Electric Potential well

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  1. Zishi's Avatar
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    Electric Potential well
    Question:
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    I'm facing a problem in part (b)(i). When using the formula V = \dfrac{Q}{4\pi\epsilon_0 r}, shouldn't we take the value of r to be 0.52x10-10 m? Mark scheme takes the value of r to be the diameter of circle, i.e 1.04x10-10 m.

    Also, why doesn't reading the value of potential from the graph give the correct answer? :confused:
    Last edited by Zishi; 18-04-2012 at 18:08.
  2. FireGarden's Avatar
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    • Posts: 385
    Re: Electric Potential well
    ..I don't know why the mark scheme would say that, since clearly the radius is the distance between the electron and nucleus. How weird . I would have thought your answer is correct.
  3. Stonebridge's Avatar
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    Re: Electric Potential well
    1) The value of V should be read directly from the graph at r = 0.52 x 10-10 m if it's the diameter that is given as 1.04x10-10
    2) The field strength at that point is the (negative) potential gradient at that point.
  4. Zishi's Avatar
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    Re: Electric Potential well
    (Original post by Stonebridge)
    1) The value of V should be read directly from the graph at r = 0.52 x 10-10 m if it's the diameter that is given as 1.04x10-10
    2) The field strength at that point is the (negative) potential gradient at that point.
    Right. But why can't we apply the formula for Potential in this question?
  5. Stonebridge's Avatar
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    Re: Electric Potential well
    (Original post by Zishi)
    Right. But why can't we apply the formula for Potential in this question?
    Why do you need to?
  6. Zishi's Avatar
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    Re: Electric Potential well
    (Original post by Stonebridge)
    Why do you need to?
    I don't need to, but just wondering that it should also give the same answer as given by the graph. . .
  7. Stonebridge's Avatar
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    Re: Electric Potential well
    If you plug in the correct values for the other quantities it should.
    The graph itself would have been plotted from the equation.
    If it doesn't give the same value that would be odd, but it is irrelevant in (what I can see of) this question as there is no need to use the formula.
    You are expected to use the graph.
  8. Zishi's Avatar
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    Re: Electric Potential well
    (Original post by Stonebridge)
    If you plug in the correct values for the other quantities it should.
    The graph itself would have been plotted from the equation.
    If it doesn't give the same value that would be odd, but it is irrelevant in (what I can see of) this question as there is no need to use the formula.
    You are expected to use the graph.
    Hmm. Anyway, it's giving a different answer. Many thanks.
  9. bmqib's Avatar
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    Re: Electric Potential well
    (Original post by Zishi)
    Question:
    Click image for larger version. 

Name:	Image0116.jpg 
Views:	19 
Size:	490.6 KB 
ID:	142645

    I'm facing a problem in part (b)(i). When using the formula V = \dfrac{Q}{4\pi\epsilon_0 r}, shouldn't we take the value of r to be 0.52x10-10 m? Mark scheme takes the value of r to be the diameter of circle, i.e 1.04x10-10 m.

    Also, why doesn't reading the value of potential from the graph give the correct answer? :confused:
    Is the answer to the 3c. -8.33x10^10 ?
  10. Zishi's Avatar
    • Peer Of The TSR Realm
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    Re: Electric Potential well
    (Original post by bmqib)
    Is the answer to the 3c. -8.33x10^10 ?
    Yeah.
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