course PHY202 G¶¤”ÞŸc¤”ŸS嬰”£ÔÑáî}Åb¿ô¤öÜassignment #023
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21:12:08 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 --> WHen held close to each other the two original pieces of tape attract each other. When two new pieces of tape are introduced after being prepared in the same manner, one will attract one of the original pieces and repel the other original piece. The other new piece will have the opposite effect on the two original pieces of tape. THis illustrates two types of charge as seen in magnets. Opposites charges attract each other and like charges repel each other. Therefore the tape illustrates this as they attract and repel each other. confidence assessment: 3
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21:28:22 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 forces between the two charged particles are equal and opposite therefore placing them at a 180 degree angle with one another. confidence assessment: 3
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21:53:55 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 --> Each piece of tape has a large number of points of attractive forces that are acting on each other instead of a single point and since neither piece of tape is confined to a point then this experience does not prove anything about actual point charges. confidence assessment: 3
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22:16:15 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, then the tape at point A will be pulled in the direction of B. If the pieces repel, then the tape at B is pushed away from A. confidence assessment: 3
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22:39:23 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 --> THe magnitude of vector AB_v depends on the magnitude of the vector BA_v because the direction is defined by the unit vector. The force exerted from A toward B is the product of the magnitude of the force. THe distance between A and B compares with the magnitude of each vector as in the equation F = (k (A*B))/r^2 where r = distance between A & B. confidence assessment: 3
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22:40:51 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 --> ok confidence assessment: 3
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22:54:06 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 F on given charge A is calculated using the equation F = (k(A*a1))/r^2 where k = 9* 10^9 Nm^2/C^2, a1 = charge at point of origin, r = distance. If the force of A and a1 are opposite, then they will attract (directly toward origin). If they are like, they will repel (directly away from origin). The angle is 180 degrees from each other. confidence assessment: 3
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22:54:20 ** 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 --> ok self critique assessment: 2
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22:58:32 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 --> arctan(y/x) = direction if q1 is positive if q1 is negative, then add 180 degrees so direction is opposite confidence assessment: 3
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22:58:46 ** 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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