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AQA Physics PHYA4 - Thursday 11th June 2015 [Exam Discussion Thread] Watch

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    (Original post by CD223)
    Anyone likely to be able to take home with them a MC copy of this year's paper after the exam so we can discuss answers? Or is that too illegal?:L


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    Loads of people did that for Chemistry 1 and 2 so don't think its illegal
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    (Original post by sykik)
    Loads of people did that for Chemistry 1 and 2 so don't think its illegal
    Written is different but MC is impossible to debate!


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    (Original post by CD223)
    Anyone likely to be able to take home with them a MC copy of this year's paper after the exam so we can discuss answers? Or is that too illegal?:L


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    What does MC mean here?


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    Multiple choice i reckon i just got it
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    (Original post by sykik)
    Loads of people did that for Chemistry 1 and 2 so don't think its illegal
    How can you take that?


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    Predictions for 6 marker? I normally get leave them out on past papers to be honest


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    (Original post by Protoxylic)
    The ratio of the charge to the distance is the same for both charges. Namely; 4uC/d=16uC/(120mm-d)
    You just need to rearrange this to find d.

    Another way of looking at it is that the potential due to the 4uC charge at that point 'd' is equal and opposite (in sign) to the potential due to the 16uC charge. So 4uC/d + -16uC/(120mm-d) = 0 And rearrange to find d.
    Thank you!!! it was just i forgot what you do with the 120mm, but thats helped
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    (Original post by EmiratesCaptain)
    Thank you!!! it was just i forgot what you do with the 120mm, but thats helped
    Np, anytime
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    (Original post by Mehrdad jafari)
    How can you take that?


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    take?
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    (Original post by sykik)
    take?
    I know teachers can actually take the paper right after the exam as my teacher did and he discussed the answers with us after the exam


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    (Original post by EmiratesCaptain)
    Can anyone remind me how to work out Q13 in this paper?
    http://filestore.aqa.org.uk/subjects...W-QP-JUN10.PDF
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    The resultant electric potential would be when the charge along the line is 0C.

    And that position is a fifth away from the positive charge on the left or 4/5 away from the negative charge on the right.

    I'm going to consider it from the +ve charge on the left.
    So all we have to do is 120 ÷ 5 = 24

    Hope that helps
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    (Original post by CD223)
    C3 went really well - thanks for asking I posted an unofficial mark scheme and most people seemed to agree so I'm happy - just need to replicate that in Core 4!


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    Ahh well done mate! hopefully i can do the same
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    Hey guys.

    I was wondering if you guys can help me with Q7 Jun '02 in the following image.

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    I unserstand that the answer could either A or B because p.d. has to be 1.5 V but how do you know if E will be multiplied by 1.5 or 1.5²?
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    (Original post by Disney0702)
    Hey guys.

    I was wondering if you guys can help me with Q7 Jun '02 in the following image.

    Name:  Screenshot_2015-06-05-20-37-11~01.png
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    I unserstand that the answer could either A or B because p.d. has to be 1.5 V but how do you know if E will be multiplied by 1.5 or 1.5²?
    Im pretty sure that the energy stored is proportional to the voltage squared, sorry in advance if this is incorrect
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    (Original post by Disney0702)
    Hey guys.

    I was wondering if you guys can help me with Q7 Jun '02 in the following image.

    Name:  Screenshot_2015-06-05-20-37-11~01.png
Views: 65
Size:  181.0 KB

    I unserstand that the answer could either A or B because p.d. has to be 1.5 V but how do you know if E will be multiplied by 1.5 or 1.5²?
    

E = \dfrac{CV^2}{2}

    

\therefore E \propto V^2

    That means that when V becomes 1.5V, E becomes 2.25E. Does that help?


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    (Original post by Disney0702)
    Hey guys.

    I was wondering if you guys can help me with Q7 Jun '02 in the following image.

    Name:  Screenshot_2015-06-05-20-37-11~01.png
Views: 65
Size:  181.0 KB

    I unserstand that the answer could either A or B because p.d. has to be 1.5 V but how do you know if E will be multiplied by 1.5 or 1.5²?
    I think its 1.5E
    Becuase it says charge is increased by 50% and doesnt say anything about capacitane

    3 formulad are
    1) E=0.5QV
    2) E=Q^2/2C
    3) E=(C*V^2)0.5


    So only formula without C is no. 1 nothing is squared so i am guessing answer is A.
    Whats the answer btw


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    (Original post by a.a.k)
    I think its 1.5E
    Becuase it says charge is increased by 50% and doesnt say anything about capacitane

    3 formulad are
    1) E=0.5QV
    2) E=Q^2/2C
    3) E=(C*V^2)0.5


    So only formula without C is no. 1 nothing is squared so i am guessing answer is A.
    Whats the answer btw


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    (Original post by a.a.k)
    I think its 1.5E
    Becuase it says charge is increased by 50% and doesnt say anything about capacitane

    3 formulad are
    1) E=0.5QV
    2) E=Q^2/C
    3) E=(C*V^2)0.5


    So only formula without C is no. 1 nothing is squared so i am guessing answer is A.
    Whats the answer btw


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    The answer is B.


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    (Original post by CD223)
    The answer is B.


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    Dont know how

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    (Original post by CD223)
    

E = \dfrac{CV^2}{2}

    

\therefore E \propto V^2

    That means that when V becomes 1.5V, E becomes 2.25E. Does that help?


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