course PHY 232 ¤ëZÒ—›©ßbÐÅwPÊŇ{ßÓ±²‚ÿýúassignment #002 002. Physics II 01-19-2008
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09:14:50 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 experience demonstrates that there are attractive and repeling charges. The pieces are originally neutrally charged, but the process of ripping them apart redistributes the charge amongst the pieces allowing us to see the imbalance of charge by seeing the pieces attracted or repelled. confidence assessment: 3
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09:21:17 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 --> As the pieces of tape were positioned, you could see that the pieces seemed as if a force was acting upon the tape along a plane perpindecular to the pieces of tape. confidence assessment: 3
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09:22:14 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 actual charges that are interacting here, not just one. The actual interaction of all of these charges is more complicated than what it may appear. This just happens to demonstrate the how point charges work. confidence assessment: 3
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09:43:04 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 pieces attract, then the unit vector AB_u is the direction of the tape at A is being pulled toward B. Similarly, the the unit vector BA_u is the direction of the tape at B. If the pieces repel, then the unit vector AB_u is the direction of the tape at B is being pushed away. Similarly, the tape at A is being pused away in the direction of unit vector BA_u. confidence assessment: 3
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09:50:45 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 --> The total distance between the points will be equal to the sum of the magnitudes of the vectors since they are acting in only one direction. confidence assessment: 3
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10:24:04 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 --> Yes, my preceeding answer was incorrect. For some reason, I was thinking that the vectors were displacement vectors, which they clearly are not.
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11:50:01 ** 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 --> I accidentally hit the Next Question button when I meant to hit the Enter Response Button. The magnitude of the force is calculated by: |F| = k * q1 * q2 / r^2, where k is the constant 9 * 10^9 N m^2/C^2, r is the distance between the points and q1 and q2 are the magnitude of the charges. The direction is found by the arctan (y/x), where y and x are the position of the point not at the origin. If x is negative, then 180 degrees need to be added. self critique assessment: 3
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12:44:21 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 electric field at the point is given by E = F/q where q is the point in the x-y coordinate plane. F can be found by F = k * Q * q / r^2 where Q is the charge at at the origin and q is the charge at the point on the x-y coordinate plane. r is the distance between the point charges. k is the constant 9 * 10^9 (approx.) E = F/q, so E*q = k * Q * q / r^2, so E = k * Q / r^2 The direction can be found easily in this case by finding arctan(y/x) and adding 180 degrees in x is negative. confidence assessment: 3
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12:44:31 ** 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 --> self critique assessment: 3
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