course PHY202 TÏ®ëóÇâ¥éj‹_Õ`‰åÍfÝZë½³Ìzz½˜ÛÝ‚Ÿ¼¼assignment #020
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17:43:31 Query introductory set #1, 1-8 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 --> The magnitude is found by the formula F=(kq1q2)/r^2 where k is the constant 9x10^9 , q1 and q2 are the charges and r is the distance between the charges. The direction is determined by the the charges, like charges repell and unlike charges attract. You also find the direction by usin the unit vector and the x & y components of each charge to fine the resultant direction. The angle of the vector is found using arctan(y/x)
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17:43:48 ** 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 --> ok
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17:45:17 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 manitude and direction of the electric field is found in the same manner as the F.
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17:46:00 ** 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 --> That right E=F/Q
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•ébáÏÉ™ô¬ŽçëWΓ‰‡·¿¿¶ÐŽ˜¥ÄÀ assignment #021 Ù‰ÈÑÜÖûj±ÀÕØØ”ÞýÚÚäÞ¦ü| Physics II 08-03-2006
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17:49:10 Query introductory set #1, 9-16 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.
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RESPONSE --> A volt is the measurement of work per unit charge, measured in J/C. So, Volt=W/Q
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17:49:23 ** 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 work, 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). **
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RESPONSE --> OK
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17:54:31 Explain how to find the potential difference between two points given the magnitude and direction of the uniform electric field between those points.
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RESPONSE --> The potiental difference Volt is equal to V=W/Q. Work ids difind as F*d. E is equal to F/Q so E='deltaW/(Q*('deltadistance)). So 'dV=E*Q
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17:54:48 ** 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. **
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RESPONSE --> Yes
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20:46:20 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.
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RESPONSE --> Work is Force*distance. The electric field, E=F/q, So Favg/Q=E
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20:46:30 ** 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 **
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RESPONSE --> OK
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20:49:56 In your own words explain the meaning of the electric field.
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RESPONSE --> Electric field is the force between two charges or the force exerted outward on a test point charge
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20:50:05 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 --> ok
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20:51:58 In your own words explain the meaning of voltage.
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RESPONSE --> Voltage is the term used to describe the electric field and can be considered the pushing mechanism for current through a circuit. The diference in potential at two points measured in J/C
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20:52:06 ** Voltage is the work done per unit of charge in moving charge from one point to another. **
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RESPONSE --> ok
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