course Phy 202 ?i????????assignment #002
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12:23:26 In your own words explain how the introductory experience with scotch tape illustrates the existence of two types of charge.
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RESPONSE --> If you place the sticky side of the tape to the non sticky side of another piece of tape and pull them apart you can see they repell each other. If you take a two more pieces of tape and repeat the above you will find that the two pieces repell one another but each one attracts a different one of the orginial two pieces. So like charges repell one another and opposite charges attract one another. confidence assessment: 2
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12:44:39 In your own words explain how the introductory experience with scotch tape supports the idea that the force between two charged particles acts along a straight line through those particles, either attracting the forces along this line or repelling along this line.
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RESPONSE --> The pieces of tape are closer at the top and repell each other further at the bottom so and attractive and replusive forces attract and repell along the line of the tape. confidence assessment: 1
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12:48:51 In your own words explain why this experience doesn't really prove anything about actual point charges, since neither piece of tape is confined to a point.
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RESPONSE --> To find a point charge you must have (x,y) coordinates and a fixed point. The tape has charges all along it. To find point charges it must be isolated and fixed. confidence assessment: 1
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12:56:58 If one piece of tape is centered at point A and the other at point B, then let AB_v stand for the vector whose initial point is A and whose terminal point is B, and let AB_u stand for a vector of magnitude 1 whose direction is the same as that of AB_u. Similarlylet BA_v and BA_u stand for analogous vectors from B to A. Vectors of length 1 are called unit vectors. If the pieces attract, then in the direction of which of the two unit vectors is the tape at point A pushed or pulled? If the pieces repel, then in the direction of which of the two unit vectors is the tape at B pushed or pulled? Explain.
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RESPONSE --> If attracted the tape at point at is pulled in the direction of AB_u (from point A toward point B). Vector BA_u is pushed toward B. If they repel at tape B the BA_u vector is pushed and the AB_u vector is pulled. Because the force is going away from A through B in the direction of AB_u. confidence assessment: 2
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13:07:33 Using the notation of the preceding question, which you should have noted on paper (keep brief running notes as you do qa's and queries so you can answer 'followup questions' like this), how does the magnitude of the vector AB_v depend on the magnitude of BA_v, and how does the magnitude of each vector compare with the distance between A and B?
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RESPONSE --> If repelled both AB_v and BA_v increase and decreaes if attracted. The distance between A and B remains constant but the vector lengths shorten and length with attraction and replusion. confidence assessment: 1
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13:12:53 Using the notation of the preceding question, how is the force experienced by the two pieces of tape influenced by the magnitude of AB_v or BA_v?`aThe expected answer is that the force exerted by two charges on one another is inversely proportional to the square of the distance between them. So as the magnitudes of the vectors, which are equal to the separation, increases the force decreases with the square of the distance; and/or if the magnitude decreases the force increases in the same proportinality. The two pieces of tape are not point charges, so this is not strictly so in this case, as some parts of the tape are closer than to the other tape than other parts.
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RESPONSE --> As magnitude increase (distance of separation) distance force decreases. Because diffrence parts of the tape are in diffrence locations this does not strictly apply. confidence assessment: 2
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13:20:03 Query introductory set #1, 1-5 Explain how we calculate the magnitude and direction of the electrostatic force on a given charge at a given point of the x-y plane point due to a given point charge at the origin.
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RESPONSE --> Magnitude can be found by finding the x and y vectors magnitude between the two charges. These vectors are then placed in the Pathagorean Theorem to find the magnitude of the vector. The angle is found by taking the inverse tangent of the vy divided by the vx. The angle then tells us the the direction by allowing us to know what quardrant it is in. confidence assessment: 1
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13:22:45 ** 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 force is one of repulsion if q1 and Q are of like sign, attraction if the signs are unlike. The force is therefore directly away from the origin (if q1 and Q are of like sign) or directly toward the origin (if q1 and Q are of unlike sign). To find the direction of this displacement vector we find arctan(y / x), adding 180 deg if x is negative. If the q1 and Q are like charges then this is the direction of the field. If q1 and Q are unlike then the direction of the field is opposite this direction. The angle of the field would therefore be 180 degrees greater or less than this angle.**
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RESPONSE --> I found the direction correctly however I did not use the correctly state how to find the magnitude of the force on the test charge. It is F= k q1*Q/ r^2. I also did not explain that if the charges are alike it is directly away from the orign and directly toward the orign if the charges are unlike. self critique assessment: 2
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13:31:28 Explain how we calculate the magnitude and direction of the electric field at a given point of the x-y plane point due to a given point charge at the origin.
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RESPONSE --> The magnitude is found by: E= k Q / r^2 E is the electric field k is the constatn Q is the give point charge at the origin r is the distant between the given point and the orign The direction is then once again foud by arctangent y/x which gives the angle. confidence assessment: 1
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13:37:09 ** 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. **
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RESPONSE --> I failed include the test charge. Instead I called Q the charge at the orgin when it should have been the test charge and q1 should have been the charge at the origin. However though I didn't use the same notation as my variables were defined I would have gotten the correct answer. I should have noted that if the x is negative you must add 180 degrees. Also if q1 is positive the direction of the displacement vector is the same of the direction of the field, and if q1 is negative it is in the opposite direction of the field. self critique assessment: 2
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?????Q????Y????? assignment #003 003. Physics II 06-20-2007 R????e????U??O????assignment #003 003. Physics II 06-20-2007
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20:24:45 In your own words explain the meaning of the electric field.
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RESPONSE --> It is the force exerted by an object over distance on a postive test charge. This gives us the equation E=F/q E = kQq/r^2 confidence assessment: 2
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20:26:11 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 did not mention magnitude or direction which are both defining aspects of an electric field. self critique assessment: 2
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20:39:03 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 --> (1m)^2 + (1m)^2 = 2m^2 sqrt(2m^2) = 1.4m = r mag = (k*q1*Q)/r^2 mag= [(9E9 Nm^2/C^2)(-6E-6C)(6E-6C)/(1.4m)^2 mag = -.1653 N Direction is opposite and away from the orgin. confidence assessment: 1
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20:43:07 ** 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 --> My mistake was using the wrong C values. I drew the diagram correctly but thoughtless used the - 6 microC value in place of the 6micro C for both. I also did not find the forces on Fy and Fx using trig. self critique assessment: 1
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20:43:28 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 --> skip confidence assessment:
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20:43:57 ** 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 --> the problem for for university physics self critique assessment: 3
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20:44:28 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 --> university phyics.. skip confidence assessment: 3
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20:44:49 ** 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 --> ok self critique assessment: 3
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20:45:11 ** The total charge on the annulus is the product Q = sigma * A = sigma * (pi R2^2 - pi R1^2). To find the field at distance x along the x axis, due to the charge in the annulus, we first find the field due to a thin ring of charge: The charge in a thin ring of radius r and ring thickness `dr is the product `dQ = 2 pi r `dr * sigma of ring area and area density. From any small segment of this ring the electric field at a point of the x axis would be directed at angle arctan( r / x ) with respect to the x axis. By either formal trigonometry or similar triangles we see that the component parallel to the x axis will be in the proportion x / sqrt(x^2 + r^2) to the magnitude of the field from this small segment. By symmetry only the xcomponent of the field will remain when we sum over the entire ring. So the field due to the ring will be in the same proportion to the expression k `dQ / (x^2 + r^2). Thus the field due to this thin ring will be magnitude of field due to thin ring: k `dQ / (x^2 + r^2) * x / sqrt (x^2 + r^2) = 2 pi k r `dr * x / (x^2 + r^2)^(3/2). Summing over all such thin rings, which run from r = R1 to r = R2, we obtain the integral magnitude of field = integral ( 2 pi k r x /(x^2 + r^2)^(3/2) with respect to r, from R1 to R2). Evaluating the integral we find that magnitude of field = 2* pi k *x* | 1 /sqrt(x^2 + r1^2) - 1 / sqrt(x^2 + r2^2) | The direction of the field is along the x axis. If the point is close to the origin then x is close to 0 and x / sqrt(x^2 + r^2) is approximately equal to x / r, for any r much larger than x. This is because the derivative of x / sqrt(x^2 + r^2) with respect to x is r^2 / (x^2+r^2)^(3/2), which for x = 0 is just 1/r, having no x dependence. So at small displacement `dx from the origin the field strength will just be some constant multiple of `dx. **
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RESPONSE --> ok self critique assessment: 3
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