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Electric Potential well

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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:
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    ..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.
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    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.
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    (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?
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    (Original post by Zishi)
    Right. But why can't we apply the formula for Potential in this question?
    Why do you need to?
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    (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. . .
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    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.
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    (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.
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    (Original post by Zishi)
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    Click image for larger version. 

Name:	Image0116.jpg 
Views:	31 
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 ?
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    (Original post by bmqib)
    Is the answer to the 3c. -8.33x10^10 ?
    Yeah.

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