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

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Original post by IWantSomeMushu
Anybody got any tips on how to tackle problems like this?


Think of it like a ratio of 1:4 charge. You'd have 5 "charge parts" in total. There's a distance of 120, so 1 "charge part" is equal to 24mm (120 divided by 5). So the answer must be a multiple of 24 and be closer to +4 since that is the weaker charge. Hence the answer is A.

Sorry if that isn't very clear but that's how I think about those types of questions hahah
Original post by CD223
That's what I make it :smile: how's revision going?


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Not too bad thanks, just going over things I struggle with. Although the exam could be hit or miss for me, I seem to either get a C or A* on past papers! How's your revision going?


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How do you do this please help:

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Original post by Klaxoii
Not too bad thanks, just going over things I struggle with. Although the exam could be hit or miss for me, I seem to either get a C or A* on past papers! How's your revision going?


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I agree! What grade are you aiming for?:smile:


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Original post by Bealzibub
How do you do this please help:



3 questions (marks) AQA want you to really work for its Q15, Q17 and Q20.
Only pursue these types of crazy difficult questions if you want close to full UMS.

As for this question I have noooooo idea someone helppppp <3
Original post by gcsestuff
The answer to this is zero? ImageUploadedByStudent Room1433945590.035900.jpg
I still don't understand


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I got C, not sure what the right answer is though.
Original post by Bealzibub
How do you do this please help:



Dunno but looks like the answer should be C or D because you get 50% close to one charge and 150% away from the other. Resultant force must be higher than before but I dunno how to calculate for working out if itis C or D
Need help on this too..

Change in potential is = -30V

work done = e * (-30V) = -30eV and force towards Y.

answer is A.. how?

Original post by Bealzibub
How do you do this please help:



Answer is C


ImageUploadedByStudent Room1433947780.819058.jpg


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Can someone define tesla


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Original post by Klaxoii
ImageUploadedByStudent Room1433946767.452770.jpg

Hope this helps


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Yeah it helped!
Thanks again!!! :smile:
Original post by Bealzibub
Need help on this too..

Change in potential is = -30V

work done = e * (-30V) = -30eV and force towards Y.

answer is A.. how?



force is not towards Y, the answers is out of A and D.

the force is away from Y OR Towards X because an electron is negative and the 100V plate is +
Original post by gcsestuff
The answer to this is zero? ImageUploadedByStudent Room1433945590.035900.jpg
I still don't understand


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Because all forces are equal and equidistant so they cancel out
Correct me someone if I'm wrong
Original post by gcsestuff
Can someone define tesla


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F = BIL

therefore, B = F/IL

> N A^-1 m^-1

force per current per metre.
Original post by CD223
I agree! What grade are you aiming for?:smile:


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As high as possible, I need a B in physics overall, so the higher this is, the less pressure I have next week. What are you going for?


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Original post by Kennethm
I got C, not sure what the right answer is though.


The answer in the mark scheme is zero. There's a few diagram with different charges. I'll take a photo to explain
ImageUploadedByStudent Room1433948082.841704.jpg


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When do you use the Right hand rule?
(edited 8 years ago)
Original post by gcsestuff
Can someone define tesla


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The magnetic flux density of a magnetic field that causes a force of 1N to act on a current of 1A in a wire of length 1m which is perpendicular to the magnetic field.

That was from a previous paper I think or the AQA text book. It's "wordy" unfortunately but quite simple at heart.
Original post by Sumz.96
When do you use the Right hand rule?


For force induced due to the motor effect? I believe


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Original post by gcsestuff
Can someone define tesla


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one weber per square metre,

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