course Phy 232
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Question: `qIn your own words explain how the introductory experience with scotch tape illustrates the existence of two types of charge.
Student Solution: The two pieces of tape were rubbed together and pulled apart. Once pulled apart, they both attracted each other. However when this was done again the same sides of the tape repelled each other, but the different sides of tape attracted each other. This illustrated that the sides of the tape had different charges plus and minus. A plus has an affinity for a negative (and vice-versa), but same charges do not.
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Self-critique (if necessary): 3
Self-critique Rating: I think I got this correct, but there is no given solution.
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Question: `qIn 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.
Student Solution: The pieces of tape that were attracted to each other each had multiple charged particles along the tape. These particles were what made the pieces of tape attracted to one another. Without this bond, the tapes would simply stand still. If the charges were opposite, the pieces of tape would repel along this line.
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Self-critique (if necessary): 3
Self-critique Rating: I think I got this correct, but there is no given solution.
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Question: `qIn 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.
Student Solution: Two pieces of tape were attached to a table. The other two pieces were held by the experimenter. Therefore, these two pieces of tape were not confined to a stationary point. The actual point charges may have been disrupted by our hands.
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Self-critique (if necessary): 3
Self-critique Rating: I think I got this correct, but there is no given solution.
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Question: `qIf 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. Similarly let 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.
Student Solution: If the charges attracted each other, the piece of tape at point A would be pulled towards the other piece of tape because it has a larger magnitude. If the charges repelled, the piece of tape at point A would be pushed away from the other piece of tape.
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Self-critique (if necessary): I think I got this correct, but there is no given solution.
Self-critique Rating: 3
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Question: `qUsing 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?
Student Answer: When AB_v and BA_v are analogous to each other, I am going to assume they have the same magnitude, distance between A and B, but in different directions. AB_v has a magnitude going from A to B, and BA_v has a magnitude going from B to A.
AB_v has a direction going from A to B, and BA_v has a direction agnitude going from B to A.
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Self-critique (if necessary): I think I got this correct, but there is no given solution.
Self-critique Rating: 3
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Question: `qUsing 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 proportionality. 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.
Student Answer: The magnitude between AB_v and BA_v =sqrt(dx^2+dy^2). Therefore, the magnitude experienced by the two pieces of tape will depend on the distance between each point. This magnitude will increase with increasing distance, and decrease with decreasing distance. Force is equal to the vector divided by the magnitude. So force will decrease with increasing distance inversely with the sqrt(dx^2+dy^2). The force will also increase with decreasing distance inversely with the sqrt(dx^2+dy^2).
Given Answer:
STUDENT COMMENT: Still not sure how to answer this question. I dont know what sort of difference there is between the magnitude vector and the unit vector.
INSTRUCTOR RESPONSE: As noted in the preceding solution, AB_v stands for the vector whose initial point is A and whose terminal point is B, and AB_u stands for a vector of magnitude 1 whose direction is the same as that of AB_u.
To get a unit vector in the direction of a given vector, you divide that vector by its magnitude. So AB_u = AB_v / | AB_v |.
The purpose of a unit vector is to represent a direction. If you multiply a unit vector by a number, you get a vector in the same direction as the unit vector, whose magnitude is the number by which you multiplied it.
The problem does not at this point ask you to actually calculate these vectors. However, as an example:
Suppose point A is (4, 5) and point B is (7, 3). Then AB_v is the vector whose initial point is A and whose terminal point is B, so that AB_v = <3, -2>, the vector with x component 3 and y component -2.
The magnitude of this vector is sqrt( 3^2 + (-2)^2 ) = sqrt(11).
Therefore AB_u = <3, -2> / sqrt(11) = (3 / sqrt(11), -2 / sqrt(11) ). That is, AB_u is the vector with x component 3 / sqrt(11) and y component -2 / sqrt(11). (Note that these components should be written with rationalized denominators as 3 sqrt(11) / 11 and -2 sqrt(11) / 11, so that AB_u = <3 sqrt(11) / 11, -2 sqrt(11) / 11 > ). The vector AB_u is a unit vector: it has magnitude 1.
The unit vector has the same direction as AB_v. If we want a vector of magnitude, say, 20 in the direction of this vector, we simply multiply the unit vector AB_u by 20 (we would obtain 20 * <3 sqrt(11) / 11, -2 sqrt(11) / 11 > = <60 sqrt(11) / 11, -40 sqrt(11) / 11 >).
If you do not completely understand this by reading it here, you should of course sketch this situation and identify all these quantities in your sketch.
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Self-critique (if necessary): I got the basic idea of the question correct. It helped a lot in your answer when you went through with actual numbers. I was able to calculate with your numbers and see exactly what my answer meant.
Self-critique Rating: 3
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Question: `qQuery 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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Your solution:
The magnitude of the electrostatic force, F=k*q1*Q/r^2
If q1 is positive, the direction is positive or away from origin
If q1 is negative, the direction is negative or going towards the origin
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Given Solution:
`a** 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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Self-critique (if necessary): I got it correct
Self-critique Rating: 3
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Question: `qExplain 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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Your solution:
The magnitude of the electric field’s force, F=k*q1*Q/r^2
F/Q=k*q1/r^2
If q1 is positive, the direction is positive or away from orgin
If q1 is negative, the direction is negative or going towards the origin
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Given Solution:
`a** 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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Self-critique (if necessary): 3
Self-critique Rating: I got the answer correct. I am glad your answer went into more detail.
&#This looks good. Let me know if you have any questions. &#