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    (Original post by ayvaak)
    Yeah me too. Are you thinking of studying physics next year?

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    It was a close call between physics and maths, but I decided to study Maths because it's versatile and I don't know what i want to do as a career yet.
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    anyone got a copy of the 2015 paper which doesn't print with "draft exemplar" over the page?

    EDIT: nvm, saved the paper and then printed it and it's fine.
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    Could anyone help please - I'm stuck on question 4 of the unit 4 paper here:
    http://qualifications.pearson.com/co...ce-physics.pdf
    The question's buried fairly deep so I'll try to explain: There are four graphs shown and the question asks which shows the relationship between Ywhen it is the electric field strength between two parallel plateswith a constant potential difference across them and X is thedistance apart of the plates?
    I thought it would be a straight line graph with a negative gradient but the answers show that it's D, a graph of exponential decay. I understand that Y should never become 0, but can't work out why it's an exponential graph because there are no powers in E=V/d?

    Thanks!
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    (Original post by candycake)
    Could anyone help please - I'm stuck on question 4 of the unit 4 paper here:
    http://qualifications.pearson.com/co...ce-physics.pdf
    The question's buried fairly deep so I'll try to explain: There are four graphs shown and the question asks which shows the relationship between Ywhen it is the electric field strength between two parallel plateswith a constant potential difference across them and X is thedistance apart of the plates?
    I thought it would be a straight line graph with a negative gradient but the answers show that it's D, a graph of exponential decay. I understand that Y should never become 0, but can't work out why it's an exponential graph because there are no powers in E=V/d?

    Thanks!
    That's not an exponential graph, that's an inverse/reciprocal graph. If you look at a graph of y = 1/x you'll see :
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    (Original post by candycake)
    Could anyone help please - I'm stuck on question 4 of the unit 4 paper here:
    http://qualifications.pearson.com/co...ce-physics.pdf
    The question's buried fairly deep so I'll try to explain: There are four graphs shown and the question asks which shows the relationship betweenYwhen it is the electric field strength between two parallel plateswith a constant potential difference across them and X is thedistance apart of the plates?
    I thought it would be a straight line graph with a negative gradient but the answers show that it's D, a graph of exponential decay. I understand that Y should never become 0, but can't work out why it's an exponential graph because there are no powers in E=V/d?

    Thanks!
    D is not an exponential decay graph, it's a y=1/x graph which is correct because as you say E=V/d where V is constant, therefore Y=c/X where Y is E, d is X and c is a constant.
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    (Original post by klosovic)
    D is not an exponential decay graph, it's a y=1/x graph which is correct because as you say E=V/d where V is constant, therefore Y=c/X where Y is E, d is X and c is a constant.
    (Original post by pineneedles)
    That's not an exponential graph, that's an inverse/reciprocal graph. If you look at a graph of y = 1/x you'll see :
    Name:  1466335312101.jpg
Views: 58
Size:  10.7 KB


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    Thank you so much! I had totally forgotten that (Unfortunately my maths is getting a little rusty now...!)
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    http://qualifications.pearson.com/co...ce-physics.pdfCould someone please help with Q 15ii of this Unit 4 paper. Im really stuck on as to why the voltage would be 1.6 initially?Thanks in advance
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    Could anyone help please - For question 15bii) same paper as above, (page 58), how has the original value of current been calculated? The mark scheme uses a value of 1.6V but I can't see where that has come from? Sorry - I'm sure I'm asking really obvious questions

    Edit:
    (Original post by ayvaak)
    http://qualifications.pearson.com/co...ce-physics.pdfCould someone please help with Q 15ii of this Unit 4 paper. Im really stuck on as to why the voltage would be 1.6 initially?Thanks in advance
    Oops - didn't see your post, I've asked the same!
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    Also what does one lot of RC represent? just to confirm its not time taken for the charge to be halved
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    (Original post by ayvaak)
    Also what does one lot of RC represent? just to confirm its not time taken for the charge to be halved
    RC is the time constant i.e. time for voltage/charge/current to reduce to 37%
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    (Original post by candycake)
    Could anyone help please - For question 15bii) same paper as above, (page 58), how has the original value of current been calculated? The mark scheme uses a value of 1.6V but I can't see where that has come from? Sorry - I'm sure I'm asking really obvious questions

    Edit:

    Oops - didn't see your post, I've asked the same!
    I got an initial value of 0.75 A, but apparently thats wrong and I'm not sure why either
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    (Original post by Mowerharvey)
    I got an initial value of 0.75 A, but apparently thats wrong and I'm not sure why either
    That's what I got too - perhaps there's a mistake in the mark scheme? Sample assessments are prone to errors sometimes.

    Also - did I see you're going to the UEA? That's one of my choices too. Would be nice to know someone there! My brother's just finished his degree there and loved it
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    Could someone pls explain 18i) and 18ii) on the SAM?

    http://qualifications.pearson.com/co...ce-physics.pdf
    http://qualifications.pearson.com/co...ce-physics.pdf
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    Two questions:
    1. Why isn't the resultant force 0?
    2. Why is the graph not B?
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    (Original post by target21859)
    Two questions:
    1. Why isn't the resultant force 0?
    2. Why is the graph not B?
    I believe for 2 its because electric field strength is a vector, and as the charge is negative then E will be negative also. Is the answer D?
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    (Original post by candycake)
    That's what I got too - perhaps there's a mistake in the mark scheme? Sample assessments are prone to errors sometimes.

    Also - did I see you're going to the UEA? That's one of my choices too. Would be nice to know someone there! My brother's just finished his degree there and loved it
    I assume that its a mistake because I do not understand why they would use 1.6 v

    Yes I am trying to get into UEA, it looks like a great uni to go to! Good to hear that your brother likes it too. What course are you thinking of doing there?
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    (Original post by Mowerharvey)
    I believe for 2 its because electric field strength is a vector, and as the charge is negative then E will be negative also. Is the answer D?
    Yes thanks that explains it. What about 1?
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    (Original post by target21859)
    Yes thanks that explains it. What about 1?
    For 1 the charged particle experiences an electrostatic force of attraction and an electrostatic force of repulsion, which both act in the same direction so they don't cancel out at x. If both of the charges Q were the same, then the resultant force at x would be 0 because the forces would act in opposite directions.

    I don't know if the resultant will be a maximum or minimum though
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    (Original post by Mowerharvey)
    For 1 the charged particle experiences an electrostatic force of attraction and an electrostatic force of repulsion, which both act in the same direction so they don't cancel out at x. If both of the charges Q were the same, then the resultant force at x would be 0 because the forces would act in opposite directions.

    I don't know if the resultant will be a maximum or minimum though
    It's a minimum.
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    (Original post by Mowerharvey)
    I assume that its a mistake because I do not understand why they would use 1.6 v

    Yes I am trying to get into UEA, it looks like a great uni to go to! Good to hear that your brother likes it too. What course are you thinking of doing there?
    English lit - the course looks great and really versatile! Did you apply for maths?
 
 
 
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