course Phy 202 1:45 am July 17 Question: `qIn your own words explain the meaning of the electric field.
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Given Solution: `aSTUDENT 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: 3 ********************************************* Question: `qQuery 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Charges that are 1 m apart are attractive. Using Coulomb’s Law we have (9*10^9 Nm^2)*(6*10^-6 C)*(6*10^-6 C)/ (1m^2)= 324*10^-3 N or .324 N Charges across the diagonal are separated by 1.414 m. They are repulsive. (9*10^9 Nm^2)*(6*10^-6 C)*(6*10^06 C)/ (1.414m^2)= 162*10^-3 N or .162 N So, we have a lower left hand charge of .324 N to the right, .324 N upward and 162 N at 225 degrees. Fy= .162N* sin(225degrees)= -.115 N Fx= .162N* cos(225 degrees)= -.115 N The total force in the x direction= -.115 N+ .324 N= .21 N The total force in the y direction= .-115N+ .324 N= .21 N Thus the net force magnitude= sqrt`(.21 N)^2+ (.21 N)^2= .29 N The angle= tan^-1(y/x)= tan^-1 (.21N/.21N)= tan^-1(1)= 45 degrees
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Given Solution: `a** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: 3