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

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    Anyone got a "mathsy" explanation for this answer? I just used an educated guess and multiplied by 4/3, but I wanna know why it works.

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    (Original post by CD223)
    Anyone got a "mathsy" explanation for this answer? I just used an educated guess and multiplied by 4/3, but I wanna know why it works.

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    I worked out the factor ratio of the potential at the centre between the charges. Hope this satisfy your curiosity
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    Edit: oops, the last bit meant to be 2/15(25)
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    (Original post by Mehrdad jafari)
    I worked out the factor ratio of the potential at the centre between the charges. Hope this satisfy your curiosity
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    Edit: oops, the last bit meant to be 2/15(25)
    Sorry for appearing silly, but why is the Q ratio 1/5?


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    (Original post by CD223)
    Sorry for appearing silly, but why is the Q ratio 1/5?


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    That's the result of factorising the equations of the potential of both charges
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    (Original post by Mehrdad jafari)
    That's the result of factorising the equations of the potential of both charges
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    But the total distance between the charges is 4.0m. And potential is a scalar quantity so you add potentials?


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    (Original post by CD223)
    But the total distance between the charges is 4.0m. And potential is a scalar quantity so you add potentials?


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    I thought you would add potentials of different charges( you still subtract them but with one charge being negative results in adding them)


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    Here is the pure maths, but to do it this way in the exam takes far too long.
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    (Original post by Mehrdad jafari)
    I thought you would add potentials of different charges( you still subtract them but with one charge being negative results in adding them)


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    In this case you add because both are positive.


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    (Original post by EHR223)
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    Here is the pure maths, but to do it this way in the exam takes far too long.
    This is the only way that actually makes sense to me - thanks for sharing.

    Does anyone know how one would do it with ratios?



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    (Original post by CD223)
    In this case you add because both are positive.


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    Oh right, even i have difficulty with the idea of potential and how to deal with them.
    So adding the potentials would mean that more work must be done to move a positive charge from infinity to that point than to move it to the middle between the charges?


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    (Original post by CD223)
    In this case you add because both are positive.


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    But you would still add them if one was positive and one was negative, although one potential would be a negative value right? you never have to subtract?
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    (Original post by EHR223)
    But you would still add them if one was positive and one was negative, although one potential would be a negative value right? you never have to subtract?
    Yeah. Potentials are scalar addition, field strengths are vector addition.

    You always add potentials, but some might be negative.


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    JUNE 2014 MC Q16

    Anyone fancy having a stab at explaining this?

    I thought energy was scalar and therefore as no work is done along equipotentials, the work done along PQRS would be twice PQ.

    Why am I wrong?


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    (Original post by CD223)
    JUNE 2014 MC Q16

    Anyone fancy having a stab at explaining this?

    I thought energy was scalar and therefore as no work is done along equipotentials, the work done along PQRS would be twice PQ.

    Why am I wrong?


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    You're right, energy is a scalar but work done is a vector quantity. Energy is required to move a charge from P to Q but equal amount of energy is released ( energy is gained from some unknown place), when the charge is moved from R to S, or vice versa as the sign of the charge is not known


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    (Original post by Mehrdad jafari)
    You're right, energy is a scalar but work done is a vector quantity. Energy is required to move a charge from P to Q but equal amount of energy is released ( energy is gained from some unknown place), when the charge is moved from R to S, or vice versa as the sign of the charge is not known


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    Oh I see. So energy is the "magnitude" of work done?


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    (Original post by CD223)
    Oh I see. So energy is the "magnitude" of work done?


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    Yeah


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    (Original post by CD223)
    Oh I see. So energy is the "magnitude" of work done?


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    You know what, actually work done is also a scalar, sorry for misleading you. But in the case of the question the overall work done across PQRS is zero because there is mo displacement in the direction of the force


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    (Original post by Mehrdad jafari)
    You know what, actually work done is also a scalar, sorry for misleading you. But in the case of the question the overall work done across PQRS is zero because there is mo displacement in the direction of the force


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    I see. No worries. Thanks.


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    (Original post by CD223)
    I see. No worries. Thanks.


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    It's ok. By the way, I will email my friend tomorrow to let me know when he's got the data. Once i get them i will upload them on here


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    (Original post by CD223)
    This is the only way that actually makes sense to me - thanks for sharing.

    Does anyone know how one would do it with ratios?



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    The main idea is to use only as much info as you need. You can solve this problem simply by knowing that V\propto \frac{Q}{r}. You don't even need to know the constant of proportionality.

    So letting the constant of proportionality be k and letting the two charges be x, we have V_{total}=k\frac{x}{2}+k\frac{x}  {2}=25 (where we put r=2), i.e. kx=25.

    In the second arrangement, V_{total}=k\frac{x}{1}+k\frac{x}  {3}=\frac{4}{3}kx=\frac{4}{3}(25  )=33.33...=33.3\ (2\ \text{s.f.})
 
 
 
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