Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    9
    ReputationRep:
    Question:
    Name:  Image0116.jpg
Views: 69
Size:  490.6 KB

    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:
    Offline

    3
    ReputationRep:
    ..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.
    Offline

    10
    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.
    • Thread Starter
    Offline

    9
    ReputationRep:
    (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?
    Offline

    10
    (Original post by Zishi)
    Right. But why can't we apply the formula for Potential in this question?
    Why do you need to?
    • Thread Starter
    Offline

    9
    ReputationRep:
    (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. . .
    Offline

    10
    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.
    • Thread Starter
    Offline

    9
    ReputationRep:
    (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.
    Offline

    1
    ReputationRep:
    (Original post by Zishi)
    Question:
    Name:  Image0116.jpg
Views: 69
Size:  490.6 KB

    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 ?
    • Thread Starter
    Offline

    9
    ReputationRep:
    (Original post by bmqib)
    Is the answer to the 3c. -8.33x10^10 ?
    Yeah.
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    Would you rather give up salt or pepper?
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

    Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

    Write a reply...
    Reply
    Hide
    Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.