#$&* course PHY 202 024. Query 24 was submitted 15 July 2011 around 11:10 AM. 024. Query 24
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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. ** STUDENT COMMENT: Faraday explain that it reached out from the charge, so would that be a concentration? It seems to me that the concentration would be near the center of the charge and the field around it would be more like radiation extending outward weakening with distance. INSTRUCTOR RESPONSE That's a good, and very important, intuitive conception of nature of the electric field around a point charge. However the meaning of the field is the force per unit charge. If you know the magnitude and direction of the field and the charge, you can find the magnitude and direction of the force on that charge. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qExplain how we calculate the magnitude and direction of the electric field at a given point of the x-y plane due to a given point charge at the origin. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Q is F = k q1 * Q / r^2, represents the magnitude of the force on a test charge where q1 is the charge at the origin. F / Q = k q1 / r^2, represents the electric field 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). So 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. 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 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. ** STUDENT QUESTION Why is it just Q and not Q2? INSTRUCTOR RESPONSE q1 is a charge that's actually present. Q is a 'test charge' that really isn't there. We calculate the effect q1 has on this point by calculating what the force would be if a charge Q was placed at the point in question. This situation can and will be expanded to a number of actual charges, e.g., q1, q2, ..., qn, at specific points. If we want to find the field at some point, we imagine a 'test charge' Q at that point and figure out the force exerted on it by all the actual charges q1, q2, ..., qn. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& 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: The magnitude and direction of the force on the positive charge at the lower left-hand corner can be 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 F = 324 * 10^-3 N F = .324 N Charges across a diagonal are like and separated by `sqrt(2) m = 1.414 m, approx, and exert repulsive forces of 0.162 N. 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 F = 162 * 10^-3 N F = .162 N. The charge at the lower left-hand corner therefore experiences a force of .324 N to the right, a force of .324 N straight upward and a force of 0.162 N 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 = 0.162 N sin(225 deg) Fy = -0.115 N, approx, Fx = 0.162 N cos(225 deg) Fx = -0.115 N. The total force in the x direction is: -0.115 N + 0.324 N =0 .21 N, approx. The total force in the y direction is: -0.115 N + 0.324 N = 0.21 N, approx. Thus the net force has magnitude: sqrt( (0.21 N)^2 + (0.21 N)^2) = 0.29 N at an angle of tan^-1( 0.21 N / 0.21 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 0.324 N to the left, 0.324 N straight up, and 0.162 N down and to the right. The net force is found by standard vector methods to be about 0.29 N up and to the left. ** confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 net force acts at an angle 45 deg with the x-axis, towards the center of the square. Similarly, the other forces on the charges at the ends 2, 3 and 4 respectively will have the same magnitude, since the four charges are similar in the magnitude and are placed at the similar distances. The forces will all be directed towards the center of the square. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!