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FYI: School A must Topic assume that the force "signal" travels infinitely fast, 6.2 Extended to react to relative motion between the masses. B – Electric field FYI: School B assumes that the force "signal" is already in place in the spacenormal surrounding the source of the force.force, Thus thetension force signal The force, the friction anddoes drag NOTall haveclassified to keep traveling the receiving mass. occurring where are as to contact forces, two eachthat other. FYI: objects Relativity contact has determined the absolute fastest ANY signal can The gravitational electric force, propogate is c = 3.0108force, m/s, the and speedthe of light. however, do not need objects to be in contact (or even close proximity). These two forces are sometimes called action at a distance forces. There are two schools of thought on action at a distance. School A: The masses know where each other are at all times, and the force is instantaneously felt by both masses at all times. School B: The masses deform space itself, and the force is simply a reaction to the local space, rather than the distant mass. Topic 6.2 Extended B – Electric field Thus School B is currently the "correct" view. If, for example, the gravitational forces were truly "action at a distance," the following would occur: SUN c c The planet and the sun could not "communicate" quickly enough to keep the planet in a stable orbit. FYI: You have probably seen such a model at the museum. A coin is Topicto be 6.2 Extended allowed to roll, and it appears in orbit. B – isn't Electric field FYI: Of course, if the slope just right to match the tangential speed, the coin will spiral into the central maw. Instead, look at the School B view of the space surrounding the sun: The planet "knows" which way the the the to "roll" because of local curvature of space surrounding sun. Long distance communication is not needed. FYI: The view of School B is called the FIELD VIEW. Topic 6.2 Extended B – Electric field Of course, if the mass isn't moving, it will roll downhill if placed on the grid: m0 m0 Depending on where we place the "test" mass m0, it will roll differently. Question: Why is the arrow at the location of the first test mass smaller than that at the second one? Topic 6.2 Extended B – Electric field We can assign an arrow to each position surrounding the sun, representing the direction the test mass will go, and how big a force the test mass feels. We represent fields with vectors of scale length. In the case of the gravitational field, the field vectors all point toward the center: Topic 6.2 Extended B – Electric field If we view our gravitational field arrows from above, we get a picture that looks like this: What the field arrows tell us is the magnitude and the direction of the force on a particle placed anywhere in space: The blue particle will feel a "downward" force. The red particle will feel SUN FYI: We don't even have to draw the object that is creating the field. a "leftward" force whose magnitude is LESS than that of the blue particle. 6.2 SINKS. Extended FYI: Inward-pointing Topic fields are called Think of the field arrows as water flowing into B a hole. – Electric field FYI: Test charges are by convention POSITIVE. Therefore, field Consider, now, charge a vectors For masses, weshow the direction around aall charge a POSITIVE would negative charge: have is an attractive want to go if placed in the field. force. SUN - All of the field If we place a very small arrows point inward. POSITIVE test charge in the vicinity of our negative charge, it will be attracted to the center: We can map out the field vectors just as we did for the sun. Topic 6.2 Outward-pointing fieldsan areELECTRIC calledExtended SOURCES. of the FYI: We call such a system DIPOLE. Think A dipole hasfield two arrows(charges) as water which flowing of a fountain. poles in sign. field Bareout –opposite Electric Question: T or F: Gravitational fields are sinks. FYI: The ELECTRIC DIPOLE consists ofalways a source and a sink. We can place a negative Question:isT or F: Electric fields are always sinks. charge positive. charge here. Now suppose our + + - If we place a very small The fields of the two POSITIVE test charge in the vicinity of our positive charge, it will be repelled from the center: We can map out the field vectors just as we did for the negative charge. charges will interact: Keep in mind that the field lines show the direction a POSITIVE charge would like to travel if placed in the field. A negative charge will do the opposite: Topic 6.2 Extended B – Electric field You may have noticed that in the last field diagram some of our arrows were unbroken. We can draw solid field arrows as long as we note that THE CLOSER THE FIELD LINES ARE TO ONE ANOTHER, THE STRONGER THE FIELD. + - FYI: There are some simple rules for drawing these solid "electric lines of force." Rule 1: The closer together the electric lines of force are, the stronger the electric field. strong Rule 2: Electric lines of force originate on positive charges and end on negative charges. weak Rule 3: The number of lines of force entering or leaving a charge is proportional to the magnitude of that charge. Topic 6.2 Extended Question: Which charge is the strongest negative one? D Question: Which charge weakest positive one? E B –is the Electric field Question: Which charge is the strongest positive one? A Determine the sign and the magnitude of the Question: by Which charge at is the weakest negative one? ofC force. charges looking the electric lines A B C D E F Topic 6.2 Extended B – Electric field So how do we define the magnitude of the electric field vector E? Here's how: E = F on q q0 Electric field definition o where q0 is the positive test charge Think of the electric field as the force per unit charge. To obtain a more useful form for E, consider the force F between a test charge q0 and an arbitrary kq0q charge q: F = r2 kq F on q = kq0q E = then = q0 q0r2 r2 so that kq Electric field in space E = r2 surrounding a point charge o Topic 6.2 Extended B – Electric field What is the magnitude of the electric field strength two meters from a +100 C charge? Use kq E = r2 9109·10010-6 = 22 = 225000 n/C What is the force acting on a +5 C charge placed at this position? From E = so that F on q q0 o we see that F = qE = 510-6·225000 = 1.125 n F = qE Force on a charge placed in an electric field Topic 6.2 Extended B – Electric field What is the electric field at the chargeless corner of the 2-meter by 2-meter square? +10 C A EB EA EC C C B Start by labeling the charges (organize your effort): kq kq kq |E | = |EA| = |E | = C B r2 r2 r2 9109·1010-6 9109·1010-6 9109·1010-6 = = = 22 22 ( 22+22 )2 = 22500 n/C = 22500 n/C = 11250 n/C +10 C -10 Now sketch in the field vectors: Now sum up the field vectors: E = EA + EB + EC = 22500i + 11250 cos 45i + 11250 sin 45j - 22500j = 30455i - 14545j (n/C) Topic 6.2 Extended FYI: The dipole electric field drops off by 1/x3 rather than inverse square like a "monopole." B – Electric field What is the electric field along the bisector of a distant dipole having a charge separation d? A dipole is an equal positive and negative charge in close proximity to one another: +q r d 2 d 2 x r -q kq r2 Note that the horizontal components cancel, and the vertical components add: |ETOTAL| = 2|E| sin . Both field vectors have the same length: But sin = (d/2)/r = d/2r so that |E| = kqd r3 But r = (d/2)2 + x2 . Since x >> d, r x and we have |ETOTAL| = 2|E| sin = EDIPOLE = kqd x3 Electric field far from a dipole