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

Anyone likely to be able to take home with them a MC copy of this year's paper after the exam so we can discuss answers? Or is that too illegal?:L


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Loads of people did that for Chemistry 1 and 2 so don't think its illegal

Anyone likely to be able to take home with them a MC copy of this year's paper after the exam so we can discuss answers? Or is that too illegal?:L


Posted from TSR Mobile

Loads of people did that for Chemistry 1 and 2 so don't think its illegal

The ratio of the charge to the distance is the same for both charges. Namely; 4uC/d=16uC/(120mm-d)
You just need to rearrange this to find d.

Another way of looking at it is that the potential due to the 4uC charge at that point 'd' is equal and opposite (in sign) to the potential due to the 16uC charge. So 4uC/d + -16uC/(120mm-d) = 0 And rearrange to find d.

Thank you!!! it was just i forgot what you do with the 120mm, but thats helped

How can you take that?


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take?

Can anyone remind me how to work out Q13 in this paper?
http://filestore.aqa.org.uk/subjects/AQA-PHYA4-1-W-QP-JUN10.PDF

C3 went really well - thanks for asking I posted an unofficial mark scheme and most people seemed to agree so I'm happy - just need to replicate that in Core 4!


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Hey guys.

I was wondering if you guys can help me with Q7 Jun '02 in the following image.



I unserstand that the answer could either A or B because p.d. has to be 1.5 V but how do you know if E will be multiplied by 1.5 or 1.5²?

Hey guys.

I was wondering if you guys can help me with Q7 Jun '02 in the following image.



I unserstand that the answer could either A or B because p.d. has to be 1.5 V but how do you know if E will be multiplied by 1.5 or 1.5²?

Hey guys.

I was wondering if you guys can help me with Q7 Jun '02 in the following image.



I unserstand that the answer could either A or B because p.d. has to be 1.5 V but how do you know if E will be multiplied by 1.5 or 1.5²?

I think its 1.5E
Becuase it says charge is increased by 50% and doesnt say anything about capacitane

3 formulad are
1) E=0.5QV
2) E=Q^2/2C
3) E=(C*V^2)0.5


So only formula without C is no. 1 nothing is squared so i am guessing answer is A.
Whats the answer btw


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I think its 1.5E
Becuase it says charge is increased by 50% and doesnt say anything about capacitane

3 formulad are
1) E=0.5QV
2) E=Q^2/C
3) E=(C*V^2)0.5


So only formula without C is no. 1 nothing is squared so i am guessing answer is A.
Whats the answer btw


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The answer is B.


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$[br]E = \dfrac{CV^2}{2}[br]$

$[br]\therefore E \propto V^2[br]$

That means that when V becomes 1.5V, E becomes 2.25E. Does that help?


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