course Phy 202 ‡‚Nç{¯ƒN˜ûÎPÑרt›ûü“DõŸÝzassignment #002
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23:33:34 In your own words explain how the introductory experience with scotch tape illustrates the existence of two types of charge.
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RESPONSE --> The movement of the tape shows that there are two opposing charges acting upon the tape and the attraction or repulsion of that tape allows it movement. confidence assessment: 2
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23:34:41 In your own words explain how the introductory experience with scotch tape supports the idea that the force between two charged particles acts along a straight line through those particles, either attracting the forces along this line or repelling along this line.
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RESPONSE --> The tape only moves in one direction between the two charges. the shortest distance between two points is a straight line, so if the tape moves along essentially a straight line, then there are two particles. confidence assessment: 3
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23:35:42 In your own words explain why this experience doesn't really prove anything about actual point charges, since neither piece of tape is confined to a point.
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RESPONSE --> there are many things influencing the tape itself. just because it moves in the direction we believe there to be a charge does not mean it has to be so. the tape may be in fact reacting to some unknown factor confidence assessment: 2
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23:38:58 If one piece of tape is centered at point A and the other at point B, then let AB_v stand for the vector whose initial point is A and whose terminal point is B, and let AB_u stand for a vector of magnitude 1 whose direction is the same as that of AB_u. Similarlylet BA_v and BA_u stand for analogous vectors from B to A. Vectors of length 1 are called unit vectors. If the pieces attract, then in the direction of which of the two unit vectors is the tape at point A pushed or pulled? If the pieces repel, then in the direction of which of the two unit vectors is the tape at B pushed or pulled? Explain.
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RESPONSE --> If the two pieces attract, the tape at point A will be pulled towards the tape at point be in the direction of AB_u. If they repel, tape B will also travel in the direction of AB_u, away from tape A.
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23:41:14 Using the notation of the preceding question, which you should have noted on paper (keep brief running notes as you do qa's and queries so you can answer 'followup questions' like this), how does the magnitude of the vector AB_v depend on the magnitude of BA_v, and how does the magnitude of each vector compare with the distance between A and B?
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RESPONSE --> If one charge is much more powerful than the other, then it will dominate the displacement of the tape. If tape A is more powerful, and attracts tape B more strongly than tape B attracts tape A, then tape B will travel further in the BA_u direction because A is stronger. If the two charges repel each other, then one may repel more strongly than the other influencing the displacement of the opposing tape. A stronger charge will be active over a longer distance than a weaker charge. confidence assessment: 1
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23:42:37 Using the notation of the preceding question, how is the force experienced by the two pieces of tape influenced by the magnitude of AB_v or BA_v?`aThe expected answer is that the force exerted by two charges on one another is inversely proportional to the square of the distance between them. So as the magnitudes of the vectors, which are equal to the separation, increases the force decreases with the square of the distance; and/or if the magnitude decreases the force increases in the same proportinality. The two pieces of tape are not point charges, so this is not strictly so in this case, as some parts of the tape are closer than to the other tape than other parts.
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RESPONSE --> ok, inversely proportional to the square of the distance between them. as the magnitudes increase, the forces decreases with the square of the distance confidence assessment: 3
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23:43:22 Query introductory set #1, 1-5 Explain how we calculate the magnitude and direction of the electrostatic force on a given charge at a given point of the x-y plane point due to a given point charge at the origin.
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RESPONSE --> using Coulomb's law: k|(q1)(q2)|/r^2. confidence assessment: 3
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23:44:17 ** The magnitude of the force on a test charge Q is F = k q1 * Q / r^2, where q1 is the charge at the origin. The force is one of repulsion if q1 and Q are of like sign, attraction if the signs are unlike. The force is therefore directly away from the origin (if q1 and Q are of like sign) or directly toward the origin (if q1 and Q are of unlike sign). To find the direction of this displacement vector we find arctan(y / x), adding 180 deg if x is negative. If the q1 and Q are like charges then this is the direction of the field. If q1 and Q are unlike then the direction of the field is opposite this direction. The angle of the field would therefore be 180 degrees greater or less than this angle.**
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RESPONSE --> if like signs, they are repulsive, if unlike, attractive. self critique assessment: 2
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23:46:51 Explain how we calculate the magnitude and direction of the electric field at a given point of the x-y plane point due to a given point charge at the origin.
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RESPONSE --> the direction of the electric field is calculated by arctan (y/x). if this value is negative, we add 180 degrees. the magnitude is found using coulomb's law: k|q1*q2|/r^2 where q2 is the point at the origin confidence assessment: 3
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23:48:09 ** The magnitude of the force on a test charge Q is F = k q1 * Q / r^2, where q1 is the charge at the origin. The electric field is therefore F / Q = k q1 / r^2. The direction is the direction of the force experienced by a positive test charge. The electric field is therefore directly away from the origin (if q1 is positive) or directly toward the origin (if q1 is negative). The direction of the electric field is in the direction of the displacement vector from the origin to the point if q1 is positive, and opposite to this direction if q1 is negative. To find the direction of this displacement vector we find arctan(y / x), adding 180 deg if x is negative. If q1 is positive then this is the direction of the field. If q1 is negative then the direction of the field is opposite this direction, 180 degrees more or less than the calculated angle. **
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RESPONSE --> the electric field would therefore be F/q1=kq2/r^2 (for my example q2 was the point at the origin) self critique assessment: 2
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®éUãéðî€m¡ìÊ`¡‚{õÛ¢ÊÂyÀbÜI퇲 assignment #003 003. Physics II 03-26-2007
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00:32:58 In your own words explain the meaning of the electric field.
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RESPONSE --> the electric field is the area in space and time that the charge of a particle is exerting its charge. other particles and charges may be affected by this field once in ""range"". confidence assessment: 1
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00:34:23 STUDENT RESPONSE AND INSTRUCTOR COMMENT: electric field is the concentration of the electrical force ** That's a pretty good way to put it. Very specifically electric field measures the magnitude and direction of the electrical force, per unit of charge, that would be experienced by a charge placed at a given point. **
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RESPONSE --> the magnitude and direction of the electical force per unit charge self critique assessment: 2
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00:34:43 Query Principles of Physics and General Physics problem 16.15 charges 6 microC on diagonal corners, -6 microC on other diagonal corners of 1 m square; force on each. What is the magnitude and direction of the force on the positive charge at the lower left-hand corner?
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RESPONSE --> the charges one meter apart are attracted with a magnitude of .324 N. The diagonal charges are 1.414 meters apart, and exert repulsive forces of .162 N. The total force in the x direction is .21N. same in y. the net force is .29N at a 45 degree angle. confidence assessment: 2
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