course Phy 122 14:44:25In 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 tape experiment shows how when you stick the two pieces of tape together and then rip them apart, the tape begins to show the electric charges. You can see the charges when you again stick teo pieces of tape together and then rip them a part, then holding them close to the ones which you had previously done that to. This shows the interaction of two point charges. confidence assessment: 2
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14:44:42 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 --> This is shown by the scotch tape because either the entire strip of tape attracts or repels an object. There are many point charges, but they are all the same so the entire piece either attracts or repels the object. confidence assessment: 2
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14:45:01 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 --> It does not show anything for an actual point charge because a single point is not singled out. This experiement shows the interaction amoung many point charges because they all have an effect on eachother. confidence assessment: 3
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14:45:19 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 --> Vector A would be pushed or pulled to the together, the two pieces of tape would be unalike, thus causing them to be attracted,and vector B would be pushed or pulled to the away because the two forces are unlike thus repelling eachother. confidence assessment: 2
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14:45:38 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 compare with 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 force is influenced by the magnitude, because if they are equal in size just not direction, it will influence the force which they act on eachother. confidence assessment: 2
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14:46:05 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?
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RESPONSE --> The force is influenced by the magnitude, because if they are equal in size just not direction, it will influence the force which they act on eachother. confidence assessment: 1
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14:46:41 The 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
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RESPONSE --> square distance increases then the force decreases or vice versa. The force exerted by the 2 charges is inversely proportional to the square distance between them. self critique assessment: 2
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14:47:01 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 --> Since some parts of the tape are closer to the other parts of the tape, this making them not point charges. self critique assessment: 2
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14:50:38 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 --> 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. If Q and q1 are alike in sign, then it is repelled, if unlike, then attract. The force is therefore directly away from the origin when they are alike in sign or directly toward the origin when unalike in sign.To find the direction of this displacement vector we find arctan(y / x), adding 180 deg if x is negative. confidence assessment: 3
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14:50:45 ** 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 --> self critique assessment:
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14:52:15 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 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 F / Q = k q1 / r^2. The direction is the direction of the force experienced by a positive test charge. It is directly away from origin if q1 is positive and towards if q1 is negative confidence assessment: 2
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14:53:07 ** 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 --> When finding direct use arctan (y/x) and add to 19=80 if x is negative. Other than that I think I got the rest. self critique assessment: 2
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