#$&* course Phy 232 7/18 7 Question: `qIn your own words explain how the introductory experience with scotch tape illustrates the existence of two types of charge.
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Given Solution: If the pieces attract, then the tape at point A is pulled toward point B. • The vectors AB_v and AB_u point from A to B. • Of these the vector AB_u is the unit vector. • So the tape at A experiences a force in the direction of the vector AB_u. If the pieces repel, then the tape at point B is pushed away from point A. • The direction of the force is therefore from A towards B. • The direction is therefore that of the vector AB_u. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* OK ********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: Vectors AB_v and BA_v are equal in magnitude and opposite in direction. Furthermore, this magnitude is equal to the distance between points A and B. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The distance from A to B is the same as the distance from B to A. So the vectors AB_v and BA_v have the same length. Each vector therefore has magnitude equal to the distance between A and B. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* OK ********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: The force between two point charges is equal to (k*q1*q2)/r^2 where r is the distance between the two points. Therefore, the force experienced by the two pieces of tape is inversely proportional to the squared magnitude of AB_v and BA_v. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The 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 distance between A and B (i.e., to the separation between the two pieces of tape), increases the force decreases with the square of the distance (similarly if the distance decreases the force increases in the same proportion). This answer isn't exactly correct, since the two pieces of tape are not point charges. Since some parts of tape A are closer to B than other parts of tape A, and vice versa, the inverse square relationship applies only in approximation for actual pieces of tape. 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 >). You don't need to understand this example at this point, but if you wish to understand it you should probably sketch this situation and identify all these quantities in your sketch. STUDENT RESPONSE AB_v/(AB_v)absolute value So AB_u would either be 1 or -1. INSTRUCTOR COMMENT AB_u and BA_v are vectors, so they have both magnitude and direction. Your answer is correct if everything is in one dimension (e.g., if all charges are on the x axis and all forces directed along the x axis). In one dimension direction can be specified by + and - signs. In two dimensions + and - signs are not sufficient. In two dimensions the direction of a vector is generally specified by its angle as measured counterclockwise from the positive x axis. In two dimensions, the magnitude of AB_u would be the same as that of BA_u, but the vectors would be in the opposite direction. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): My answer was basically correct; however, I did not address the fact the distance is only an approximation for the pieces of tape since they are not real point charges. ------------------------------------------------ Self-critique rating #$&* 3 ********************************************* 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The magnitude of the force can be found using the formula abs(force) = [k*abs(q1*q2)]/r^2. In order to find this distance r, the Pythagorean formula is used with the given x and y coordinates. Finally, the direction of the force can be found by taking the inverse tangent of y/x. If the charges are alike, they will repel and the force of the charge at the origin on the second charge will be in this direction. However, if they are alike, they will attract and the direction will be arctan(y/x) + 180 degrees. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The magnitude of the force on a test charge Q is F = k q1 * Q / r^2, where q1 is the charge at the origin. The force is one of repulsion if q1 and Q are of like sign, attraction if the signs are unlike. The force is therefore directly away from the origin (if q1 and Q are of like sign) or directly toward the origin (if q1 and Q are of unlike sign). To find the direction of this displacement vector we find arctan(y / x), adding 180 deg if x is negative. If the q1 and Q are like charges then this is the direction of the field. If q1 and Q are unlike then the direction of the field is opposite this direction. The angle of the field would therefore be 180 degrees greater or less than this angle.** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* OK "