course phy202
aug 10, 2009, 10:30pm
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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.
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Your Solution:
This gives us a visible example of repulsive/attractive forces on two objects. It appears when ripped apart each side of the tape developed different charges. As another piece of tape was ripped the same thing happened and it is interesting to view the reaction of forces between the 4 pieces. Despite not knowing which pieces were + / - it was obvious where the repulsion and attraction occurred. The tape initially had and equal charge throughout it when in one piece but disruption of the charges causes them to separate.
confidence rating:
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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.
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Your Solution:
The charges must exist at 180 degrees of each other in order to exert equal and opposite forces on one another. This is accomplished by the arrangement of x and y coordinates of both forces being equal and opposite either in repulsion or attraction despite the angle made by the forces.
confidence rating:
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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.
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Your Solution:
The pieces contain many point charges and would be hard to pinpoint an interaction between 2 charges.
confidence rating:
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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. 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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Your Solution:
If attracted A will be pulled towards B therefore moving to the right. If the pieces repel then B will also be pushed to the right. The the AB vector breaks down into y components only therefore giving it horizontal movement along a 180 degree axis. If the points repel then the force will be equal and opposite causing B to be pushed away from A on an 180 degree axis.
confidence rating:
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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?
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Your Solution:
The greater the magnitude the greater the distance. This means that the magnitudes of the vectors are equal resulting in equal Forces.
confidence rating:
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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 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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Your Solution:
F = kQq/r^2
The greater the F the less the magnitude but if the distance increases between them then the F will decrease resulting in greater magnitude.
confidence rating:
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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):
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Self-critique Rating:
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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:
F = k Qq/r^2
If the charges are alike = repulsion
If the charges are different = attraction
As if solving for vector components we would use arctan (y/x)
confidence rating:
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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 left out some additional rule to follow such as adding 180 if needed
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Self-critique Rating:2
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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:
E = F / q
E = k(Qq/r^2) /q = kQ/r^2 where the direction is the direction of the Force on a tiny positive test charge. If Q is + then the direction is towards the point from the origin (direction of the displacement vector) and away from the origin if the Q is -.
confidence rating:
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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):
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Self-critique Rating:
&#Let me know if you have questions. &#