course Phy 122 In the following introductory problem, what does the change in potential energy tell me? How can I find it if I don't know the change in KE? How is it different from the magnitude of the force? PROBLEM: Estimate, based on the force at the initial and final points, the work required to move the second charge to the point (0.0059m, 0, 0) assuming a stationary charge of 61 microC located at the origin, if the second charge of 87 microC is initially located at the point (.01m, 0, 0). What is the approximate potential energy chanage associated with moving the second charge between these points? SOLUTION: F at initial point = 47760N F at second point = 1372000N Fave = 924800N `ds = -0.0041m W = 3792J PE = ???
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18:09:46 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 I held the two pieces of tape next to each other first, there was no response. I then placed the sticky side of one strip to the smooth side of the other and ripped them apart. Afterwards, I held the pieces of tape close and they were attracted to each other. The instructions said they would repel but even after I conducted the experiment a second time, the pieces were still attracted. This strong attraction shows that there were opposite charges (negative and positive) present on the individual pieces of tape. confidence assessment: 3
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18:15:14 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 --> It was easy to see that the opposite sides of tape (sticky and smooth) were attracted to each other after they were charged, whereas, the two sticky sides or two smooth sides repelled each other. The attraction charges were present along the ENTIRE LENGTH of both pieces of tape. It was hard to test the repelling forces because the attraction was so much stronger that my piece of tape turned over. confidence assessment: 3
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18:17:36 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 was found that the entire length of the strips were charged by testing attraction and repulsion forces. Therefore, we cannot designate one particular point on either strip as the point charge. confidence assessment: 3
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18:24:14 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 are attracted, the tape at point A will be pulled in the direction of the AB unit vector which is toward B. Since the charges are opposite, the attracting force will pull the tape at A towards the tape at B. If the pieces are repelled, the tape at point B will be pushed in the direction of the AB unit vector which is away from A. The repelling force will puch the tape at B away from the like charged piece at A. confidence assessment: 2
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18:33:26 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 magnitudes of the two vectors should be the same. To show this we can calculate the magnitude of the force for each vector: Fab = k (Qa)(Qb) / r^2 Fba = k (Qb)(Qa) / r^2 These two equations will give the same magnitude, so Fab = Fba. The distance between the charges of A and B is reflected as r in the above equation. If the two point charges are close together, the value of r is small. Therefore the magnitude of the force on that point is greater. As the distance between the point charges becomes greater, the r value gets bigger and the magnitude of the force on those points becomes smaller. confidence assessment: 3
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18:37:03 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 greater the magnitude, the greater the force that either repels or attracts the two objects. The direction however, depends on the charge. The two pieces might slightly repel each other if the charges are the same and the magnitude is small. On the other hand, they may greatly attract each other is the charges are opposite and the magnitude is great. Also, parts of the tape may experience a greater charge than other areas on the same tape due to their distance from the other strip of tape. Therefore, we cannot say that the magnitude is uniform for the entire strip. confidence assessment: 2
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18:41:59 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 --> Magnitude of the force on a point is calculated using Coloumb's Law: F= [k(q1)(q2)]/r^2 The direction is determined by looking at the two point charges given. If the charges are opposite, the direction is attracting. If the charges are the same, the direction is repelling. The factor q1 is the origin. If the charges are repelling (same charges), the vector will direct away from q1 and if the charges are the opposite (attractive), the vector will direct toward q1. confidence assessment: 3
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18:45:25 ** 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 failed to mention how to find the direction of the displacement vector. I understand that if we know the x and y components of the force vector, and the two charges are alike, we can calculate the direction (theta) by the following equation: `theta=tan^-1(y/x) If x is negative, 180 degrees should be added to the resulting angle. If the charges are of opposite sign, the direction is exactly 180 degrees opposite of the angle that is calculated with the above equation. self critique assessment: 3
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18:48:32 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 --> As it was stated in the previous question, the magnitude of the force can be calculated by the equation (q1 is the charge at the origin and q2 is the test charge): F= [k(q1)(q2)]/r^2 According to the text (page 450), the electric field can be calculated as: E = F/q2 = k(q1)/r^2 Therfore, if the point charge is negative, the direction will be toward the origin and away from it if q1 is positive. The direction of the vector can be calculated if we know the x and y components using the following equation: `theta=tan^-1(y/x) The same rules apply that were stated in the previous answer. If x is negative 180 degrees must be added. If the charges are opposite, the direction is opposite the one that is calculated. confidence assessment: 3
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18:57:56 ** 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 --> I feel that I adequately covered this topic. I looked up the information that I needed in the textbook in order to give an accurate answer. I did not state things as concisely or clearly as the solution. self critique assessment: 3
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