Join TSR now and get all your revision questions answeredSign up now

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

    Offline

    2
    ReputationRep:
    I made one for fun:

    The electric potential at a point one third along the line connecting two charges, starting from the first charge, is x volts. The electric potential one third along the same line, starting from the second charge, is y volts. What is the electric potential at the midpoint of this line (in terms of x and y)?
    Offline

    2
    ReputationRep:
    Another:

    Three charges are placed at the vertices of an equilateral triangle. The electric potentials at the midpoints of the sides of the triangle are x volts, y volts and z volts. Find an expression for the electric potential at the center of the triangle.
    Offline

    2
    ReputationRep:
    Does anyone have tips for the exam?


    Posted from TSR Mobile
    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by Mehrdad jafari)
    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
    Oh brilliant, thank you!

    If I was to PM you my email, could you send me it that way too? (It's just I have work all day today so won't really be on TSR!) - just don't wanna lose the post if it does just uploaded


    Posted from TSR Mobile
    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by PotterPhysics)
    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.})
    That makes sense. Thanks for sharing


    Posted from TSR Mobile
    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by PotterPhysics)
    Another:

    Three charges are placed at the vertices of an equilateral triangle. The electric potentials at the midpoints of the sides of the triangle are x volts, y volts and z volts. Find an expression for the electric potential at the center of the triangle.
    How would you go about this?


    Posted from TSR Mobile
    Offline

    16
    ReputationRep:
    (Original post by CD223)
    Oh brilliant, thank you!

    If I was to PM you my email, could you send me it that way too? (It's just I have work all day today so won't really be on TSR!) - just don't wanna lose the post if it does just uploaded


    Posted from TSR Mobile
    Yeah, no problem


    Posted from TSR Mobile
    Offline

    2
    ReputationRep:
    What are you guys generally scoring on the multiple choice section?!


    Posted from TSR Mobile
    Offline

    2
    ReputationRep:
    (Original post by CD223)
    That makes sense. Thanks for sharing


    Posted from TSR Mobile
    No problem!

    (Original post by CD223)
    How would you go about this?


    Posted from TSR Mobile
    Leaning on a generalization of my first solution. We let the three charges be q_1,q_2,q_3 and let the midpoints between (q_1,q_2),(q_2,q_3),(q_3,q_1) be x,y,z respectively. Then letting (wlog) the sides of the triangle be 2, and as before letting the constant for V\propto \frac{Q}{r} be k, we have k(\frac{q_1}{1}+\frac{q_2}{1})=x  \Leftrightarrow k(q_1+q_2)=x, and similarly k(q_2+q_3)=y and k(q_3+q_1)=z. Hence adding up we have 2k(q_1+q_2+q_3)=x+y+z, giving q_1+q_2+q_3=\frac{x+y+z}{2k}. (Call this equation 1)

    Now, since the figure is an equilateral triangle, its area is \frac{s^2\sqrt{3}}{4} where s is its side length (I leave this as an exercise to the reader), therefore since our triangle has s=2 then plainly its area is \sqrt{3}. Then from the formula \frac{s^3}{4R}=\frac{s^2\sqrt{3}  }{4} (where R is the circumradius, and both sides give the area of the triangle) we have R=\frac{2}{\sqrt{3}}. Therefore, at the center of the triangle the potential is k\frac{q_1}{R}+k\frac{q_2}{R}+k\  frac{q_3}{R}=\frac{k\sqrt{3}}{2}  (q_1+q_2+q_3). Then substituting equation (1) into this last expression yields the answer \frac{k\sqrt{3}}{2}(\frac{x+y+z}  {2k})=\frac{\sqrt{3}}{4}(x+y+z).
    Offline

    10
    ReputationRep:
    (Original post by zaybun)
    Does anyone have tips for the exam?


    Posted from TSR Mobile
    Keep an eye on how much time you're spending on each half - you shouldn't be exceeding the recommended 45 minutes on the multiple choice I'd say.

    In the multiple choice half, look for proportionality in questions. Also look for contradicting statements (particularly on true or false type questions) - they put these in to speed it up, but they're often missed.

    Don't forget to leave time to fill in the multiple choice answer sheet! You might forget about this because in mocks you probably just circle the answer on the paper, but there is in fact a separate sheet to fill in.

    Obviously never leave multiple choice questions blank - you've got a one in four chance of getting a question right even if you've got absolutely no clue (if you've got the time though, aim to make educated guesses where possible!).

    Decide whether doing multiple choice or written part is best for you, and whether starting or finishing with six markers works better. Remember that you don't have to do the paper in order! (May seem obvious, but a lot of people try and do everything in order and lose a lot of time when they get stuck).

    If you get stuck, move on and come back to it. Get all your guaranteed "I can definitely do this right now" marks before you try and figure out anything that's confusing you.
    Offline

    10
    ReputationRep:
    (Original post by Fred Cantoni)
    What are you guys generally scoring on the multiple choice section?!


    Posted from TSR Mobile
    Initially I was stuck getting 18 or 19 on every paper, but now in the region of 22+ you?
    Offline

    1
    ReputationRep:
    Name:  Screen Shot 2558-05-25 at 13.16.14.png
Views: 118
Size:  94.2 KB

    guys I really need your help. I don't know why C is correct and the other answers are invalid Name:  Screen Shot 2558-05-25 at 13.16.14.png
Views: 118
Size:  94.2 KB
    Offline

    1
    ReputationRep:
    Name:  Screen Shot 2558-05-25 at 13.35.02.png
Views: 66
Size:  130.3 KB

    for part ciii) why didn't they use the v that is found by v=wr r is the radius but they used v for the mid point m
    Offline

    1
    ReputationRep:
    (Original post by boombozx)
    Name:  Screen Shot 2558-05-25 at 13.16.14.png
Views: 118
Size:  94.2 KB

    guys I really need your help. I don't know why C is correct and the other answers are invalid Name:  Screen Shot 2558-05-25 at 13.16.14.png
Views: 118
Size:  94.2 KB
    I think that the distractor here is D. The reason why D does not induce an E.M.F may be because there is no change in magnetic flux through the coil when it is rotated in this particular direction, wheras if it is rotated in the other direction (C), there is a change in magnetic flux; hence an EMF is induced according to Faraday's law.
    Offline

    2
    ReputationRep:
    Name:  image.jpg
Views: 78
Size:  337.0 KB

    Anyone?
    Offline

    2
    ReputationRep:
    (Original post by saad97)
    Name:  image.jpg
Views: 78
Size:  337.0 KB

    Anyone?
    Think about how the mass is related to g, and what the mass of a planet is.


    Name:  ImageUploadedByStudent Room1432559379.580389.jpg
Views: 59
Size:  102.2 KB


    Posted from TSR Mobile
    Offline

    2
    ReputationRep:
    (Original post by Jimmy20002012)
    Think about how the mass is related to g, and what the mass of a planet is.


    Name:  ImageUploadedByStudent Room1432559379.580389.jpg
Views: 59
Size:  102.2 KB


    Posted from TSR Mobile
    Thank you! I knew the two equations needed but forgot that V=4/3pir^3 -.-
    Thanks anyway
    Offline

    16
    ReputationRep:
    (Original post by boombozx)
    Name:  Screen Shot 2558-05-25 at 13.35.02.png
Views: 66
Size:  130.3 KB

    for part ciii) why didn't they use the v that is found by v=wr r is the radius but they used v for the mid point m
    Because M is the midpoint of the radius of the disc, so you need to consider half of the radius when calculating v=wr


    Posted from TSR Mobile
    Offline

    2
    ReputationRep:
    (Original post by Lau14)
    Initially I was stuck getting 18 or 19 on every paper, but now in the region of 22+ you?
    Sort of 20-23, was just wondering if that was good enough, or if everyone was getting 25!


    Posted from TSR Mobile
    Offline

    2
    ReputationRep:
    I was doing a past paper, last quesiton on june 2011 and it said that in a transformer a high voltage introduces insulation problems and raises safety issues.
    I dont understand how a high voltage can affect insulation can anyone explain?
 
 
 
Poll
Which web browser do you use?

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Quick reply
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.