course Phy 202 ±[ZSC}}assignment #002
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16:44:21 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 --> One of the ends was positively charged and one of the ends was negatively charged. The like charged ends repelled each other and the oppisitely charged ends attracted each other. This was a very interesting experiment. confidence assessment: 3
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16:46:06 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 --> I am not sure how to answer this question. confidence assessment: 0
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16:47:32 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 --> This experiment was conducted just to give us an idea of how positive and negative charges interact with each other. The experiment was not done to prove anything about actual point charges. confidence assessment: 3
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16:51:12 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 they attract, A is pulled in the direction of unit vector BA_u. If they repel, B is pushed in the direction of unit vector AB_u. confidence assessment: 2
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16:55:23 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 --> Vector are made up of a magnitude and a direction. Vectors are equal if they have the same magnitude and direction. Since these two vectors have a different direction, they are not equal. confidence assessment: 0
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16:56:58 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 --> As the magnitude increases, the force decreases and as the magnitude decreases, the force increases. confidence assessment: 2
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17:00:16 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 --> We would use the Pythagorean Theorem to find the direction and Coulomb's Law to find the magnitude. confidence assessment: 3
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17:01:35 ** 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 understand self critique assessment: 3
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17:02:33 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 --> We use the Pythagorean Theorem to calculate direction and then Coulomb's Law to find the magnitude. confidence assessment: 3
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17:02:55 ** 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 --> yes self critique assessment: 3
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