The Student Room Group

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

Scroll to see replies

Reply 1800
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!

ImageUploadedByStudent Room1432410439.594264.jpg

Question 18
The answer is C.

[br]V=QC[br][br]V = \dfrac{Q}{C}[br]

and

[br]E=12CV2[br][br]E= \dfrac{1}{2} CV^2[br]

Therefore
[br]EαV2[br][br]E \alpha V^2[br]

and

[br]VαQ[br][br]V \alpha Q[br]

If the charge Q is increased by 50%, it comes 1.5Q1.5Q. As such, as V is proportional to Q, V becomes 1.5V1.5V.

This follows that, as E is proportional to V2V^2, E becomes (1.5)2E\left(1.5\right)^{2} E which is the same as 2.25E2.25E.
(edited 8 years ago)
Original post by CD223
Question 15
The answer is C. Excuse the "May" - using my college planner haha!

ImageUploadedByStudent Room1432410439.594264.jpg

Question 18
The answer is C.

[br]V=QC[br][br]V = \dfrac{Q}{C}[br]

and

[br]E=12CV2[br][br]E= \dfrac{1}{2} CV^2[br]

Therefore
[br]EαV2[br][br]E \alpha V^2[br]

and

[br]VαQ[br][br]V \alpha Q[br]

If the charge Q is increased by 50%, it comes 1.5Q1.5Q. As such, as V is proportional to Q, V becomes 1.5V1.5V.

This follows that, as E is proportional to V2V^2, E becomes (1.5)2E\left(1.5\right)^{2} E which is the same as 2.25E2.25E.


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.
Reply 1802
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.


Posted from TSR Mobile
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:

[br]Qq4πϵ0r2[br][br]\dfrac{Qq}{4\pi {\epsilon_0} r^2}[br]

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:

[br]Qq4πϵ0r2[br][br]\dfrac{Qq}{4\pi {\epsilon_0} r^2}[br]

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


Hello :smile:

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
Reply 1804
Original post by Disney0702
Hello :smile:

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 :smile:


Posted from TSR Mobile
Original post by CD223
Lol I'm making mistakes all over the place! Thanks for the correction. Yes, it is electric field lines :smile:


Posted from TSR Mobile


Ah its alright :wink:
Reply 1806
Original post by Disney0702
Ah its alright :wink:


Got carried away because I thought I could actually answer an electric/magnetic fields question... Alas I cannot :wink:


Posted from TSR Mobile
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=dts=\frac{d}{t} (speed = distance / time) to work out the time: d=0.16d=0.16 and s=8.0×105ms1s=8.0\times 10^5\text{ms}^{-1}, hence t=ds=2.0×107st=\frac{d}{s}=2.0\times 10^{-7}\text{s}. From this analysis we see that A,B,D,\bf A,B,D, are all correct so C\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=4200E=4200, so F=4200×4.8×1019=2.016×1015=2.0×1015N (2 s.f.)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 C\bf C is incorrect.

Question 18) As CD basically showed, we have E=12QVE=\frac{1}{2}QV and Q=CVQ=CV. Then because CC is constant the second equation implies QVQ\propto V. Thus increasing QQ by a factor of 1.51.5 (which is equivalent to increasing by 50%) will cause the same increase in VV. Therefore E=12QVE=\frac{1}{2}QV will increase by a factor of 1.5×1.5=2.251.5\times 1.5=2.25. So the answer is C\bf C.
(edited 8 years ago)
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 vv, therefore v=108kmh=108103m3600s=30ms=30ms1v=\frac{108km}{h}=\frac{108\cdot 10^3m}{3600s}=\frac{30m}{s}=30ms^{-1}.

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

Thus ω=vr=300.2=150rads1B\omega=\frac{v}{r}=\frac{30}{0.2}=150rads^{-1}\rightarrow \boxed{\mathbf{B}}
(edited 8 years ago)
Original post by AR_95


and this one

They have a lot of JJ'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 δW=QδV\delta W=Q\delta V, so dividing through by some length LL gives δWL=QδVL\frac{\delta W}{L}=Q\frac{\delta V}{L}.

Now, E=VdE=\frac{V}{d} so we know that the units of δVL\frac{\delta V}{L} are the same as field strength. The units of δWL\frac{\delta W}{L} are just Jm1Jm^{-1} and the units of QQ are CC.

So we can think of the formula WL=QVL\frac{W}{L}=Q\frac{V}{L} as WL=QE\frac{W}{L}=QE where E is field strength. Therefore E=WQLE=\frac{W}{QL}, and plugging in the units on the right hand side we get that unitsoffieldstrength=Jm1C1units of field strength = Jm^{-1}C^{-1}, which contradicts D\bf D. So the answer is D\bf D.
(edited 8 years ago)
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.
(edited 8 years ago)
Original post by PotterPhysics
They have a lot of JJ'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 δW=QδV\delta W=Q\delta V, so dividing through by some length LL gives δWL=QδVL\frac{\delta W}{L}=Q\frac{\delta V}{L}.

Now, E=VdE=\frac{V}{d} so we know that the units of δVL\frac{\delta V}{L} are the same as field strength. The units of δWL\frac{\delta W}{L} are just Jm1Jm^{-1} and the units of QQ are CC.

So we can think of the formula WL=QVL\frac{W}{L}=Q\frac{V}{L} as WL=QE\frac{W}{L}=QE where E is field strength. Therefore E=WQLE=\frac{W}{QL}, and plugging in the units on the right hand side we get that unitsoffieldstrength=Jm1C1units of field strength = Jm^{-1}C^{-1}, which contradicts D\bf D. So the answer is D\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.
Reply 1812
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:

ImageUploadedByStudent Room1432489480.197388.jpg

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


Posted from TSR Mobile
Original post by CD223
Lol my method was to write out each implied formula like so:

ImageUploadedByStudent Room1432489480.197388.jpg

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


Posted from TSR Mobile


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 :/
Reply 1814
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.


Posted from TSR Mobile
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.


Posted from TSR Mobile


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 :tongue:
Reply 1816
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 :tongue:


I hate how hard some questions can be. There's no uniform difficulty from year to year haha.


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


Posted from TSR Mobile


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.
Reply 1818
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.


Posted from TSR Mobile
Original post by CD223
Agreed. It's annoying how it takes ~45 minutes for 25 marks.


Posted from TSR Mobile


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

Quick Reply

Latest

Trending

Trending