course Phy 122 ???m?I???????assignment #003003.
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23:24:46 `questionNumber 30000 In your own words explain the meaning of the electric field.
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RESPONSE --> The electric field found at a certain point can be defined as the force per unit of charge found by a test charge at that particular point. confidence assessment: 2
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23:25:19 `questionNumber 30000 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. **
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RESPONSE --> I failed to mention that electric field measures magnitude and direction of the electrical force. self critique assessment: 2
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23:47:13 `questionNumber 30000 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?
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RESPONSE --> The illustration shows that there are three other charges on any one of the four charges (one on the left, one on the right, and one diagonally). First we must find the magnitude and direction of the forces between each charge. Since the two charges adjacent to the left bottom charge are the same. These charges are equal and opposite the point charge, therfore, the force is attractive. The charge found diagonally is equal and the same as the point charge and therefore, will be repelled. However, since the force is an absolute value, the only value in calculation that will differ is the distance between points. The distance between each of the two adjacent points is the same, r =1.0m. The distance from our point and the diagonal charge can be found by the Pythagorean Theorem: r = `sqrt(1m^2 + 1m^2) = 1.414m Fattract = [(9x10^9Nm^2/C^2)(6x10^-6C)(6x10^-6C)] / (1m^2) = 0.324 N Frepel = [(9x10^9Nm^2/C^2)(6x10^-6C)(6x10^-6C)] / (1.414m^2) = 0.162 N We can conclude that this positive charge in the lower left-hand corner experiences forces of 0.324 N to the right and up. It also experiences a force of 0.162N diagonally. The angle of this force is 225 degrees from the x-axis. Now we can calculate the components of the diagonal force. Fx=(0.162N)cos(225) = -0.1145 N Fy=(0.162N)sin(225) = -0.1145 N Total Fx= 0.324N + -0.1145N = 0.209 N Total Fy= 0.324 + -0.1145N = 0.209 N Magnitude = `sqrt(Fx^2 + Fy^2) = `sqrt(0.209N^2 + 0.209N^2) = 0.2956 N Direction (`theta) = tan^ -1(.209/.209) = 45 degrees confidence assessment: 3
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23:47:24 `questionNumber 30000 ** 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. **
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RESPONSE --> Ok. self critique assessment: 3
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23:47:55 `questionNumber 30000 query university physics 21.68 (22.52 10th edition) 5 nC at the origin, -2 nC at (4 cm, 0). If 6 nC are placed at (4cm, 3cm), what are the components of the resulting force?
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RESPONSE --> I am not a university of physics student. confidence assessment:
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23:47:59 `questionNumber 30000 ** The 5 nC charge lies at distance sqrt( 4^2 + 3^2) cm from the 6 nC charge, which is 3 cm from the charge at (4 cm ,0). The force exerted on the 6 nC charge by the two respective forces have magnitudes .00011 N and .00012 N respectively. The x and y components of the force exerted by the charge at the origin are in the proportions 4/5 and 3/5 to the total charge, giving respective components .000086 N and .000065 N. The force exerted by the charge at (4 cm, 0) is in the negative y direction. So the y component of the resultant are .000065 N - .00012 N = -.000055 N and its x component is .000086 N. **
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RESPONSE --> self critique assessment:
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23:48:14 `questionNumber 30000 Query univ phy 21.76 (10th edition 22.60) quadrupole (q at (0,a), (0, -a), -2q at origin). For y > a what is the magnitude and direction of the electric field at (0, y)?
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RESPONSE --> I am not a university of physics student confidence assessment:
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23:48:18 `questionNumber 30000 ** The magnitude of the field due to the charge at a point is k q / r^2. For a point at coordinate y on the y axis, for y > a, we have distances r = y-a, y+a and y. The charges at these distances are respectively q, q and -2q. So the field is k*q/(y - a)^2 + k*q/(y + a)^2 - 2k*q/y^2 = 2*k*q*(y^2 + a^2)/((y + a)^2*(y - a)^2) - 2*k*q/y^2 = 2*k*q* [(y^2 + a^2)* y^2 - (y+a)^2 ( y-a)^2) ] / ( y^2 (y + a)^2*(y - a)^2) = 2*k*q* [y^4 + a^2 y^2 - (y^2 - a^2)^2 ] / ( y^2 (y + a)^2*(y - a)^2) = 2*k*q* [y^4 + a^2 y^2 - y^4 + 2 y^2 a^2 - a^4 ] / ( y^2 (y + a)^2*(y - a)^2) = 2*k*q* [ 3 a^2 y^2 - a^4 ] / ( y^2 (y + a)^2*(y - a)^2) . For large y the denominator is close to y^6 and the a^4 in the numerator is insignifant compared to a^2 y^2 sothe expression becomes 6 k q a^2 / y^4, which is inversely proportional to y^4. **
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RESPONSE --> self critique assessment:
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