course phy 232 }HҌ~֭{FۚQassignment #003
......!!!!!!!!...................................
18:02:08 In your own words explain the meaning of the electric field.
......!!!!!!!!...................................
RESPONSE --> 3 confidence assessment:
.................................................
......!!!!!!!!...................................
18:02:15 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 --> 3 self critique assessment: 3
.................................................
......!!!!!!!!...................................
18:02:19 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 --> 3 confidence assessment:
.................................................
......!!!!!!!!...................................
18:02:29 ** 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 --> 5 self critique assessment:
.................................................
......!!!!!!!!...................................
18:02:33 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?
......!!!!!!!!...................................
RESPONSE --> 4 confidence assessment:
.................................................
......!!!!!!!!...................................
18:02:40 ** 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. **
......!!!!!!!!...................................
RESPONSE --> 4 self critique assessment:
.................................................
......!!!!!!!!...................................
18:03:46 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)?
......!!!!!!!!...................................
RESPONSE --> 4 confidence assessment:
.................................................
......!!!!!!!!...................................
18:04:09 ** 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. **
......!!!!!!!!...................................
RESPONSE --> e self critique assessment:
.................................................
......!!!!!!!!...................................
18:04:22 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?
......!!!!!!!!...................................
RESPONSE --> 4 confidence assessment:
.................................................
^zlfajPx assignment #004 004. Physics II 06-15-2008
......!!!!!!!!...................................
19:57:49 Query introductory set #1, 10-17 Explain how to find the potential difference in volts between two given points on the x axis, due to a given charge at the origin.
......!!!!!!!!...................................
RESPONSE --> First, you find the force between the origin charge and the two points separately then find the force between the two points and multiple by the distance to get the work then divide bythe charge of the field to get the volts or J/C. confidence assessment: 3
.................................................
......!!!!!!!!...................................
20:11:19 ** Potential difference is the work per Coulomb of charge moved between the two points. To find this work you can multiply the average force on a Coulomb of charge by the displacement from the first point to the second. You can find an approximate average force by finding the force on a 1 Coulomb test charge at the two points and averaging the two forces. Multiplying this ave force by the displacement gives an approximate potential difference. Since the force is not a linear function of distance from the given charge, if the ratio of the two distances from the test charge is not small the approximation won't be particularly good. The approximation can be improved to any desired level of accuracy by partitioning the displacement between charges into smaller intervals of displacement and calculating the work done over each. The total work required is found by adding up the contributions from all the subintervals. University Physics students should understand how this process yields the exact w ""ork, which is the integral of the force function F(x) = k Q / x^2 between the two x values, yielding total work W = k * Q * 1 Coulomb ( 1 / x1 - 1 / x2) and potential difference V = k * Q ( 1 / x1 - 1 / x2). **
......!!!!!!!!...................................
RESPONSE --> I stated in the response that a volt is J/C but it is not. It is J/cm which means that it is missing the displacement.So it would be much easier to find the two forces on each point then average them then multiply by the displacement. self critique assessment: 1
.................................................
......!!!!!!!!...................................
20:15:59 Explain how to find the potential difference between two points given the magnitude and direction of the uniform electric field between those points.
......!!!!!!!!...................................
RESPONSE --> Given E, the magnitude of the uniform electric field, F=E*q so the magnitude of the field times the charge would give the force between them. Once this force is obtained for both points the average can be taken and multiplied by the displacement to get the difference between the two points in J. confidence assessment: 1
.................................................
......!!!!!!!!...................................
20:17:10 ** The work per Coulomb done between the two points is equal to the product of the electric field E and the displacement `dr. Thus for constant field E we have V = E * `dr. **
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 3
.................................................
......!!!!!!!!...................................
20:21:39 Explain how to find the average electric field between two points given a specific charge and the work done on the charge by the electric field as the charge moves between the points.
......!!!!!!!!...................................
RESPONSE --> First, calculate the average displacement between the two points then divide the work by it to get the average force. E=F/q so then divide the force by the charge. confidence assessment: 3
.................................................
......!!!!!!!!...................................
20:22:16 ** You get ave force from work and distance: F_ave = `dW / `ds. You get ave electric field from work and charge: E_ave = F / q. An alternative: Find potential difference `dV = `dW / q. Ave electric field is Eave = `dV / `ds **
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 3
.................................................
......!!!!!!!!...................................
20:27:33 In your own words explain the meaning of voltage.
......!!!!!!!!...................................
RESPONSE --> The voltage is specifically the work divided by the coulombs per unit of charge. It is like the potential energy in an electric field. confidence assessment: 2
.................................................
......!!!!!!!!...................................
20:27:52 ** Voltage is the work done per unit of charge in moving charge from one point to another. **
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 3
.................................................
"