course Phy 202 I realize I'm very far behind on work. If you could advise me on how to best go about finishing on time (or whether you give extensions or not) I would greatly appreciate it. Thank you, Mr. Smith. ???O????????assignment #002002.
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18:26:21 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 fact that the pieces of tape sometimes attract and repel other times indicates that there is some difference in the charges of the tape. confidence assessment: 2
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18:32:27 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 demonstrated by the fact that the tape appears to make an angle that could be evenly bisected by a straight line when they repel one another. The pieces that are attracted to one another are attracted to form the line that would theoretically bisect the two that repel. confidence assessment: 1
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18:34:06 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 --> There is no way to confine the charges to a point on a piece of tape, or to any surface for that matter. There are many charges on the tape that combine to result in the charge of the entire piece. confidence assessment: 2
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18:44:10 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 --> The tape at point A is pulled in the direction of AB_u. If there are attractive forces between the pieces of tape, it will be pulled toward the piece of tape at point B,which indicates that it is pulled in the director of the unit vector AB. If they repel, the tape at B is pushed in the direction of AB_u. If they repel one another then the one at point B will be pushed away from A. confidence assessment: 3
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18:45:34 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 AB_v, BA_v, and the distance between A and B should be equal. confidence assessment: 1
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18:48:34 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 --> As the magnitude of AB_v or BA_v decreases, the force of each charge should increase. The force should decrease as the magnitude increases. These values are inversely proportional (this was stated after the question but I tried to put it in my own words). confidence assessment: 3
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19:01:55 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 electrostatic force is calculated by calculating the resultant vector between the differences between the x & y values in the coordinate set to get the distance between the charges. This is then used to calculate the force using Coulomb's Law (F = k * q1 * q2 / r^2). To find the direction, the differences between the x and y values are put into the equation
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19:08:14 ** 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 got the first part correct, but I did not indicate that the direction would be related to the vector. I missed everything about the angle, but if you take the inverse tangent of y/x (opposite/adjacent sides), you can find the angle of the direction, and the difference (or lack of difference) between the signs of the charges indicates whether this is the final direction or if it is actually the opposite of this direction. self critique assessment: 3
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19:46:04 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 --> First the force has to be found on a ""test charge"" at the point. The force is found using the same equation as before: F = k * q1 * q2 / r^2 using the test charge and the given point charge at the origin. Then the magnitude of the electric field can be calculated using the formula E = F12/q which is equal to kq1/r^2. The direction of the force can be determined by taking the inverse tangent of the difference between y values / the difference between x values. confidence assessment: 2
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19:47:09 07-25-2007 19:47:09 ** 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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NOTES -------> I think I got it right but I'm going to save this as notes.
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