course phy 122 cËĺцöîÉ°‡¯}Á“ã~ø¥«þ}™ß€×Ý“èassignment #002
......!!!!!!!!...................................
14:35:09 In your own words explain how the introductory experience with scotch tape illustrates the existence of two types of charge.
......!!!!!!!!...................................
RESPONSE --> the experience with scotch tape shows that there are repellant anfd atttactive types of force, and there appears tobe two different types. from the tape experiemtn, the conclusion can be drawn that the two forces work in either one of two ways. the first possibility is that like force attracts like force, and the other possibility is tht like does not attract (there is no way to tell in the experiment. confidence assessment: 2
.................................................
......!!!!!!!!...................................
14:36:50 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 --> the fact that the scotch tape repeled in a stragith line ad tended to move away in a striaght line. (in some cases it moved in a curve, but in midair tests, th tape moved perpendicurally away or towards the other peice. confidence assessment: 3
.................................................
......!!!!!!!!...................................
14:37:43 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 --> my understanding is that the tape cannot move and thus does not give a relistick point example. confidence assessment: 1
.................................................
......!!!!!!!!...................................
14:41:42 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 peices attract, than the peice of tape at point at is pulled in the diection of AB_u, but if the peices repel, than the tape at point B is pushed in the direction of AB_u. this is because in attraction, tape a is pulled, but in repulsion, the tape at b is pushed in the same direction that A had been pulled. confidence assessment: 3
.................................................
......!!!!!!!!...................................
14:44:14 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 AB_v depends on the magnitude of BA_v in that the magnitude is not a measure of a singe source for the equations sake, but rather it is a masuere of th total of the force, and it can then be used to find wheather it is attractive or repulsive. Part of the magnitude is also the distance betwene the two points. confidence assessment: 2
.................................................
......!!!!!!!!...................................
14:46:20 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
......!!!!!!!!...................................
RESPONSE --> it looks to be an incomplete question, but square of the square of the distance increases, than the force must proprtionally increase. confidence assessment: 2
.................................................
......!!!!!!!!...................................
14:46:45 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 --> ok confidence assessment: 2
.................................................
......!!!!!!!!...................................
14:48:54 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 --> the first thing that is done is the tkae the actual distance betwene the points. once this is done, label that R. the actuall equations, minus unit labels is magnitude= (9*10^9*charge one*charge two)/R^2 confidence assessment: 3
.................................................
......!!!!!!!!...................................
14:49:46 ** 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: 3
.................................................
......!!!!!!!!...................................
14:53:25 Explain 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.
......!!!!!!!!...................................
RESPONSE --> the magniude can be calculated by first finding R, than by using the m=(9*10^9* charge one* charge 2)/R^2 this give magnitude, but does not give direction. the direction can be found by using M tan ^-1 (y/x), which yeilds a degree notation. this degree may or may not be poitnin in the right direction. if the charges are the same sign, than they are repelent, and you need to either add or subtract 180 from/ to the angle to get the right one. confidence assessment: 2
.................................................
......!!!!!!!!...................................
14:53:47 ** 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. **
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
"