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Proton Acceleration - Do you Agree? |
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| May25-05, 07:05 PM | #1 |
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Proton Acceleration - Do you Agree?
A proton, initially at rest, is accelerated from plate A to plate B, and aquires 1.92 * 10-17 J of kinetic energy.
i. Which plate is positive and which is negative? ii. What is the potential difference? iii. Sketch the correct direction of E between the plates. Does the proton move with or against the field E? Does this agree with the gain (or loss) calculated in ii?? Explain? My Solution: i. Plate A is positive Plate B is negative ii. q * delta V = 1.92 * 10-17 J Delta V = -1.92 * 10-17 / 1.602 * 10-19 C = -1.1985 * 10^2 V The potential difference is -1.1985 * 10^2 V This is a loss in potential iii. The proton moves with the field and this agrees with my previous calculation as an electron moving with a field is like a mass falling downward within Earth's gravitational fieldd. There is no need for an external force so there we witness a loss of potential. Does this make sense to everyone? |
| May25-05, 07:12 PM | #2 |
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That last line is a bit confusing.
In (ii) the potential between the plates isnt affected by the proton's gain in kinetic energy. The proton's final kinetic energy is due to the presence of the potential difference. The proton experienced a change in PE when flowing from A to B, this change in PE results in the change in KE. |
| May25-05, 11:45 PM | #3 |
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1)Ok for your first question, as the proton gains energy moving from A to B ... B should have a higher potential, that is B should be positive.
2) Potential difference can be calculated: [latex]qV= 1.92 x 10^-17[/latex] Calculate V 3)Electric field lines go from +ve to -ve plates and the proton moves against the field. The proton has to do work in moving against the field ...so that much energy is stored in it which is the gain in energy. |
| May26-05, 04:02 PM | #4 |
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Proton Acceleration - Do you Agree? |
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