course Phy 202
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21:16:28 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 --> You would multiply the average force on a Coulomb of charge by the displacement from the first point to the second. ave force mulitplied by the displacement gives an approx. potential difference.
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21:16:52 ** 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, i seem to understand this
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21:17:13 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 --> Given the magnitude and direction of the uniform electric field between those points one would use work per Coulomb = E * dr
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21:17:29 ** 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 --> This is what i would do
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21:18:21 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 --> Use Fave = `dW / `ds for the ave force and then get the electric field by dividing F/q
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21:18:43 ** You get ave force from work and distance: Fave = `dW / `ds. You get ave electric field from work and charge: Eave = F / q. An alternative: Find potential difference `dV = `dW / q. Ave electric field is Eave = `dV / `ds **
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RESPONSE --> OK i see how to use both ways
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21:18:59 In your own words explain the meaning of the electric field.
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RESPONSE --> The electric field is composed of an electrical force and what the electrical force is
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21:19:27 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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21:21:31 In your own words explain the meaning of voltage.
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RESPONSE --> V = 2k*omega, this is what i know from the experiment
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21:22:15 ** Voltage is the work done per unit of charge in moving charge from one point to another. **
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RESPONSE --> I believe this is the what the equation means
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21:23:49 Query experiment 16 current flow and energy You should have submitted your writeup of this experiment. However, answer the following questions. First question: Why is it that when the circuit formed has the least resistance the generator is hardest to crank?
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RESPONSE --> Possibly due to a greater power, or more current flow running through the circuit
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21:24:39 ** Less resistance implies more current, meaning more charge per unit of time and therefore more work per unit of time. To perform this work at a constant cranking rate (hence a constant voltage, and also a constant distance per unit of time) requires more force (since the distance in a unit of time is constant, in order to do work at a greater rate the force must be greater). **
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RESPONSE --> OK this is what i thought, with having more current, the force is increased to crank the generator
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21:25:10 Second question: Why do you think it is that the generator is harder to crank with two bulbs in parallel then with the same to bulbs and series?
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RESPONSE --> There are more leads to take into account in the parallel
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21:26:01 ** In a parallel circuit the full voltage of the generator is applied to both branches of the circuit, since both are directly connected to the generator. In a series circuit, the circuit doesn't split and the voltage is divided (not equally unless the bulbs have equal resistance) between the two bulbs. So in the parallel circuit both bulbs experience greater voltage, and hence greater current, than in the series circuit. **
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RESPONSE --> This is what i mean, the parallel has more leads, or the voltage is branched out. In the series, the circuit doesn't split and the voltage is divided.
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