course phy202 ??????????????assignment #002
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09:02:14 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 scotch tape experiment involved sticking pieces of scotch tape together so that they would express both positively and negatively charged sides. Four total pieces of tape were made to allow the positive and negative side of each to be determined. Thus, it demonstrated how opposite charges attract and like charges repel, similar to the forces we can observe by playing with magnets. confidence assessment: 3
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09:05:16 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 charged particles on the surfaces of the pieces of tape either attracted or repelled each other along a straight line when held together. Because these particles were distributed across the surface of the tape, the tape itself appeared to be have a charged surface that either attracted or repelled in a straight line. confidence assessment: 3
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09:07:13 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 --> As I mentioned in my last response, the charges are spread across the surface of the tape. Thus, we can only observe the behavior of the surface of the tape as a whole. To really study point charges, we would need an experiment in which individual point charges could be isolated and exposed to each other under observable circumstances. confidence assessment: 3
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09:13:31 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, the tape at point A is pulled toward vector AB_v, that is the vector whose initial point is A and whose terminal point is B. If the pieces repel, the tape at point B is pushed toward vector -BA_v, that is the vector whose initial point is B and whose terminal point is farther away from point A. confidence assessment: 3
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09:16:05 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 magnitude of the vector AB_v is equal to the magnitude of BA_v when an attractive or repulsive force exists between points A and B. The difference is whether the direction of the two vectors are positive or negative, that is, toward each other or away from each other. confidence assessment: 3
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09:17:53 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 --> F=1/(AB_v)^2 or F=1/(BA_v)^2. confidence assessment: 3
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09:21:55 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 --> F=kq1q2/r^2. r=`sqrt((x2-x1)^2 + (y2-y1)^2). Attractive if opposite charges, repulsive if like charges. confidence assessment: 3
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09:23:38 ** 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 mentioned how to determine if the forces were attractive or repulsive, but I neglected to mention that we find arctan(y / x), adding 180 deg if x is negative to specifically determine the angle at which the vector points. self critique assessment: 2
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09:26:32 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 --> F=kq1q2/r^2. r=`sqrt((x2-x1)^2 + (y2-y1)^2). Angle=arctan(y/x). Together if opposite charges, away if like charges. confidence assessment: 3
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09:26:44 ** 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 --> ok self critique assessment: 3
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