course phy ???????y??y???assignment #003003.
......!!!!!!!!...................................
22:06:00 In your own words explain the meaning of the electric field.
......!!!!!!!!...................................
RESPONSE --> electric feild is the feild of force that exists in all directs around all charged objects. the word feild refers to the point that it is not directeed at a given vector, but rather is omnidirectional. confidence assessment: 2
.................................................
......!!!!!!!!...................................
22:06:20 STUDENT RESPONSE AND INSTRUCTOR COMMENT: electric field is the concentration of the electrical force ** That's a pretty good way to put it. Very specifically electric field measures the magnitude and direction of the electrical force, per unit of charge, that would be experienced by a charge placed at a given point. **
......!!!!!!!!...................................
RESPONSE --> reasonable self critique assessment: 2
.................................................
......!!!!!!!!...................................
00:05:30 Query Principles of Physics and General Physics problem 16.15 charges 6 microC on diagonal corners, -6 microC on other diagonal corners of 1 m square; force on each. What is the magnitude and direction of the force on the positive charge at the lower left-hand corner?
......!!!!!!!!...................................
RESPONSE --> an interesting question, one that uses a fair bit of thought, but here goes components; 6 MicroC, (1,1) -6microC, (0,1) -6MicroC, (1,0) f1=(9*10^9,*6*10^-6, 6*10^-6)/2=.162 f2=(9*10^9,*6*10^-6, -6*10^-6)/1= -.324 f3=(9*10^9,*6*10^-6, -6*10^-6)/1= -.324 x and y components can be found by assigning f2 to y and f3 to x. that leaves f1. F1 can be assigned by spliting it halfway. one half goes to x and the other to y. amazingly enough, fx and fy are the same, so magnitude is fx^2, times two yeilds q, magnitude is the square root of q. the direction of the force can be found by tan (fy/fx), which yeild 45 degrees. confidence assessment: 3
.................................................
???????????? assignment #003 003. Physics II 02-01-2007
......!!!!!!!!...................................
09:34:56 In your own words explain the meaning of the electric field.
......!!!!!!!!...................................
RESPONSE --> ok confidence assessment: 2
.................................................
......!!!!!!!!...................................
09:35:01 STUDENT RESPONSE AND INSTRUCTOR COMMENT: electric field is the concentration of the electrical force ** That's a pretty good way to put it. Very specifically electric field measures the magnitude and direction of the electrical force, per unit of charge, that would be experienced by a charge placed at a given point. **
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 2
.................................................
......!!!!!!!!...................................
09:35:10 Query Principles of Physics and General Physics problem 16.15 charges 6 microC on diagonal corners, -6 microC on other diagonal corners of 1 m square; force on each. What is the magnitude and direction of the force on the positive charge at the lower left-hand corner?
......!!!!!!!!...................................
RESPONSE --> ok confidence assessment: 3
.................................................
......!!!!!!!!...................................
09:35:25 ** The charges which lie 1 meter apart are unlike and therefore exert attractive forces; these forces are each .324 Newtons. This is calculated using Coulomb's Law: F = 9 * 10^9 N m^2/C^2 * ( 6 * 10^-6 C) * ( 6 * 10^-6 C) / ( 1 m)^2 = 324 * 10^-3 N = .324 N. Charges across a diagonal are like and separated by `sqrt(2) meters = 1.414 meters, approx, and exert repulsive forces of .162 Newtons. This repulsive force is calculated using Coulomb's Law: F = 9 * 10^9 N m^2/C^2 * ( 6 * 10^-6 C) * ( 6* 10^-6 C) / ( 1.414 m)^2 = 162 * 10^-3 N = .162 N. The charge at the lower left-hand corner therefore experiences a force of .324 Newtons to the right, a force of .324 Newtons straight upward and a force of .162 Newtons at 45 deg down and to the left (at angle 225 deg with respect to the standard positive x axis, which we take as directed toward the right). This latter force has components Fy = .162 N sin(225 deg) = -.115 N, approx, and Fx = .162 N cos(225 deg) = -.115 N. The total force in the x direction is therefore -.115 N + .324 N = .21 N, approx; the total force in the y direction is -.115 N + .324 N = .21 N, approx. Thus the net force has magnitude `sqrt( (.21 N)^2 + (.21 N)^2) = .29 N at an angle of tan^-1( .21 N / .21 N) = tan^-1(1) = 45 deg. The magnitude and direction of the force on the negative charge at the lower right-hand corner is obtained by a similar analysis, which would show that this charge experiences forces of .324 N to the left, .324 N straight up, and .162 N down and to the right. The net force is found by standard vector methods to be about .29 N up and to the left. **
......!!!!!!!!...................................
RESPONSE --> "