course phy 232 What is a quadrupole? I looked them up but it seems confusing. I was also wondering what exact is on the test? I have looked at it but the questions keep changing so im not sure of the extent of the material covered. I am also trying to follow along with the queries with my book but some of the problems I cannot find? I think the numbers are wrong! The notes also skip big portions in the book. It's confusing. ‘É—„Ê×´â‰ùäÉçTwÿ±ÒªúåHâÑdÝŠassignment #003
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13:34:28 In your own words explain the meaning of the electric field.
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RESPONSE --> An electric field is the space around a charge in which other charges are affected. confidence assessment: 3
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13:34:51 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 --> self critique assessment: 3
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14:02:32 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 top right corner is a, left top corner is b, the right bottom is c and the left bottom is d. The magnitude of the force due to point a is = k*qa*qc/(rab)^2= 9x10^9* -6x10^-6*6x10^-6 / 1 = -.324N The magnitude due to the force at d is also -.324 The magnitude of the force at b is .164 useing the same equal with a larger radius sqrt(2). With the x-axis at point c the angle is 45 degrees and the vector is (-1,-1). confidence assessment: 1
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14:04:14 ** 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 --> self critique assessment: 3
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16:21:45 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 --> By placing the 5nC at the origin, to put the second charge at [4,0] requires force= k*qi*q2/r^2 force = -5.625x10^-5 N. If 6nC is placed at [4,3], there will be two resulting forces and then x and y components of each.The force between the second and third points using coulomb's law is -1.2x10^-4 N. The force between the first and third points is 1.08x10^-4 N. The x-components are -1.08x10^-4*cos(36.87)=8.64x10^-5 N. The y-components are 1.08x10^-4sin(36.87)+-1.2x10^-4 = - 5.52x10^-5 N. confidence assessment: 2
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16:22:24 ** 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: 3
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17:29:54 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 --> The quadrupole at these points has an dipole perpendicular to the lineof charges on the y-axis so a force is applied in the x direction. So, the magnitude of the electric field at (0,y) should be equal to coulomb's law k*q*q/(y-a)^2. confidence assessment: 1
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17:37:20 ** 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 --> I forgot to include that the force on the point[0,y] is not only due to the charge at[0.a] but also due to the the charge at the origin and the other charge. All of them together have their effect on the new charge. The equal used to find the charge at a point is k q / r^2. Where are here is the difference between y and a. self critique assessment: 1
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17:40:23 query univ 22.102 annulus in yz plane inner radius R1 outer R2, charge density `sigma.What is a total electric charge on the annulus?
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RESPONSE --> The total energy is equal to E = sigma/ [2epilson] where sigma is the charge density. From here, I have no idea where to go, I found the problem in the book but could not find the previous one and had no idea what to do. My book number do not match with the numbers given, I have the 12th edition too. This problem is 21.102. This section doen't even contain 22 does it? confidence assessment: 0
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