74.9 V We need to find the distance from P to q1. r r1,2 sin 2.5 m sin 33o 1.36 m 2 1 9 Nm 0.793 x 103V V2 8.99 x 10 0.12 x 106 C 2 C 1.36 m V V1 V2 74.9 V 793 V 793 V 870 V (b) Finding the field is easy too. First we find the field strength produced by each charge at point P. These will be vectors, so after we find them we have to add them up using vector addition. The two electric field vectors look something like this: 302 E2 P 3.0 m q1 E1 E1 2.5 m E r1,2 r E2 q2 E V d E2 E1 793 V 1.36 m V1 d1 74.9 V 3.0 m 583 25.0 V m V m Now we can add the vectors. We need to look at the old x and y component deal. E2 We’ll have up and to the left as the positive direction. E2 sin E2 cos Ex E1 E2 cos E y E2 sin E1 V V 25 sin 33o 13.7 m m V V V 25.0 583 608 m m m E y E2 sin For y: For x: E1 E2 cos Now we can find the magnitude of the electric field strength: 2 E Ex Ex 2 2 2 V V 608 13.7 m m 610 V m For the direction of the vector, we can use the tangent function with the x and y components. tan Ey Ex V m V 608 m 13.7 1.3o 303 AP Physics - Capacitance We’ve already looked at capacitors when we talked about Leyden jars. The Leyden jar being, as mentioned before, the very first capacitor. Capacitor technology has evolved quite a bit since the days of Ben Franklin and they are no longer made with jars and metal foil. A capacitor is simply two conductors separated by an insulator. The insulator is given a fancy name - dielectric. The dielectric can be almost anything that doesn’t conduct electricity – paper, plastic, rubber, wax, air, etc. Capacitors can be very small or very large. Most of them are made up from metal foils separated by thin plastic or paper sheets. These things are then rolled up and put into little metal cans (for the bigger ones) or covered with a glob of plastic or ceramic if they are small. Capacitors store charge and are very useful in electronics. They are rated on their capacitance, which is the ratio of stored charge to the potential difference. C Q V Equation for capacitance. C is the capacitance, Q is the stored charge, and V is the potential difference across the two plates. The unit for capacitance is the farad whose symbol is F. The farad is named after a famous electricity physicist (whose discoveries we have not yet gotten to, but, have no fear, we will), Michael Faraday. A farad is a Coulomb per volt. 1 F 1 C V The farad is a very large amount of capacitance, so when you look at a capacitor, you don’t see farads listed, mostly you see F and pF. 1 F 106 F 1 pF 1012 F A 3.0 pF capacitor is connected to a 12 V battery. What is the charge on the capacitor? C Q V Q CV 3.0 x 1012 F 12 V 3.6 x 1011C 304 Finding Capacitance: The equation for capacitance when the two plates are separated by air is: C 0 A d Where C is the capacitance, 0 is the permittivity of free space, A is the area of the capacitor’s plates, and d is the distance between the plates. The value for the permittivity of free space, which has to do with how well lines of force travel through air is: 0 8.85 x 10 C2 12 N m2 A parallel plate capacitor has an area of 2.00 cm2. Plate separation is 0.100 mm. capacitance? First we will convert the area to square meters: 2 2 4 2 2.00 cm 2 2.00 x 10 m 10 cm 1m Next we will convert the plate separation to meters. 1m 3 1.00 mm 3 1.00 x 10 m 10 mm Now we can use the capacitance equation to figure out the answer. A C 0 d 2 2.00 x 104 m 2 12 C 8.85 x 10 2 3 N m 1.00 x 10 m C 17.7 x 1013 F 1.77 x 1012 F 1.77 pF Equations are very powerful things. We can look at them and make predictions and figure out what is really going on. We can do this with the capacitance equation. C 0 A d What happens to capacitance if the area of the plates increases or gets smaller? What happens if the plates are moved closer together? Moved farther apart? 305 Energy Stored in Capacitor: Capacitors store energy. We can think of them as potential energy devices. When you first put a capacitor into a circuit, the device has no charge. If electricity is allowed to flow, electrons will begin to charge up the capacitor. It takes time for the charge to build up. Eventually the capacitor will have all the charge that it can hold and electricity will stop flowing. The capacitor will have maximum potential energy at this point and store its maximum amount of Uncharged capacitor Capacitor charging Electricity flows Charged capacitor no electricity flows charge. When electrons first begin to flow, they pile up on one of the plates. This plate becomes more and more negative as the electrons collect. The electrons on the plate across the dielectric are repelled by the strong negative charge on the oppostite plate, so they begin to flow. The current direction is the same on either side of the capacitor. As more and more electrons collect on the negative plate, fewer and fewer free electrons remain on the other plate. It has a positive charge that gets bigger and bigger as time goes by. As the negative charge increases, the flow of electrons slows down. The flow of electrons on the other side of the capacitor slows down as well as there ain’t many electrons left on the postive plate to flow. Eventually the negative plate has as many electrons as it can possibly hold and the flow of electrons stops. The capacitor is completely charged. This is exactly what happened with the Leyden jar. The charged capacitor has a large amount of charge. If we give the thing a path for the electrons to flow, all the charge will immediately go rushing out. Remember the spark when the Leyden jar discharged? Capacitors are used to deliver big amounts of sudden charge. They are used in flash units for camera. A small little battery puts charge into a large capacitor. When you take a picture, all the stored charge is immediately released to a high voltage xenon bulb that makes a very bright flash of light. The potential energy stored in the capacitor is the product of the charge and the voltage (potential difference). The average potential energy in the capacitor is half of the total amount of energy it will have when fully charged, so we can write: 1 U QV 2 Average potential energy stored in capacitor. We know that the charge is related to the capacitance by: 306 C Q V Q CV We can plug this into the potential energy equation: 1 U QV 2 1 CV V 2 1 CV 2 2 So we have two equations we can use for potential energy stored in a capacitor: 1 U QV 2 1 CV 2 2 This is the form of the equations that you will see on the equation sheets for the AP Physics Test. A 12.5 F capacitor is connected across a 9.0 V battery. Find the potential energy stored in the thing. 1 U CV 2 2 1 2 12.5 x 1012 F 9.0 V 2 5.1 x 1010 J 307 AP Physics – Charge Wrap up Quite a few complicated equations for you to play with in this unit. Here them babies is: F 1 q1q2 4 0 r 2 This is good old Coulomb’s law. You use it to calculate the force exerted by two charges on each other. The 1 term is basically a constant. 4 0 It’s called Coulomb’s constant. (Usually most authorities assign the thing, Coulomb’s constant, a single symbol, k, but the College Board in its tremendous wisdom does not do that. So we have this very awkward multi-value term.) E F q This is the definition of the electric field. E is the field strength. F is the force exerted on a test charge q divided by the test charge. An important point - q is not the charge that created the field. U E qV 1 q1q2 4 0 r This equation gives you the electric potential energy. E Avg V d This equation relates the potential difference V with the distance d a charge is moved within an electric field E. V 1 4 0 q ri i i This equation provides us with the potential difference V caused at a distance r from a field established by a charge q. Potential difference is a scalar, so it is the simple algebraic sum of all the potential differences that are acting on a point from difference charges. The main difficulty with doing electric fields, forces, potential energy, and potential difference is that there are a lot of terms and they can be confusing. So its really 308 important to know all this stuff cold. Somebody says, “Potential energy in a field.” And you are immediately thinking U qV . Next, as has been our custom, are the things that the ever-wise College Board folks who produce the AP curriculum require you, the proud student, to do. A. Electrostatics 1. Charge, Field, and Potential a. You should understand the concept of electric field so you can: (1) Define it in terms of the force on a test charge. This is just an application of E F q . (2) Calculate the magnitude and direction of the force on a positive or negative charge placed in a specified field. Similar to the one above. The force is found using E F . The direction is figured out q logically using the old like charges repel unlike charges attract deal. (3) Given a diagram on which an electric field is represented by flux lines, determine the direction of the field at a given point, identify locations where the field is strong and where it is weak, and identify where positive or negative charges must be present. Simple stuff. Basically pie. The direction of the force is the direction of the force on a positive test charge and is tangent to the line of force. The arrows point in the direction of the force for a positive charge. The field is strong where the lines of force are close together. The force is weak where the lines of force are far apart. The lines of force come out of a charged body. If the lines are pointing away, the body is positive if the lines of force are pointing towards the body it’s negative. (4) Analyze the motion of a particle of specified charge and mass in a uniform electric field. Same kind of stuff as the one above, except we throw in Newton’s laws. So you find the force using Coulomb’s law, then figure out the acceleration using the second law (F = ma), and then the velocity. You can also use conservation of energy. You can also use energy. The potential energy of the field is going to be equal to the kinetic energy of the particle. We did several such problems. b. You should understand the concept of electric potential so you can: 309 (1) Calculate the electrical work done on a positive or negative charge that moves through a specified potential difference. The idea here is that the work done when moving a charge is equal to the change in its potential energy. You use the equations for potential energy. (2) Given a sketch of equipotentials for a charge configuration, determine the direction and approximate magnitude of the electric field at various positions. This would be a set of curves around a charge. Any point on the curve has the same potential difference as any other point on the curve. Pretty simple stuff. Use the lines of force to figure out the direction. (3) Apply conservation of energy to determine the speed of a charged particle that has been accelerated through a specified potential difference. Find the potential energy, assume it is all turned into kinetic energy and then solve for the velocity. 1 K mv 2 . We did several of these problems. 2 (4) Calculate the potential difference between two points in a uniform electric field, and state which is at the higher potential. Use the equation for electric field strength, potential difference, and distance - E Avg V . Calculate the potential for each point. Then recall that potential difference d is a scalar, so you just add them algebraically to find their difference. 2. Coulomb’s Law and Field and Potential of Point Charges a. You should understand Coulomb’s Law and the principle of superposition so you can: (1) Determine the force that acts between specified point charges, and describe the electric field of a single point charge. You use Coulomb’s law to find the force. The electric field around a single point charge is simple – arrows point out for a positive charge and inward for a negative charge. (2) Use vector addition to determine the electric field produced by two or more point charges. This is where you add up the fields between the charges. This is a vector addition deal, so you have to resolve the vectors into x and y components. Remember the joy of that? b. You should know the potential function for a point charge so you can: (1) Determine the electric potential in the vicinity of one or more point charges. 310 This is where you use the equation for the potential difference in a field, good old V 1 4 0 q ri i . Anyway, pretty simple stuff. The potential difference is a scalar, so i you just add ‘em up. 3. Fields and Potentials of Other Charge Distributions a. You should know the fields of highly symmetric charge distributions so you can: (1) Describe the electric field of: (a) Parallel charged plates. E Basically the field looks like this: To describe the field, the Physics Kahuna would say that the lines of force are perpendicular to the surface of the plate and are all parallel and evenly spaced. Therefore the field is uniform (except at the ends). The arrows should point from positive to negative. B. Conductors 1. Electrostatics with Conductors a. You should understand the nature of electric fields in and around conductors so you can: (1) Explain the mechanics responsible for the absence of electric field inside a conductor, and why all excess charge must reside on the surface of the conductor. The Physics Kahuna did a lovely job of explaining all this in the handout. See the section on electric fields. (2) Explain why a conductor must be an equipotential, and apply this principle in analyzing what happens when conductors are joined by wires. All points in a conductor have the same potential difference. This is because the charge is free to move about – there is nothing to stop their moving about as much as they like. Join wires together and they will still have the same potential difference. (3) Determine the direction of the force on a charged particle brought near an uncharged or grounded conductor. This simply has to do with the old basic law of static electricity like charges repel and unlike charges attract. A positively charged particle will be attracted to the grounded conductor, 311 which it will polarize. The same thing will happen to a negatively charged particle. It too will polarize the conductor. Pretty much the same thing will happen to the uncharged conductor. b. You should be able to describe and sketch a graph of the electric field and potential inside and outside a charged conducting sphere. The electric field surrounding a charged conducting sphere (positive charge let’s say) would look like this: Positively charged object A negative charge on the sphere would be the same except that the arrows would be towards the sphere. The electric field and potential within the sphere would be zero – no field, no lines of force. c. You should understand induced charge and electrostatic shielding so you can: (1) Describe qualitatively the process of charging by induction. You have to admit that the old Physics Kahuna did a bang up job of explaining all this to his beloved students. See the handout. (2) Determine the direction of the force on a charged particle brought near an uncharged or grounded conductor. Didn’t we do this one already? Good thing that you don’t have to do much isn’t it? 312 Nasty Old AP Test Questions: From 2001: Four charged particles are held fixed at the corners of a square of side s. All the charges have the same magnitude Q, but two are positive and two are negative. In Arrangement 1, shown to the right, charges of the same sign are at opposite corners. Express your answers to parts (a) and (b) in terms of the given quantities and fundamental constants. (a) For Arrangement 1, determine the following. i. The electrostatic potential at the center of the square. (This means potential difference.) 1 q V 4 o r Potential difference from one charge: Finding r: 2r 2 s 2 s 2 1 q V 4 o s 2 r s 2 2 Q 4 o s Potential difference at a point is the algebraic sum of all the potentials acting at that point; potential difference is a scalar: 2 Q 2 Q 2 Q 2 Q Vtot 4 s 4 s 4 o o o s 4 o s 0 a +Q The potential difference is zero. The four values add up to zero. ii. The magnitude of the electric field at the center of the square. The electric field from each of the charges is a vector. The vectors appear as shown. You can see that the sum of these vectors is zero. Therefore, the electric field in the center of the square is zero. The bottom two charged particles are now switched to form Arrangement 2, shown to the right, in which the positively charged particles are on the left and the negatively charged particles are on the right. c -Q Ed Eb Ec Ea -Q +Q +Q d s s +Q b -Q 313 s s -Q (b) For Arrangement 2, determine the following. i. The electrostatic potential at the center of the square. 1 Q 1 Q 1 Q 1 Q Vtot 2 2 2 2 4 o s 4 o s 4 o s 4 o s 0 Once again, the voltage in the center is zero. ii. Find the magnitude of the electric field at the at the center of the square. E A ED 2 EB EC 2 E E A ED EB EC But a -Q +Q b So: 2 EB EC E Eb 2 To find E, we need the voltage: V E d -Q +Q c Ea d Ec E Ed We already figured out the voltage in the first part of the problem, so we can plug that in. Remember that the distance was: r s 2 EB V d 2 Q Vb 4 o s 1 Q 1 2Q 2 4 o s 2 4 o s s 2 Q 2 o s 2 EC will have same magnitude. 314 E 2 EB EC 2Q E 2 2 2 o s 2 2 Q Q 2 2 2 o s 2 2 o s Q 2 2 o s 2 2 2 Q2 2 o s 2 4 Q 2 o s 2 (c) In which of the two arrangements would more work be required to remove the particle at the upper right corner from its present position to a distance a long way away from the arrangement? ___Arrangement 1 ___Arrangement 2 Justify your answer. Arrangement 1 would require more work to move the upper right particle a great distance away. The Physics Kahuna’s reasoning? The removal of a negative particle is resisted because it’s close to a positive charge and it is helped because of its close distance to a negative charge. Arrangement 2 moves a positive charge farther away from the upper right particle and moves a negative charge closer to the upper right particle, both actions reducing the work needed to remove the upper right particle. Therefore, the arrangement 1 will require more work. 315 From 1999: In a television set, electrons are first accelerated from rest through a potential difference in an electron gun. They then pass through deflecting plates before striking the screen. a. Determine the potential difference through which the electrons must be accelerated in the electron gun in order to have a speed of 6.0 x 107 m/s when they enter the deflecting plates. UE K mv 2 1 2 qV mv V 2q 2 2 m 9.11 10 kg 6 107 s 2 1.6 1019 C 31 1.0 104V The pair of horizontal plates shown below is used to deflect electrons up or down in the television set by placing a potential difference across them. The plates have length 0.040 m and separation 0.012 m, and the right edge of the plates is 0.50 m from the screen. A potential difference of 200. V is applied across the plates, and the electrons are deflected toward the top screen. Assume that the electrons enter horizontally midway between the plates with a speed of 6.0x107 m/s and that fringing at the edges of the plates and gravity are negligible. b. Which plate in the pair must be at the higher potential for the electrons to be deflected upward? Check the appropriate box below. Justify your answer. Upper plate Lower plate The electrons are attracted to the upper positive plate. The positive plate must be at the higher potential and the negative plate is ground. c. Considering only an electron’s motion as it moves through the space between the plates, compute the following. i. The time required for the electron to move through the plates. v x t t x 0.040 m v 6 107 m s 6.7 1010 s ii. The vertical displacement of the electron while it is between the plates. E qV a md 1 y at 2 2 F q E F qE 1.6 10 9.1110 V d ma q kg 0.012 m m 6.7 10 s s 19 C 200 V 2.9 1015 31 1 2.9 1015 2 10 2 V d 2 m s2 6.51 104 m 316 Show why it is a reasonable assumption to neglect gravity in part (c). The force of electricity is billions and billions of times stronger. Gravity is negligible by comparison. e. Still neglecting gravity, describe the path of the electrons from the time they leave the plates until they strike the screen. State a reason for your answer. Moves in a straight line due to inertia. The electric field no longer acts on the particle. Gravity is too weak to cause a measurable deflection. d. From 1998: A wall has a negative charge distribution producing a uniform horizontal electric field. A small plastic ball of mass 0.010 kg, carrying a charge of –80.0 μC, is suspended by an uncharged, nonconducting thread 0.30 m long. The thread is attached to the wall and the ball hangs in equilibrium, as shown below, in the electric and gravitational fields. The electric force on the ball has a magnitude of 0.032 N. t FE w a. On the diagram below, draw and label the forces acting on the ball. See above right b. Calculate the magnitude of the electric field at the ball’s location due to the charged wall, and state its direction relative to the coordinate axis shown. E F q 0.032 N 80 106 C 400 N C The negative x direction. The electric field direction is the direction a positive test charge would go. Since the negative charge moves right, the field must be to the left. c. Determine the perpendicular distance from the wall to the center of the ball. FE 0.032 N m w mg 0.010 kg 9.8 2 0.098 N s t FE 2 w2 0.032 N 2 0.098 N 2 t l FE r 0.103 N w 317 The angle made by the tension force with the vertical is the same as the angle formed by the string with the vertical. This is shown in the drawing. We can solve for using trig. Once we know we can find r since we know l. d. sin FE t sin r l 0.032 N 0.103 N r l cos 18.1o 0.30 m sin 18.1o 0.093 m The string is now cut. i. Calculate the magnitude of the resulting acceleration of the ball, and state its direction relative to the coordinate axis shown. The tension disappears, but w and FE are still present. The net force will be the sum of these two forces. Since they are perpendicular, we can use the good old Pythagorean theorem to find the net force. F FE 2 w2 0.032 N 2 0.098 N 2 0.103 N We now use the second law to find the acceleration acting on the ball: F ma a F m 9.8 m s 2 tan 3.2 m s 2 0.103 N 0.010 kg 10.3 m s2 72o , below the x axis ii. Describe the resulting path of the ball. Straight to the right and down. If t is missing from the FBD two forces are still acting. So the net force is down and to the right FE w 318 From 1996: Robert Millikan received a Nobel Prize for determining the charge on the electron. To do this, he set up a potential difference between two horizontal parallel metal plates. He then sprayed drops of oil between the plates and adjusted the potential difference until drops of a certain size remained suspended at rest between the plates, as shown to the right. Suppose that when the potential difference between the plates is adjusted until the electric field is 10,000 N/C downward, a certain drop with a mass of 3.27 x 10-16 kg remains suspended. a. What is the magnitude of the charge on this drop? E F q F Eq F ma m 3.27 x 1016 kg 9.8 2 s q N 104 C b. Eq ma q ma E 32.05 x 1016 N N 104 C 3.20 x 1019 C The electric field is downward, but the electric force on the drop is upward. Explain why. The electric field is oriented in reference to a positive test charge. A positive charge would move downward. However, a negative charge would experience the opposite effect, or a force upward. The drop must be negatively charged. c. E Avg d. If the distance between the plates is 0.01 m, what is the potential difference between the plates? V d ignore the s ign E V d V Ed 10 000 V 0.010 m m 100 V The oil in the drop slowly evaporates while the drop is being observed, but the charge on the drop remains the same. Indicate whether the drop remains at rest, moves upward, or moves downward. Explain briefly. Originally the drop did not move since gravity and the electric force were in equilibrium. But, if the drop looses mass then the force of gravity gets smaller. Since the charge does not change the upward electric force, is greater than the downward force. This unbalanced force will cause the drop to accelerate upward. 319