course PHY 232 Û¾¤¿ërÞßìÑwãÕQ”ÝبÌõGä¤Å‰ÑC›{assignment #002
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00:07:31 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 --> Each piece of tape receives either positive or negative point charges when attached to different objects. The point charges add up to give the piece of tape to have an overall charge. When two pieces of tape have the same charges, they attract and when they have opposite charges, they repell. distributed on each piece of tape. confidence assessment: 3
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00:07:40 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 point charges all line up to give the piece of tape an overall charge. For pieces of tape that have opposite charges, a force draws the two pieces of tape to stick together which can be seen. confidence assessment: 2
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00:07:51 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 --> Point charges can not be seen with the eye so pieces of tape do not provide a realistic view of a point charge. However, a piece of tape has lots of point charges throughout it and in the experiment, tape is only representing a point charge to make it easier for others to understand. confidence assessment: 3
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00:07:59 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 --> Point charges with opposite directions will attract. Therefore, if the pieces attract, then the tape at point A will be pulled toward point b. If the pieces repel, then the tape at point A will push away from point B. confidence assessment: 3
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00:08:06 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 --> Because they are unit vectors, the magnitude is 1. The force is directly proportional to the two charges multiplied together. To calculate the force of one point charge on another point charge, you dividide the charges by the distance between the points squared. This means that the force is inversely proportional to the square of the distance. confidence assessment: 3
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00:08:14 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 --> Right, the distance between the two pieces of tape vary so the force between the point charges in each of the pieces of tape will differ. confidence assessment: 3
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00:08:26 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 find the force, you must use the equation F=k*q1*q2/r^2. k is always 9*10^9 q1 is the first point charge q2 is the second point charge and r is the distance between the points. To find r when a coordinate is given, we must find the distance to the orgin using the equation r=sqrt(x^2+y^2). (if the first point were not at the origin we would have to subtract the second point from the other x and y points) confidence assessment: 3
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00:08:33 ** 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 --> yes self critique assessment: 3
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00:08:43 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 force is k*q1*q2/r^2 and the electric field is force/q so the electric field = k*q1/r^2 To find direction, we must look at the signs of the charges. If q1 is negative, we have a negative direction and if q2 is positive, we have a positive direction. confidence assessment: 3
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00:08:50 ** 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 --> exactly self critique assessment: 3
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