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

1. why is field strength 1560 and not 1875?
2. (Original post by chughes17)
why is field strength 1560 and not 1875?
Potential was 625V and r was 0.4m. E=V/r gives 1560 Vm^-1

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3. (Original post by CD223)
Potential was 625V and r was 0.4m. E=V/r gives 1560 Vm^-1

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1/0.4=2.5 on the graph right? I might be being stupid but i'm reading it as 750 for V?

edit: that's for 3 lmao how stupid. 1 mark or 0?
4. (Original post by CD223)
Potential was 625V and r was 0.4m. E=V/r gives 1560 Vm^-1

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shouldnt it be minus?
5. I managed to get the paper today, would anyone like me to post pictures of any pages/questions?
6. (Original post by CD223)
Potential was 625V and r was 0.4m. E=V/r gives 1560 Vm^-1

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Why can we not use the electric field formula for a point charge?

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7. (Original post by TashJacobs)
I managed to get the paper today, would anyone like me to post pictures of any pages/questions?
Would be great if you could upload the multi choice part!
8. (Original post by TashJacobs)
I managed to get the paper today, would anyone like me to post pictures of any pages/questions?
Possible to take pictures of the Multiple Choice Section? Kind of find it hard to remember which answer I put down when I do not know the other three options that were available (trying to check what I put down against the mark scheme). Thanks!
9. (Original post by Salt and Peppa)
Possible to take pictures of the Multiple Choice Section? Kind of find it hard to remember which answer I put down when I do not know the other three options that were available (trying to check what I put down against the mark scheme). Thanks!

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10. (Original post by Vincent Song)
shouldnt it be minus?
I suppose it should yes, although I'm not sure whether that will be a marking point.

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11. (Original post by CD223)
I suppose it should yes, although I'm not sure whether that will be a marking point.

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I used the electric field strength equation and I substituted the value of Q ,which was negative, in that equation and then I got -1560.
12. (Original post by Vincent Song)
I used the electric field strength equation and I substituted the value of Q ,which was negative, in that equation and then I got -1560.
Damn, which charge did you use??

Edit: sorry, did you use the field strength for a point charge?

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13. (Original post by Mehrdad jafari)
Damn, which charge did you use??

Edit: sorry, did you use the field strength for a point charge?

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-27.8nC if that helps.
14. How fussy are examiners on minus signs? Neither my electric field strength nor work done was negative :/

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15. (Original post by CD223)
How fussy are examiners on minus signs? Neither my electric field strength nor work done was negative :/

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I don't think if mine was either . If in the previous past papers the minus sign is in a bracket then the mark is not deducted

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16. (Original post by Vincent Song)
-27.8nC if that helps.
Sorry, where did you get that?
Edit: i got it, but i used the electric field formula for a point charge which gave me a different answer

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17. (Original post by CD223)
How fussy are examiners on minus signs? Neither my electric field strength nor work done was negative :/

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The work done should be positive I think.
18. (Original post by Vincent Song)
The work done should be positive I think.
Oh really? I'm still unsure. To be honest I guess the numerical answer being right is the main thing.

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19. The work done should be positive because it was the work done on a positive (60nC) charge
20. (Original post by Mehrdad jafari)
Sorry, where did you get that?
Edit: i got it, but i used the electric field formula for a point charge which gave me a different answer

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.
I got that value in 2c(i). Which formula? Is it E=Q/(4pie sth r^2)?

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