course Phy 202 ???????????]?|??assignment #002
......!!!!!!!!...................................
10:01:53 In your own words explain how the introductory experience with scotch tape illustrates the existence of two types of charge.
......!!!!!!!!...................................
RESPONSE --> The introductory experience with the tape didn't work for me as well as it did in the class note videos, but I could clearly see that they became charged once pulled apart and that the charge created made the pieces either attract or repel depending on which sides of the tape were facing towards each other. confidence assessment: 2
.................................................
......!!!!!!!!...................................
10:02:13 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.
......!!!!!!!!...................................
RESPONSE --> When the sides facing together were oppositely charged they came together and when the sides that had like charges were facing together, they repelled. If you placed the middle sections of the tape together they stick or repel form the center. If you placed the ends of the tapes together the very ends would interact. So, the idea that the charges act through a straight line is supported by this because the charges were equally distributed along the surfaces of the tape pieces and when the charges acted on each other it seemed to do it form only one place at a time and not just the whole piece sticking. confidence assessment: 2
.................................................
......!!!!!!!!...................................
10:02:35 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.
......!!!!!!!!...................................
RESPONSE --> This experience doesn't relate to actual point charges because, as stated before , the charges are all distributed along the length of the tape and go in all directions until they come into contact or near each other. Point charges have the entire sum of all the surrounding charges coming together at the central point. A smaller piece of tape would attempt to simulate this but it doesn't quite portray the correct concept. confidence assessment: 2
.................................................
......!!!!!!!!...................................
10:02:57 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.
......!!!!!!!!...................................
RESPONSE --> If the pieces attract then the tape at point A will be pulled towards B in the direction along the unit vector AB_u = 1. If the pieces repel then the tape at point B will be pushed outwards in the direction along the unit vector AB_u = 1 as well. confidence assessment: 2
.................................................
......!!!!!!!!...................................
10:03:13 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?
......!!!!!!!!...................................
RESPONSE --> The magnitude of the vector AB_v will be the same as the magnitude for vector BA_v. The magnitude of the vectors is based on the distance between them. So whatever that distance is, it can be used to calculate the magnitude of either vector. They'll have the same magnitude but the vectors will go in opposite directions. confidence assessment: 2
.................................................
......!!!!!!!!...................................
10:03:25 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.
......!!!!!!!!...................................
RESPONSE --> The force experienced by the two pieces of tape is influenced by the magnitude of AB_v or BA_v in a way that when the magnitude increases the force decreases with the square of the distance between the two pieces of tape. Likewise, if the magnitude decreases, the force increases with the square of the distance between the pieces of tape. confidence assessment: 2
.................................................
......!!!!!!!!...................................
10:14:49 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.
......!!!!!!!!...................................
RESPONSE --> To calculate the magnitude of the force on a charge at a given point on the x,y plane due to another point at the origin, we use the formula: F= k(q1* Q/ r^2). q1 is the charge at the origin, Q is the charge at the given point, and r is the distance between them. The force we calculate from this will be repulsive id the signs of the charges are like and attactive if they are unlike. If they are like then the force of Q will point away from the charge at the origin(q1) and if unlike the force will towards the charge at the origin. To calculate direction of the vector, we find the angle of the field by doing tan^(-1) * (y/x). We add 180 to the resulting angle if x is negative. If q1 and Q are like charges, then the angle calculated is the direction of the field. If they are unlike, then the direction of the field is opposite of the angle and in that case the field would be 180 degrees greater than or less than the angle we calculated. confidence assessment: 1
.................................................
......!!!!!!!!...................................
10:15:10 ** 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.**
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 2
.................................................