#$&* course Phy 232 Assignment 2 Physics II11-08-2011
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07:04:58 In your own words explain how the introductory experience with scotch tape illustrates the existence of two types of charge.
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RESPONSE --> By bringing the ""sticky"" and ""non-sticky"" side of different strips of scotch tape together and ripping them apart-the two pieces of scotch tape demonstrate how the scotch tape illustrates the existence of positive and negative electric charges. These charges are demonstrated by the repulsive and attractive forces that each piece shows every time it's both are hanged beside each other. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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18:07:45 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 scotch tape supports this idea by each strip acting as a positive and negative charge. Whether they attract or repel depends on what charges they hold. If one of the sides is positive and they other one is negative x axis, the forces will attract each other. On the other hand, if the charges on the scotch tape are both on the positive x-axis or both on the negative x -axis - they will both repel each other. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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18:12:35 In your own words explain why this experience doesn't really prove anything about actual point charges, since neither piece of tape is confined to a point.
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RESPONSE --> These does not prove anything about the actual point charges since in the ""scotch tape"" experience we were dealing with a large number of point charges distributed over it. The interaction between many charges on one piece and many charges on another is more complicated. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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18:20:44 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 the pieces attract each other, then point A is pulled in the direction of the unit vector AB_u. In addition, if the pieces repel - point B will be pushed in the direction of the unit vector AB_u. They both end up being in the same direction AB_u since for each respective question different points were asked for. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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18:34:05 Using the notation of the preceding question, which you should have noted on paper (keep brief running notes as you do qa's and queries so you can answer 'followup questions' like this), how does the magnitude of the vector AB_v depend on the magnitude of BA_v, and how does the magnitude of each vector compare with the distance between A and B?
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RESPONSE --> The directions of AB_v and BA_v are different, however, the magnitudes of both AB_v and BA_v are the same. The magnitude is the distance between both A and B and this is how the magnitude of each vector compare with the distance beetween A and B. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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18:39:19 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 --> The force experienced by the two pieces of tape are inversely proportional to the square of the distance between them. In other words, the larger the distance the smaller the force. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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20:25:51 Query introductory set #1, 1-5 Explain how we calculate the magnitude and direction of the electrostatic force on a given charge at a given point of the x-y plane point due to a given point charge at the origin.
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RESPONSE --> Throughout the problems set #1, 1-5, the magnitude of the force was calculated by utilizing the Coulomb force equation F = K |q1||q2|/r^2 and the direction of the force is parallel to the vector r12 from the second charge to the first. After determining the vector r12x and r12y, the r12 unit vector is determined. By doing this we will be able to determine the direction of r12. For example: if we have charges q1 and q2 and q1 is located at (x1,y1) and q2 is located at (x2,y2). The magnitude of the F = k|q1||q2|/r^2 = k|q1||q2|/sqrt((x2-x1)^2+(y2-y1)^2). The direction of the electrostatic force can be determined r12_vector = <(x2-x1),(y2-y1)>/sqrt((x2-x1)^2+(y2-y1)^2). The x component of the force F12x = k q1 q2 / [ (x2 - x1)^2 + (y2 - y1)^2 ] * [ (x2 - x1) / `sqrt( (x2 - x1)^2 + (y2 - y1)^2 ) ] The y direction of the force F12y = k q1 q2 / [ (x2 - x1)^2 + (y2 - y1)^2 ] * [ (y2 - y1) / `sqrt( (x2 - x1)^2 + (y2 - y1)^2 ) ] confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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20:35:48 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 of the electric field can be calculated by utilizing the following equation: E = k Q / r^2 The direction can be determined by utilizing: E_x = k Q q / ( (x2-x1)^2 + (y2 - y1)^2) * (x2 - x1) / sqrt( (x2-x1)^2 + (y2 - y1)^2) E_y = k Q q / ( (x2-x1)^2 + (y2 - y1)^2) * (y2 - y1) / `sqrt( (x2-x1)^2 + (y2 - y1)^2) The degree or angle at which the electric field is directed is equal to sqrt(E_x^2+E_y^2) confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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" Self-critique (if necessary): ------------------------------------------------ Self-critique rating:
#$&* course Phy 232 Assignment 2 Physics II11-08-2011
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07:04:58 In your own words explain how the introductory experience with scotch tape illustrates the existence of two types of charge.
......!!!!!!!!...................................
RESPONSE --> By bringing the ""sticky"" and ""non-sticky"" side of different strips of scotch tape together and ripping them apart-the two pieces of scotch tape demonstrate how the scotch tape illustrates the existence of positive and negative electric charges. These charges are demonstrated by the repulsive and attractive forces that each piece shows every time it's both are hanged beside each other. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.................................................
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18:07:45 In your own words explain how the introductory experience with scotch tape supports the idea that the force between two charged particles acts along a straight line through those particles, either attracting the forces along this line or repelling along this line.
......!!!!!!!!...................................
RESPONSE --> The scotch tape supports this idea by each strip acting as a positive and negative charge. Whether they attract or repel depends on what charges they hold. If one of the sides is positive and they other one is negative x axis, the forces will attract each other. On the other hand, if the charges on the scotch tape are both on the positive x-axis or both on the negative x -axis - they will both repel each other. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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18:12:35 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.
......!!!!!!!!...................................
RESPONSE --> These does not prove anything about the actual point charges since in the ""scotch tape"" experience we were dealing with a large number of point charges distributed over it. The interaction between many charges on one piece and many charges on another is more complicated. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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18:20:44 If one piece of tape is centered at point A and the other at point B, then let AB_v stand for the vector whose initial point is A and whose terminal point is B, and let AB_u stand for a vector of magnitude 1 whose direction is the same as that of AB_u. Similarlylet BA_v and BA_u stand for analogous vectors from B to A. Vectors of length 1 are called unit vectors. If the pieces attract, then in the direction of which of the two unit vectors is the tape at point A pushed or pulled? If the pieces repel, then in the direction of which of the two unit vectors is the tape at B pushed or pulled? Explain.
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RESPONSE --> If the pieces attract each other, then point A is pulled in the direction of the unit vector AB_u. In addition, if the pieces repel - point B will be pushed in the direction of the unit vector AB_u. They both end up being in the same direction AB_u since for each respective question different points were asked for. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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18:34:05 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 --> The directions of AB_v and BA_v are different, however, the magnitudes of both AB_v and BA_v are the same. The magnitude is the distance between both A and B and this is how the magnitude of each vector compare with the distance beetween A and B. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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18:39:19 Using the notation of the preceding question, how is the force experienced by the two pieces of tape influenced by the magnitude of AB_v or BA_v?`aThe expected answer is that the force exerted by two charges on one another is inversely proportional to the square of the distance between them. So as the magnitudes of the vectors, which are equal to the separation, increases the force decreases with the square of the distance; and/or if the magnitude decreases the force increases in the same proportinality. The two pieces of tape are not point charges, so this is not strictly so in this case, as some parts of the tape are closer than to the other tape than other parts.
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RESPONSE --> The force experienced by the two pieces of tape are inversely proportional to the square of the distance between them. In other words, the larger the distance the smaller the force. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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20:25:51 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 --> Throughout the problems set #1, 1-5, the magnitude of the force was calculated by utilizing the Coulomb force equation F = K |q1||q2|/r^2 and the direction of the force is parallel to the vector r12 from the second charge to the first. After determining the vector r12x and r12y, the r12 unit vector is determined. By doing this we will be able to determine the direction of r12. For example: if we have charges q1 and q2 and q1 is located at (x1,y1) and q2 is located at (x2,y2). The magnitude of the F = k|q1||q2|/r^2 = k|q1||q2|/sqrt((x2-x1)^2+(y2-y1)^2). The direction of the electrostatic force can be determined r12_vector = <(x2-x1),(y2-y1)>/sqrt((x2-x1)^2+(y2-y1)^2). The x component of the force F12x = k q1 q2 / [ (x2 - x1)^2 + (y2 - y1)^2 ] * [ (x2 - x1) / `sqrt( (x2 - x1)^2 + (y2 - y1)^2 ) ] The y direction of the force F12y = k q1 q2 / [ (x2 - x1)^2 + (y2 - y1)^2 ] * [ (y2 - y1) / `sqrt( (x2 - x1)^2 + (y2 - y1)^2 ) ] confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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20:35:48 Explain how we calculate the magnitude and direction of the electric field at a given point of the x-y plane point due to a given point charge at the origin.
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RESPONSE --> The magnitude of the electric field can be calculated by utilizing the following equation: E = k Q / r^2 The direction can be determined by utilizing: E_x = k Q q / ( (x2-x1)^2 + (y2 - y1)^2) * (x2 - x1) / sqrt( (x2-x1)^2 + (y2 - y1)^2) E_y = k Q q / ( (x2-x1)^2 + (y2 - y1)^2) * (y2 - y1) / `sqrt( (x2-x1)^2 + (y2 - y1)^2) The degree or angle at which the electric field is directed is equal to sqrt(E_x^2+E_y^2) confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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