course PHY 202 ??h??_??????G?assignment #002
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16:52:33 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 pieces of scotch tape are attracted to one another by a series of point charges that are scattered all over the tape. When point charger intersct with the other charger on seperate pieces of tape their effects either attract the piece of tape, if the charges are different, or repel the piece if the charges are like. confidence assessment: 3
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16:54:29 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 scotch tape pieces acted in the same magnitude of the pieces working on it. if the piece was attracted then the pulls between the two opposite chargers were equal in magnitude but opposite in direction. However, if the pieces had the same charge than both pieces of tape would exert equal forces on the other in the same lines. confidence assessment: 2
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16:55:44 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 --> The actual interaction was much more complicated than the one described seeing as the description acted upon all points acting uniformly with the same charge and strength. This is not the case the points were acting in all different planes and with different strengths. confidence assessment: 3
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17:00:26 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 than then the tape a would bu pulled in the direction of vector iof ABv and AB u. This is becuase FB would attract tape A and would pull with forces that worked along the hypotenuse of Abv and ABu. Similarly if the piece was reppelled it would mean that B was exerting a like force along the same path at the charges inpoint A. Therefore Tape A would move along a path of the hypotenus of BAv and BAu. confidence assessment: 2
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17:02:21 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 both vectors depend on the relative strength of the other. Therefore the if the Magnitude of Av_b was weaker than BA_v would exert the commanding force and the distance between A and B would lengthen If AB-v were stronger or attractive, and BA_v would exert less force and the distance would shorten. confidence assessment: 2
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17:10:57 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 --> the foreces that are exerted on the tape largely depend on the distance between them. Therefore if the magnitudes increase, which include the distance between the to poines, than the force should decrease by the square of the distance between them. confidence assessment: 3
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17:26:31 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 --> to determine the magnitude ofthe force the distance between the piont and the orgin must be established, this could be done by using the the displacement equation to get the r variable. Once r has been found the equation F= k(q) / r2 can be used to get the magnitude of the force of the charge. the direction can eb determined if the forces are attracted or repelled. if the forces are unlike thenthe direction will be opposite that of the given charge, if they are like then the direction will be parallel to the charge. confidence assessment: 1
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17:29:08 ** 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 --> the magnitude of the force is similar to the equation I used but I left out q1. howver the direction is found by by actually finding the angle than the force makes with the x axis and then subtractingg 180 if the forces are like. self critique assessment: 2
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17:31:31 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 --> Following the same line of thinking the magnitude is found by calcultaing R and substitutig it intoCoulomb's Law. Finding the direction again refers to the direction of the attractive or repulsive forces working on the point from the orgin. confidence assessment: 1
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17:32:44 ** 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 --> Ididn't understand that q1 was the charge at the orgin, howver I was on the correct train of thought. However i had a different equation written out for the electric feild. self critique assessment: 2
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