course Phys 202
10/12 11:32 am
023.
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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:
By placing the two pieces of tape together and pulling them apart, the one with the sticky end on the other pulls charges away from this one. This redistribution of charges causes one piece of tape to be slightly negative and the other slightly positive. Since these are opposite charges when you bring them close together they will attract one another. The same thing happens with the second set of tape pieces. This time however when you take the positively charged piece of tape in the second set and hold it up to the positive piece from the first set the two will repel. This is because they have the same charge. This also happens when the two negatively charged pieces are held next to one another.
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 two charges are located on the same straight line, having the same x and y axis. If they are not alike charges on this straight line they have equal and opposite forces at 180 degrees from one another and they contain all the same angles from the axis.
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:
This doesn’t really prove anything about point charges because there are too many angles and dimensions interacting in the two pieces of tape. For example if you hold one piece of tape slightly sideways it will still exert the same effect on the other piece of tape even though its not really aligned on the same axis.
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 the pieces attract, the tape at point A is pulled towards unit vector BA. If the pieces repel then B is pushed in the direction of AB unit vector. Since unlike charges attract the piece at A should be pulled towards the piece at point B which is where unit vector BA is. If the pieces are repelling they will be pushing away from each other, hence B would be pushing away from A or unit vector AB.
Confidence 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?
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Your Solution:
The magnitude of vector AB v should be equal and opposite to the magnitude of the BA v vector. The magnitude of each vector should be equal to the distance between the two points.
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: Force is equal to the product of the two magnitudes of each vector, divided by the distance between the two points squared. This is represented by Coulomb’s Law.
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): OK
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Self-critique Rating: OK
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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:
First you would find the differences in the x and y values from the origin. This would be done by taking the x value of the charge and subtracting 0 and then taking the y value and subtracting 0 since the other charge is at the origin. Then Pythagorean Theorem could be used to find the distance between the points. So the difference in x squared plus the difference in y squared would equal the distance squared. We could solve for distance by taking the square root of those sums. To calculate the magnitude of the force we use Coulombs Law.
F = k (q1q2)/r^2
If the charges are like (q1 and q2 have the same sign) the direction will be parallel to the force exerted on one charge by the other or away from the origin, and opposite or towards the origin if unlike.
To find the direction we essentially have a triangle with the x and y values and an unknown angle. To solve for this angle we know tan(theta) = x/y, so we can plug in these values to get the angle. If the x value is negative you must add 180 degrees to get the total angle.
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): OK
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Self-critique Rating: OK
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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:
We use E = F/q after calculating F with F= k *q1*q2 /r^2. When a positive test charge is placed in the field the direction of the force experienced by this charge is the direction of the electric field. The electric field is therefore directly away from the origin (if q1 is positive) or directly toward the origin (if q1 is negative).
Once again to find the direction we essentially have a triangle with the x and y values and an unknown angle. To solve for this angle we know tan(theta) = x/y, so we can plug in these values to get the angle. If the x value is negative you must add 180 degrees to get the total angle.
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): OK
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Self-critique Rating: OK
&#Very good responses. Let me know if you have questions. &#