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

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    (Original post by AR_95)
    Done for the day. Hoping someone can help me with these two
    Question 15
    The answer is C. Excuse the "May" - using my college planner haha!

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    Question 18
    The answer is C.

    

V = \dfrac{Q}{C}

    and

    

E= \dfrac{1}{2} CV^2

    Therefore
    

E \alpha V^2

    and

    

V \alpha Q

    If the charge Q is increased by 50%, it comes 1.5Q. As such, as V is proportional to Q, V becomes 1.5V.

    This follows that, as E is proportional to V^2, E becomes \left(1.5\right)^{2} E which is the same as 2.25E.
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    (Original post by CD223)
    Question 15
    The answer is C. Excuse the "May" - using my college planner haha!

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    Question 18
    The answer is C.

    

V = \dfrac{Q}{C}

    and

    

E= \dfrac{1}{2} CV^2

    Therefore
    

E \alpha V^2

    and

    

V \alpha Q

    If the charge Q is increased by 50%, it comes 1.5Q. As such, as V is proportional to Q, V becomes 1.5V.

    This follows that, as E is proportional to V^2, E becomes \left(1.5\right)^{2} E which is the same as 2.25E.
    For the electric field one, you gave the right answer but I think you misread the info on the question. You use F=EQ right, because 4200 is the electric field strength? As it's given in Vm^-1 which is the unit for E? F comes out as 2.016x10^-15 doing this so C is still right.
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    (Original post by JJBinn)
    For the electric field one, you gave the right answer but I think you misread the info on the question. You use F=EQ right, because 4200 is the electric field strength? As it's given in Vm^-1 which is the unit for E? F comes out as 2.016x10^-15 doing this so C is still right.
    Oops yeah. The formula I wrote down was correct but I subbed in the wrong numbers :L thanks.


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    (Original post by CD223)
    The direction of the magnetic field lines is the direction a free positive charge would move along.

    Hence, they point towards a negative charge in this case.

    The proton accelerates to the right as it is attracted to the negative charge. It's acceleration increases as the distance between it and the negative charge decreases, and due to the electric field strength equalling:

    

\dfrac{Qq}{4\pi {\epsilon_0} r^2}

    This means that the proton's acceleration increases as it gets closer to the negative charge.

    Conversely, the electron accelerates to the left as it is repelled by the negative charge. It's acceleration decreases as the distance between it and the negative charge increases, and due to the electric field strength equalling:

    

\dfrac{Qq}{4\pi {\epsilon_0} r^2}

    This means that the electron's acceleration decreases as it gets further away from the negative charge.
    Hello

    I was stuck on this question as well and I wanted to know if you were suppose to write electric field lines instead of magnetic lines or is it really magnetic field lines. If so, why?

    Thanks for your time
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    (Original post by Disney0702)
    Hello

    I was stuck on this question as well and I wanted to know if you were suppose to write electric field lines instead of magnetic lines or is it really magnetic field lines. If so, why?

    Thanks for your time
    Lol I'm making mistakes all over the place! Thanks for the correction. Yes, it is electric field lines


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    (Original post by CD223)
    Lol I'm making mistakes all over the place! Thanks for the correction. Yes, it is electric field lines


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    Ah its alright
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    (Original post by Disney0702)
    Ah its alright
    Got carried away because I thought I could actually answer an electric/magnetic fields question... Alas I cannot


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    (Original post by AR_95)
    Done for the day. Hoping someone can help me with these two




    Question 15) There are usually two ways to go about questions.

    Way 1: Elimination
    Clearly the force on the positive ion only acts downwards (it isn't a magnetic field -- the direction of force is fixed). Therefore there is no horizontal acceleration, so we can use s=\frac{d}{t} (speed = distance / time) to work out the time: d=0.16 and s=8.0\times 10^5\text{ms}^{-1}, hence t=\frac{d}{s}=2.0\times 10^{-7}\text{s}. From this analysis we see that \bf A,B,D, are all correct so \bf C must be incorrect.
    Another way is just playing around with the data. By definition, the electric field strength is force per unit positive charge. Therefore, field strength times charge gives the force acting on the charge. In this case, E=4200, so F=4200\times 4.8\times 10^{-19}=2.016\times 10^{-15}=2.0\times 10^{-15}N\ (2\ \text{s.f.}) (downwards), which immediately shows that option \bf C is incorrect.

    Question 18) As CD basically showed, we have E=\frac{1}{2}QV and Q=CV. Then because C is constant the second equation implies Q\propto V. Thus increasing Q by a factor of 1.5 (which is equivalent to increasing by 50%) will cause the same increase in V. Therefore E=\frac{1}{2}QV will increase by a factor of 1.5\times 1.5=2.25. So the answer is \bf C.
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    (Original post by AR_95)


    looks simple but i cant get to the answer
    [QUOTE=AR_95;56155707]
    The speed of the car is the same as the tangential (or linear) velocity of the wheels v, therefore v=\frac{108km}{h}=\frac{108\cdot 10^3m}{3600s}=\frac{30m}{s}=30ms  ^{-1}.

    Moreover, d=0.4=2r, so r=0.2.

    Thus \omega=\frac{v}{r}=\frac{30}{0.2  }=150rads^{-1}\rightarrow \boxed{\mathbf{B}}
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    (Original post by AR_95)


    and this one
    They have a lot of J's there, so it's probably best to start working from a formula that relates energy to electric field strength. This immediately reminded me of \delta W=Q\delta V, so dividing through by some length L gives \frac{\delta W}{L}=Q\frac{\delta V}{L}.

    Now, E=\frac{V}{d} so we know that the units of \frac{\delta V}{L} are the same as field strength. The units of \frac{\delta W}{L} are just Jm^{-1} and the units of Q are C.

    So we can think of the formula \frac{W}{L}=Q\frac{V}{L} as \frac{W}{L}=QE where E is field strength. Therefore E=\frac{W}{QL}, and plugging in the units on the right hand side we get that units of field strength = Jm^{-1}C^{-1}, which contradicts \bf D. So the answer is \bf D.
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    Sorry, the LaTeX on this site is really bad compared to other math forums I've been on, so I'm editing my posts to fix it.

    Edit: Whoops, didn't see CD's replies there. Ignore my last few posts now, sorry about that.
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    (Original post by PotterPhysics)
    They have a lot of J's there, so it's probably best to start working from a formula that relates energy to electric field strength. This immediately reminded me of \delta W=Q\delta V, so dividing through by some length L gives \frac{\delta W}{L}=Q\frac{\delta V}{L}.

    Now, E=\frac{V}{d} so we know that the units of \frac{\delta V}{L} are the same as field strength. The units of \frac{\delta W}{L} are just Jm^{-1} and the units of Q are C.

    So we can think of the formula \frac{W}{L}=Q\frac{V}{L} as \frac{W}{L}=QE where E is field strength. Therefore E=\frac{W}{QL}, and plugging in the units on the right hand side we get that units of field strength = Jm^{-1}C^{-1}, which contradicts \bf D. So the answer is \bf D.
    Also solvable more quickly (which is what they probably expect, as this is the multiple choice) by noticing that A and D contradict each other, so it must be one of these. E = V/d and voltage can be given as joules per couloumb (V = WQ, from the AS section), so A is correct and D is the answer.
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    (Original post by Lau14)
    Also solvable more quickly (which is what they probably expect, as this is the multiple choice) by noticing that A and D contradict each other, so it must be one of these. E = V/d and voltage can be given as joules per couloumb (V = WQ, from the AS section), so A is correct and D is the answer.
    Lol my method was to write out each implied formula like so:

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    I prefer your "elimination by contradiction" method though. Do you use that method often?


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    (Original post by CD223)
    Lol my method was to write out each implied formula like so:

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    I prefer your "elimination by contradiction" method though. Do you use that method often?


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    I use it when I see it because it's usually the fastest way to get a question done (but I'm not necessarily great at spotting them!). I do seem to end up just going through the formula sheet when they ask about units though :/
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    (Original post by Lau14)
    I use it when I see it because it's usually the fastest way to get a question done (but I'm not necessarily great at spotting them!). I do seem to end up just going through the formula sheet when they ask about units though :/
    Ah right haha! Just wondered if you ever used that in a true/false question too - must make it much easier.


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    (Original post by CD223)
    Ah right haha! Just wondered if you ever used that in a true/false question too - must make it much easier.


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    That is one of the places they're most likely to pop up! Unfortunately for me it always seems to be something like magnetic fields, so I can see one must be wrong but I have no clue which one
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    (Original post by Lau14)
    That is one of the places they're most likely to pop up! Unfortunately for me it always seems to be something like magnetic fields, so I can see one must be wrong but I have no clue which one
    I hate how hard some questions can be. There's no uniform difficulty from year to year haha.


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    (Original post by CD223)
    I hate how hard some questions can be. There's no uniform difficulty from year to year haha.


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    Yeah some years you get a pretty nice mc section with only one or two questions that you make an educated guess at, but other years you end up just picking a random letter for five questions or more! I think it more depends on how much it pushes you for time though than them being impossibly hard questions.
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    (Original post by Lau14)
    Yeah some years you get a pretty nice mc section with only one or two questions that you make an educated guess at, but other years you end up just picking a random letter for five questions or more! I think it more depends on how much it pushes you for time though than them being impossibly hard questions.
    Agreed. It's annoying how it takes ~45 minutes for 25 marks.


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    (Original post by CD223)
    Agreed. It's annoying how it takes ~45 minutes for 25 marks.


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    Yeah, it almost doesn't seem worth it but it's actually a big chunk of the paper :/
 
 
 
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