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

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    (Original post by Leonacatherine)
    Whats everyone doing for revision now?
    I've done all the papers except june 14 which I've been saving for a couple days before the exam
    At the moment I've been repeating section A past papers to get my speed up on that bit and because they reuse
    questions
    Just finished going back though the 02 papers. Now I'm going to go and re do the new papers. Then I'm just thinking about all the topics and thinking of possible questions for 6 markers etc


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    (Original post by CD223)
    Wow really? That's bad :/


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    Yeah man. He did eventually but he wasn't really into it


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    Hi guys, I'm struggling with Question 13 from the MC section of June 2010, can someone help me out? Sorry if someone else has already asked about it, I just don't have the time to go through the thread!

    The diagram shows two charges, +4x10^-6C and -16x10^-6C, 120mm apart. What is the distance from the +4x10^-6C charge to the point between the two charges where the resultant electric potential is zero?

    A) 24mm
    B) 40mm
    C) 80mm
    D) 96mm
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    (Original post by TheSmartOne)
    Hi guys, I'm struggling with Question 13 from the MC section of June 2010, can someone help me out? Sorry if someone else has already asked about it, I just don't have the time to go through the thread!

    The diagram shows two charges, +4x10^-6C and -16x10^-6C, 120mm apart. What is the distance from the +4x10^-6C charge to the point between the two charges where the resultant electric potential is zero?

    A) 24mm
    B) 40mm
    C) 80mm
    D) 96mm
    I would do it as a ratio
    so with the charges youve got a 4:16 ratio which is equal to a 1:4 ratio
    1+4 = 5, 120mm divided by 5 gives 24 mm

    This method will only work wwith linear problems though i.e potentials dont try it with field strengths or force as theyre quadratics
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    (Original post by TheSmartOne)
    Hi guys, I'm struggling with Question 13 from the MC section of June 2010, can someone help me out? Sorry if someone else has already asked about it, I just don't have the time to go through the thread!

    The diagram shows two charges, +4x10^-6C and -16x10^-6C, 120mm apart. What is the distance from the +4x10^-6C charge to the point between the two charges where the resultant electric potential is zero?

    A) 24mm
    B) 40mm
    C) 80mm
    D) 96mm
    Just as a side not if you are ever unsure you can use a bit of common sense to make an educated guess. If they were equal charges the resultant would be equal to zero at the midpoint, but because the -16x10^-6 is much larger than 4x10^-6 in magnitude you know the point where the resultant is equal to zero is going to be closer to the 4 one. This immediately eliminates C and D as theyre closer to the 16 one
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    (Original post by Leonacatherine)
    I would do it as a ratio
    so with the charges youve got a 4:16 ratio which is equal to a 1:4 ratio
    1+4 = 5, 120mm divided by 5 gives 24 mm

    This method will only work wwith linear problems though i.e potentials dont try it with field strengths or force as theyre quadratics
    You can do it with quadratics. Just square root the ratio as the negative solution can be rejected.


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    Can anyone help me with this question please?
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    (Original post by saad97)
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    Can anyone help me with this question please?
    F = qq/ 4 x pi x E0 x r^2
    Because the magnitude of q is the same qq = q squared
    4x pi is just a number and so has no units
    so Q^2/ E0 x r^2
    so d x
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    (Original post by saad97)
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    Can anyone help me with this question please?
    Coulomb's law states that, for two charges in an electric field:

    

F = \dfrac{Qq}{4 \pi \epsilon_0 d^2}

    In this case the charges are equal in magnitude, so:

    

F = \dfrac{Q^2}{4 \pi \epsilon_0 d^2}

    As 4 \pi is a constant, it can be ignored when looking at units as it doesn't have any.

    As such, removing this from the equation, this means:

    

\dfrac{Q^2}{\epsilon_0 d^2}

    Must have the same units as F.
    Does that help?


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    (Original post by Leonacatherine)
    F = qq/ 4 x pi x E0 x r^2
    Because the magnitude of q is the same qq = q squared
    4x pi is just a number and so has no units
    so Q^2/ E0 x r^2
    so d x
    (Original post by CD223)
    Coulomb's law states that, for two charges in an electric field:

    

F = \dfrac{Qq}{4 \pi \epsilon_0 d^2}

    In this case the charges are equal in magnitude, so:

    

F = \dfrac{Q^2}{4 \pi \epsilon_0 d^2}

    As 4 \pi is a constant, it can be ignored when looking at units as it doesn't have any.

    As such, removing this from the equation, this means:

    

\dfrac{Q^2}{\epsilon_0 d^2}

    Must have the same units as F.
    Does that help?


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    Thank you!
    I have another question from the same paper that I need help with please?

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    Does anyone know if we need to know much about the conservation of kinetic energy, and if there have been any questions on momentum to do with kinetic energy?
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    defintion of one tesla: magnetic field strength required to exert a force of 1N on a wite of length 1m carrying a current of 1A? Do you guys agree?
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    (Original post by Ilovemaths96)
    defintion of one tesla: magnetic field strength required to exert a force of 1N on a wite of length 1m carrying a current of 1A? Do you guys agree?
    The tesla can be defined as the force per unit length per unit current that a magnetic field exerts on a conductor in a magnetic field.


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    (Original post by DeiJaVu)
    Does anyone know if we need to know much about the conservation of kinetic energy, and if there have been any questions on momentum to do with kinetic energy?
    I dont think theres much they could ask about that is there?
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    (Original post by DeiJaVu)
    Does anyone know if we need to know much about the conservation of kinetic energy, and if there have been any questions on momentum to do with kinetic energy?
    There was a MC question (Jan 2013 Q2 I think? Not sure). It was to do with expressing the momentum of a daughter nucleus in terms of kinetic energy.


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    (Original post by CD223)
    There was a MC question (Jan 2013 Q2 I think? Not sure). It was to do with expressing the momentum of a daughter nucleus in terms of kinetic energy.


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    Okey thankyou, I'll go check that out.
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    (Original post by saad97)
    Thank you!
    I have another question from the same paper that I need help with please?

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    Does this help?

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    (Original post by DeiJaVu)
    Does anyone know if we need to know much about the conservation of kinetic energy, and if there have been any questions on momentum to do with kinetic energy?
    Kinetic energy is conserved in elastic collisions, and all collisions (including elastic) conserve momentum provided no external resultant force acts on the system. I think that's all.
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    (Original post by CD223)
    The tesla can be defined as the force per unit length per unit current that a magnetic field exerts on a conductor in a magnetic field.


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    Should we also say at right angles to the magnetic field?

    Force on 1m of wire carrying a current of 1A at right angles to the field.

    Thanks
 
 
 
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