#$&* course Phy 232 If your solution to stated problem does not match the given solution, you should self-critique per instructions athttp://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.
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Given Solution: `a** 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). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `qExplain how to find the potential difference between two points given the magnitude and direction of the uniform electric field between those points. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Given a the magnitude of an electric field, the potential difference is just the magnitude of the force times the distance or r. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `qExplain 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Work divided by charge will give potential difference. The average electrical field can be found by taking the potential difference divided by distance. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** You get ave force from work and distance: F_ave = `dW / `ds. You get ave electric field from work and charge: E_ave = F / q. An alternative: Find potential difference `dV = `dW / q. Ave electric field is Eave = `dV / `ds ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `qIn your own words explain the meaning of voltage. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Voltage is defined as the work divided by charge from one point to another, or in other words, a potential difference. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution:ok `a** Voltage is the work done per unit of charge in moving charge from one point to another. ** STUDENT SOLUTION Voltage is the difference in electric field times the distance between two points: V = E*d The bigger the difference in voltage between two points, the greater potential to drive electrical current. INSTRUCTOR COMMENT Your answer was also correct. Note that the units for voltage are Joules / Coulomb, which is consistent with the given solution (work per unit of charge). However this unit can also be expressed as N * m / C , or (N / C) * m, consistent with your statement (N / C is the unit of electric field, so this would be the product of electric field and distance). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!