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Day 9: Dipole moments and Electric Flux Motion of a Charged Particle in a Uniform Electric Field When a particle of charge q and mass m is placed in an electric field E, the electric force exerted on the charge is qE. If that is the only force exerted on the particle, it must be the net force, and it causes the particle to accelerate according to the particle under a net force model. Therefore, Fe= qE = ma and the acceleration of the particle is 𝑎= 𝑞𝐸 𝑚 Given a pair of plates one of which is + charged and the other is negatively charged. A electron is fired horizontally between the two plates as shown. Draw the electric field and the trajectory of the electron. What 2A topic does this remind you of? Electric Dipole in an Electric Field We defined an electric dipole as consisting of two charges of equal magnitude and opposite sign separated by a distance 2a as shown at right. We want to define a quantity that represents the electrical polarity of a of a system of charges which we will call the Electric Moment. The electric dipole moment is defined as the product of the magnitude of charge and the distance of separation between the charges. In this case it is the vector p directed from -q toward +q along the line joining the charges and having magnitude p = 2aq Now suppose an electric dipole is placed in a uniform electric field E and makes an angle with the field as shown at right. We identify E as the field external to the dipole, established by some other charge distribution, to distinguish it from the field due to the dipole. Each of the charges is modeled as a particle in an electric field. Make a sketch showing the force on each of these particles and include the net force on the dipole. Therefore, the net force on the dipole is zero. Will there be torque acting on this system? Remember that there is some angle between the F and the moment vector as shown at right. Write three equations: 1. The torque on the positive charge due to the electric field 2. The torque on the negative charge due to the electric field 3. The net torque on the dipole. The two forces produce a net torque on the dipole, however; the dipole is therefore described by the rigid object under a net torque model. As a result, the dipole rotates in the direction that brings the dipole moment vector into greater alignment with the field. The torque due to the force on the positive charge about an axis through O in has magnitude Fa sin , where a sin is the moment arm of F about O. This force tends to produce a clockwise rotation. The torque about O on the negative charge is also of magnitude Fa sin ; here again, the force tends to produce a clockwise rotation. Therefore, the magnitude of the net torque about O is = 2Fasin Now write this torque as a function of q, E, and a. Note several of these things are difficult to measure. However we can use our definition of the diplole moment as p = 2aq to simplify this further. = pEsin Based on this expression, it is convenient to express the torque in vector form as the cross product of the vectors p and E: We can also model the system of the dipole and the external electric field as an isolated system for energy. Let’s determine the potential energy of the system as a function of the dipole’s orientation with respect to the field. To do so, recognize that work must be done by an external agent to rotate the dipole through an angle so as to cause the dipole moment vector to become less aligned with the field. The work done is then stored as electric potential energy in the system. Notice that this potential energy is associated with a rotational configuration of the system. Previously, we have seen potential energies associated with translational configurations: an object with mass was moved in a gravitational field, a charge was moved in an electric field, or a spring was extended. Now we want the work associated with a rotation. Write an expression for the work associated with a torque: dW = d Our expression for torque previously was = pEsin Plug this into the definition above and integrate it between an angle of i and f to find an expression for the work done in rotating the dipole. Note we have written this as a change in energy due to the work energy theorem. Lets look at some convenient limits: Assume that you start with Ui = 0 at an angle of 90o. Now write an expression for Uf. Uf = -pE cos If you were particularly clever, you noticed that this is just the same as the dot product between the moment and the electric field. Lets compare this to the expression for the potential energy: mgh. Rewriting mhg, we notice that the definition of the moment matches that from potential energy, with the Electric field replacing the gravitational acceleration. In the case of the gravitational field, it is a translational potential, such that a mass will tend to return to the bottom position. For the electric field it is a rotational potential, such that the charge will tend to return to an orientation where the charges are aligned with the field. Molecules are said to be polarized when a separation exists between the average position of the negative charges and the average position of the positive charges in the molecule. In some molecules such as water, this condition is always present; such molecules are called polar molecules. Molecules that do not possess a permanent polarization are called nonpolar molecules. We can understand the permanent polarization of water by inspecting the geometry of the water molecule. The oxygen atom in the water molecule is bonded to the hydrogen atoms such that an angle of 105o is formed between the two bonds. The center of the negative charge distribution is near the oxygen atom, and the center of the positive charge distribution lies at a point midway along the line joining the hydrogen atoms. Washing with soap and water is a household scenario in which the dipole structure of water is exploited. Grease and oil are made up of nonpolar molecules, which are generally not attracted to water. Plain water is not very useful for removing this type of grime. Soap contains long molecules called surfactants. In a long molecule, the polarity characteristics of one end of the molecule can be different from those at the other end. In a surfactant molecule, one end acts like a nonpolar molecule and the other acts like a polar molecule. The nonpolar end can attach to a grease or oil molecule, and the polar end can attach to a water molecule. Therefore, the soap serves as a chain, linking the dirt and water molecules together. When the water is rinsed away, the grease and oil go with it. Shown below in the photograph at the left and in the detail at the right is a device called a Faraday cage. In the detail photograph three suspensions of aluminum foil chunks on pieces of string can be seen. Each of the suspensions include a string passing over the top of the cylindrical screen with small rectangular chunks of aluminum foil, a few centimeters in size, attached to the ends of the string. Note that one end of each string is outside the cylinder and one end of the string is inside the cylinder. The question this week involves what the foils do when the van de Graff generator is turned on, putting large negative charge onto the cylindrical screen and perhaps by conduction onto the foils. In particular, when the cylinder is charged, which, if any, of the aluminum foils will move away from the cylinder, either inward or outward. When the cylinder is charged: (a) only the inner foils will move away from the cylinder. (b) only the outer foils will move away from the cylinder. (c) both the inner and the outer foils will move away from the cylinder. (d) neither the inner nor the outer foils will move away from the cylinder. The answer is (b): only the outer foils will move away from the cylinder, as can be seen in the photograph at the right and in an mpeg video by clicking your mouse on the photograph at the right. In a device like this, the electric charge moves to the outside of the cage, leaving no net charge on the inside. Therefore no effect is felt by the foils on the inside of the cage. This works the same way as a solid metallic cylinder would work, except that using a screen allows us to see inside and observe the effect of charging the cylinder on foils both outside and inside the cage. ELECTRIC FIELD LINES AND FLUX ELECTRIC FIELD LINES You have been representing the electric field due to a configuration of electric charges by an arrow that indicates magnitude and direction; using the principles of superposition and linearity, you can determine the length and direction of the arrow for each point in space. This is the conventional representation of a “vector field.” An alternative representation of the vector field involves defining electric field lines. Unlike an electric field vector, which is an arrow with magnitude and direction, electric field lines are continuous. You can use a simulation written by Cal Tech to explore some of the properties of electric field lines for some simple situations. For this activity you will need: • 1 computer • 1 Coulomb computer simulation http://www.cco.caltech.edu/~phys1/java/phys1/EField/EField.html Activity: Simulation of Electric Field Lines from Point Charges Load the Coulomb program into the computer. Figure out how to turn off everything except electric field. a. Set a single charge of ±1, 2, 3, or 4 Coulombs somewhere on the computer screen and run the program. Sketch the lines in the space below. Show the direction of the electric field on each line by placing arrows on them. Indicate how much charge you used on the diagram. How many lines are there in the drawing? Are the lines more dense or less dense near the charge? Explain. How does the direction of the lines depend on the sign of the charge? b. Try another magnitude of charge. No need to sketch the result, but how many lines are there? Can you describe the rule for telling how many lines will come out of, or start or end on, a charge in this simulation if you know the magnitude of the charge in Coulombs? c. Repeat the exercise using two charges with the same magnitude having unlike signs. Place them at two different locations on the screen. After a few minutes sketch the lines in the space below. Indicate how much charge you used on the diagram. Comment on why the lines are more or less dense near the charges. How does the direction of the lines depend on the sign of the charge? d. Summarize the properties of electric field lines. What does the number of lines signify? What does the direction of a line at each point in space represent? What does the density of the lines reveal? ELECTRIC FLUX As you have just seen, we can think of an electrical charge as having a number of electric field lines, either converging on it or diverging from it, that is proportional to the magnitude of its charge. We can now explore the mathematics of enclosing charges with surfaces and seeing how many electric field lines pass through a given surface. Electric flux is defined as a measure of the number of electric field lines passing through a surface. In defining “flux” we are constructing a mental model of lines streaming out from the surface area surrounding each unit of charge like streams of water or rays of light. Modern physicists do not really think of charges as having anything real streaming out from them, but the mathematics that best describes the forces between charges is the same as the mathematics that describes streams of water or rays of light. So, for now, let’s explore the behavior of this model. It should be obvious that the number of field lines passing through a surface depends on how that surface is oriented relative to the lines. The orientation of a small surface of area A is usually defined as a vector that is perpendicular to the surface and has a magnitude equal to the surface area. By convention, the normal vector points away from the outside of the surface. The normal vector is pictured below for two small surfaces of area A and A' respectively. In the picture it is assumed that the outside of each surface is white and the inside grey. A diagram showing how the normal vector representing an area points from inside to outside. The inside of a surface is shaded grey. Activity: Drawing Normal Vectors Use the definition of “normal to an area” given above to draw normals to the surfaces shown below. Let the 2 length of the normal vector in cm be equal in magnitude to each area in cm . Don’t just draw arrows of arbitrary length. By convention, if an electric field line passes from the inside to the outside of a surface, we say the flux is positive. If the field line passes from the outside to the inside of a surface, the flux is negative. How does the flux through a surface depend on the angle between the normal vector to the surface and the electric field lines? In order to answer this question in a concrete way, you can make a mechanical model of some electric field lines and of a surface. What happens to the electric flux as you rotate the surface at various angles between 0 degrees (or 0 radians) and 180 degrees (or radians) with respect to the electric field vectors? To make your model you will need to arrange nails in a 10 10 array poking up at 1/4" intervals through a piece of Styrofoam. The surface can be a copper loop painted white on the “outside.” You will need: • Styrofoam or 3/8" plywood (5" 5" square) • 100 nails, approximately 4" in length (mounted on the Styrofoam or plywood) • 1 wire loop (4" 4" square) • 1 paper, 5" 5" with 1/4" graph rulings (to affix to the mounting square to help with spacing the nails) • 1 protractor Once the model is made you can perform the measurements with a protractor and enter the angle in radians and the flux into a computer data table for graphical analysis. Apparatus designed to determine how many uniformly spaced flux lines will pass through an imaginary surface area as a function of the angle between the direction of the flux lines and the normal vector representing the surface area. Activity: Flux as a Function of Surface Angle a. Use your mechanical model, a protractor, and some calculations to fill in the following data table, or you may want to feed data directly into your graphing software for later display. Hint: Interested in being lazy but clever? Use symmetry to determine the angles corresponding to negative flux without making more measurements. Degrees Radians Number of lines 49 42 35 28 21 14 7 0 -7 -14 -21 -28 -35 -42 -49 b. Plot the flux as a function of radians. Look very carefully at the data. Is it a line or a curve? What mathematical function might describe the relationship? c. Try to confirm your guess by constructing a spreadsheet model and overlay graph of the data and the mathematical function you think matches the data. Affix or sketch the plots in the space to the right of the data table. Summarize your procedures and conclusions in the following space. d. What is the definition of the vector dot product of two vectors in terms of vector magnitudes andthe angle between them? Can you relate the scalar value of the flux, , to the dot product of the vectors E and A ? In the simulation, the yellow ring is the surface across which we determine the electric flux. The flux is caused by the electric field, the green lines, caused by one or two electric charges. You can grab the edge of the yellow ring and change its shape. If you move the pointer inside the ring and press down on the mouse, you can drag the center of the ring to other positions on the screen. http://media.pearsoncmg.com/bc/aw_young_physics_11/pt2a/Media/Electricity/1107ElecFlux/Main.html Questions 1 – 4. Question 1: Flux Into or Out of an Oval Design your own experiments to see how electric charge affects the flux into or out of the oval. In your experiments, you can: • change the shape of the oval; • move the center of the oval so that it surrounds the charge or does not surround the charge; and • change the magnitude and sign of the electric charge. When finished, develop in words a qualitative rule to determine the electric flux flowing into or out of the oval. Give examples to support your statements. Question 2: Electric Flux with Two Charges In the simulation, click the "two charges" configuration. You can now adjust the sign, magnitude and separation of two electric charges. Repeat the experiments such as done in Question 1 to see if your rule applies for this two charge system. You can: • change the shape of the ring; • change the position of the center of the ring (move it all over); and • change the magnitudes and signs of the electric charges. Question 3: First way to determine electric flux Change the simulation back to "one charge." The meter indicates the electric flux Φ into or out of the oval and the net electric charge Q inside the oval. • What happens to the flux if you double or triple the positive electric charge inside the oval? • What happens to the flux if you double or triple the negative electric charge inside the oval? • Find an equation with a proportionality constant that relates the electric flux into or out of the oval and the electric charge inside the oval. Question 4: Second way to determine electric flux The green electric field lines represent the electric field surrounding the source charges. Develop a rule for the electric flux passing out of or into the oval by counting the electric field lines passing out of or into the oval. A MATHEMATICAL REPRESENTATION OF FLUX THROUGH A SURFACE One convenient way to express the relationship between angle and flux for a uniform electric field is to E use the dot product dA , so that E A . Flux is a scalar. If the electric field is not uniform or if the surface subtends different angles with respect to the electric field must calculate the flux by lines, then we breaking the surface into infinitely many infinitesimal areas dA so that d E A , and then taking the integral sum of all of the flux elements. This gives d E dA (flux through a surface) Some surfaces, like that of a sphere or that representing a rectangular box, are closed surfaces— that is, they have no holes or breaks in them. Because we want to study the amount of flux passing through closed surfaces, there is a special notation to represent the integral of E dA through a closed surface. It is represented as follows: dA E • dA (flux through a closed surface) Question 5: Find the electric flux leaving a sphere that surrounds a +3.0 ?µC point charge. We use the oval shown in Simulation 2 to help in our thinking. The definition for electric flux that we will use is: . where the integral is over a closed surface--a sphere in this example. 5a: Choosing a symmetric Gaussian surface. This integration method is almost always used with surfaces that have considerable symmetry--a sphere in this example. Grab the side of the oval ring and change it into a circle of any radius--as best you can. Center the circle about the point charge. Press down on the pointer when it is outside the circle and move the pointer onto the circle and observe the electric field. Move the pointer to all sides of the ring and observe the electric field. If it is in fact a circle centered on the point charge, the electric field reading should be fairly constant (don't worry about minor variations). If the field varies dramatically on different parts of the "circle", you will either have to reshape the circle or center it better. When done, record the electric field on the circle and leave it as is. 5b: In the flux integral, what is the direction of --a small part of the sphere's surface area? 5c: For the situation shown in the simulation, can you see why surface? for every element on the 5d: Why can you take E outside the integral--the last step below? . 5e: Why can we now simply multiply the magnitude of the electric field at the surface times the total area of the sphere's surface--the last step below? = E (4πr2). Summary of the Integral Method for Calculating Electric Flux: • The direction of dA is perpendicular to the surface and outward from its inside. • Often for the situations in which we calculate electric flux by integration, EdA = EdA because E and dA are parallel. • Often, we choose surfaces on which the electric field has the same magnitude at each point on the surface, which allows us to move the electric field outside the integral sign: . • This leaves only an integral over the area which equals the area of the surface. We get: = E (Area) Question 6: Change the charge to Q1 = 4 µC and pull the yellow ring in the simulation off to the side so that there is no charge inside it. Determine the electric flux leaving or entering the ring in the simulation using the charge inside the ring and the number of flux lines crossing the ring surface methods (Methods 1 and 2). Compare your predictions with the value shown in the simulation. Question 7: Click the "two charges" button and set Q1 = +4 µC and Q2 = -2 µC. Make the yellow oval into a big circle that surrounds Q2 but not Q1 (see Question 7 in the worksheet). Determine the electric flux leaving or entering the oval in the simulation using the charge inside the oval and the number of flux lines crossing the oval surface methods (Methods 1 and 2). Check your predictions against the flux value shown in the meter. Question 8: Change the radius of the Gaussian sphere to R = 15 mm. Use the charge inside and integral methods (Methods 1 and 3) to determine the electric flux leaving the 15 mm-radius sphere surrounding the +10 x 10-10-C charge that is uniformly distributed in the 10 mm-radius solid sphere. The electric field at the surface of the 15-mm radius sphere is 40,000 N/C (we learn later that 1 N/C = 1 V/m). Question 9: Change the charge distributed in the solid sphere to -10.0 x 10-10 C. Repeat the calculations in Question 8 only this time with the 15mm-radius sphere surrounding the -10 x 10-10 C charge. The electric field at the surface of the 15-mm radius sphere is 40,000 N/C and points into the spherical surface. Electric Flux: The Electric Flux E is the product of Component of the Electric Field Perpendicular to a Surface times the Surface Area. (SI: V.m = N.m2/C) Unit Vectors The electric flux becomes negative when the direction of the E-field and the normal to the surface are in opposite directions relative to each other, i.e. when > 90o. Flux over a Closed Surface: A Closed Surface is the surface of some Spacial Volume like a sphere or a cylinder. The volume could have any shape. = Differential Area associated with the surface. Its direction is perpendicular (or normal) to the surface. If you can find a surface over which the E-field is constant then E can be taken out of the surface integral and the electric flux reduces to a simple multiplication of E and the area of the surface. Flux through a Cube Consider a uniform electric field E oriented in the x direction in empty space. A cube of edge length l, is placed in the field, oriented as shown. Find the net electric flux through the surface of the cube. a) What faces have a zero flux? b) Write the integrals for the remaining faces. c) Sum the flux over all six faces. GAUSS’ LAW DISCOVERING GAUSS’ LAW IN FLATLAND How is the flux passing through a closed surface related to the enclosed charge? Let’s pretend we live in a twodimensional world in which all charges and electric field lines are constrained to lie in a flat two-dimensional space— of course, mathematicians call such a space a plane.1 For this project you will need: • 1 computer • 1 Coulomb software simulation† Open the Coulomb program on the computer again and set it to sketch lines for some nutty, creative mix of charges. Don’t be too creative or the lines will take forever to sketch out, limit to 3 charges of low magnitude. You should do the following: 1. Open the Coulomb simulation and place some positive and negative charges at different places on the screen. Then start the program to calculate and display the electric field lines in two dimensions. 2. Either sketch or print out the screen configuration showing the charges and the associated “E-field” lines. If you fail at this, use the image below, noting that the bottom left charge is negative and the other two are positive: * If you haven’t already read it, we recommend that you read E.A. Abbott’s book entitled Flatland; A Romance of Many Dimensions (Dover, New York, 1952). It’s a delightful piece of late nineteenth-century political satire in the guide of a mathematical spoof. † In a two-dimensional map of flux lines it is not possible to assign a fixed number of lines to a charge and to assign spacings between coming from infinity unambiguously. Thus, the line densities may not look proper in some of the Coulomb software plots. This should not matter to students completing this exercise. 3. Draw arrows on each of the lines indicating in what direction a small positive test charge would move. Note: “Small” means that the test charge does not exert enough forces on the charge distributions that create the Efield to cause the field to change noticeably. 4. Figure out what the two-dimensional equivalent of a “closed surface” ought to look like and draw several “closed surfaces” on your diagram. Having done all of this preparation, you should be ready to discover how the net number of lines passing through a surface is related to the net charge enclosed by the surface. Activity: Gauss’ Law in Flatland a. Place a replica of the charge configuration you designed and the associated field lines in the space below. b. Draw some two-dimensional closed “surfaces” in pencil in the space above. Some of them should enclose charge, and some should avoid enclosing charge. Count net flux lines coming out of each “surface.” Note: Consider lines coming out of a surface as positive and lines going into a surface as negative. The net number of lines is defined as the number of positive lines minus the number of negative lines. Charge enclosed by the arbitrary surface +q -q qnet Lines of flux in and out of the surface out in net 1 2 3 c. What is the apparent relationship between the net flux and the net charge enclosed by a twodimensional “surface”? Gauss’ Law in Three Dimensions If you were to repeat the simulated exploration you just performed in a three-dimensional space, what do you think would be the appropriate expression for Gauss’ law? Activity: Statement of Gauss’ Law a. Express the three-dimensional form of Gauss’ law in words. b. Express the law using an equation. Electric Flux over a Closed Surface = Charge enclosed by the Surface divided by o. (o = the permittivity of free space 8.854x10-12 C2/(N m2) = 1/4k) A more intuitive statement: the total number of electric field line entering or leaving a closed volume of space is directly proportional to the charge enclosed by the volume. If there is no net charge inside some volume of space then the electric flux over the surface of that volume is always equal to zero. ELECTRIC FIELDS AND CHARGES INSIDE A CONDUCTOR An electrical conductor is a material that has electrical charges in it that are free to move. If a charge in a conductor experiences an electric field, it will move under the influence of that field since it is not bound (as it would be in an insulator). Thus, we can conclude that if there are no moving charges inside a conductor, the electric field in the conductor must be zero. Let’s consider a conductor that has been touched by a charged plastic rod so that it has an excess of negative charge on it. Where does this charge go if it is free to move? Is it distributed uniformly throughout the conductor? If we know that E = 0 inside a conductor, we can use Gauss’ law to figure out where the excess charge on the conductor is located. Diagram showing how an imaginary Gaussian surface can be constructed just inside the surface of the conductor. Activity: Where Is the Excess Charge in a Metal? a. Consider a conductor with an excess charge of Q. If there is no electric field inside the conductor (by definition), then what is the amount of excess charge enclosed by the Gaussian surface just inside the surface of the conductor? Hint: Use the three-dimensional Gauss’ law here. b. If the conductor has excess charge and it can’t be inside the Gaussian surface according to Gauss’ law, then what’s the only place the charge can be? c. Given the fact that as like charges, the excess charges will repel each other, is the conclusion you reached in part b. above physically reasonable? Explain. Hint: How can each unit of excess charge that is repelling every other unit of excess charge get as far away as possible from the other excess charges on the conductor?