# OCR Physics A G485 - Frontiers of Physics - 18th June 2015Watch

3 years ago
#501
(Original post by randlemcmurphy)
G485, June 2014, Question 1b ii
Lol lets hope that is easy to see...

The point X when its closest to A the force between them is max (since X is also positive i believe) so the repulsive force is max. Also take in account of the direction, it is pushing X to the right so we take right as positive. So the graph starts from max and goes diagonally (curvy, negative gradient) towards 0 at the point where X usually was. (They mention that electric force is zero or something). As X moves away from A it comes in contact with the force from B which is directed LEFT, so negative. Graph follows from that point (still negative gradient) and ends at minimum(max).

EDIT:
The graph should not be touching the Y-axis at the top, sorry about that since that's kind like fusion -_-'. But yeah its a similar shape...
1
3 years ago
#502
(Original post by sagar448)
Lol lets hope that is easy to see...

The point X when its closest to A the force between them is max (since X is also positive i believe) so the repulsive force is max. Also take in account of the direction, it is pushing X to the right so we take right as positive. So the graph starts from max and goes diagonally (curvy, negative gradient) towards 0 at the point where X usually was. (They mention that electric force is zero or something). As X moves away from A it comes in contact with the force from B which is directed LEFT, so negative. Graph follows from that point (still negative gradient) and ends at minimum(max).
Thanks extremely helpful. I will rep you when I can! (Good luck tomorrow for G482 )
0
3 years ago
#503
(Original post by randlemcmurphy)
Thanks extremely helpful. I will rep you when I can! (Good luck tomorrow for G482 )
No worries and thank you. I edited the post don't forget to read the last line.
0
3 years ago
#504
(Original post by sagar448)
No worries and thank you. I edited the post don't forget to read the last line.
That is the only thing I was going to question! Thanks again.
0
3 years ago
#505
(Original post by sagar448)
Oh I see, yes, it is easier if you visualise it. If you want I can help you with that question?
oh i was able to do it but i just thought it was a weird question haha
0
3 years ago
#506
(Original post by L'Evil Fish)
Only done 2, A*s in both

Even with this post I think I'll get slated lol
You assume everyone is horrible on here haha!
How did you revise for this subject?
#507

You assume everyone is horrible on here haha!
How did you revise for this subject?
G485 just listened to teachers, did their questions, and just did the papers

Didn't read the revision guide for this one

Don't think I will this year
0
3 years ago
#508
(Original post by L'Evil Fish)
.r
If you get a chance would you explain 2cii on G484 Jan 2012?
3 years ago
#509
Can anyone explain the answer to this question please? Finding the electric field strength between two point charges, at a point which does not lie on the line joining the charges.
Thanks!!

0
3 years ago
#510
(Original post by ReeceG46)
Can anyone explain the answer to this question please? Finding the electric field strength between two point charges, at a point which does not lie on the line joining the charges.
Thanks!!

What's the answer? I've got 0.047 N/C and could explain to you how I got that.
EDIT just realised I've done it wrong, I'll have a look again
EDIT A bit confusing but I found the examiner's report. Basically the vertical components cancel out since they are equal and opposite. You then need to work out the electric field strength 0.33m away which is 0.033N/C. You can use cos49 = adjacent (horiz component)/0.033 to work out the horizontal component, which is 0.0217N/C. You multiply this by 2 to get the resultant because there's 2 charges. = 0.043N/C.

Does anybody know why you can do this? I thought you would need to work out the resultant of the diagonal lines, not the horizontal lines?
0
3 years ago
#511
(Original post by BrokenS0ulz)
What's the answer? I've got 0.047 N/C and could explain to you how I got that.
EDIT just realised I've done it wrong, I'll have a look again
Hi mate, the way they do it is:

Find the field strength of one of the charges at a distance 33cm (E=kQ/r^2)
Then multiply this by cos49(which is the bit I don't get? Getting the horizontal component?)
And then multiply this by 2 (I guess because the fields add to give a resultant field of double the magnitude)

0
3 years ago
#512
(Original post by ReeceG46)
Hi mate, the way they do it is:

Find the field strength of one of the charges at a distance 33cm (E=kQ/r^2)
Then multiply this by cos49(which is the bit I don't get? Getting the horizontal component?)
And then multiply this by 2 (I guess because the fields add to give a resultant field of double the magnitude)

Tried my best to explain it above, don't really get it myself. Would be good if someone could explain it!
0
3 years ago
#513
(Original post by BrokenS0ulz)
Tried my best to explain it above, don't really get it myself. Would be good if someone could explain it!
Thanks! That does make sense, about the vertical components cancelling out, but I wouldn't have guessed to do that.

I doubt we would be given one that hard though
0
3 years ago
#514
(Original post by ReeceG46)
Thanks! That does make sense, about the vertical components cancelling out, but I wouldn't have guessed to do that.

I doubt we would be given one that hard though
Yeah it's quite an odd question but it's something I can imagine ocr doing haha, hopefully someone can explain why you work out the resultant of the horizontal components.
0
3 years ago
#515
(Original post by Elcor)
For d.i) on that same question, why is the total capacitance just the sum of the capacitors? I'd have thought that because the p.d.s across them are different at t=0 that means they're in series and therefore you have to use the inverse addition formula.
I'm still confused about 1.d.i in June 2010, I understand that the capacitors are in parallel before the first one is charged, because they have the same p.d across them (0V). Then, the first one is charged and has a p.d across it. Then it's connected to the second one and you need to find the total capacitance at this instant. One has 6.3V, the other has 0V. Therefore they are in series. Yet to find the total capacitance you have to simply add them as if they were in parallel. Why?
0
3 years ago
#516
(Original post by Elcor)
I'm still confused about 1.d.i in June 2010, I understand that the capacitors are in parallel before the first one is charged, because they have the same p.d across them (0V). Then, the first one is charged and has a p.d across it. Then it's connected to the second one and you need to find the total capacitance at this instant. One has 6.3V, the other has 0V. Therefore they are in series. Yet to find the total capacitance you have to simply add them as if they were in parallel. Why?
Yeah I've noticed same problem and I think they are in series too. I realised the examiners wanted parallel however as there was only one voltage in the last question hence same voltage must have been across both.

Examiners report is patronising and says 'some candidates thought the capacitors were in series despite the question clearly stating they are in parallel'. The problem is however that statement and the description afterwards are contradictory! OCR were probably still trying to learn their own specification at this point, it was the first exam after-all and they are usually poor being the first in a series haha.

Posted from TSR Mobile
0
3 years ago
#517
(Original post by Hilton184)
Yeah I've noticed same problem and I think they are in series too. I realised the examiners wanted parallel however as there was only one voltage in the last question hence same voltage must have been across both.

Examiners report is patronising and says 'some candidates thought the capacitors were in series despite the question clearly stating they are in parallel'. The problem is however that statement and the description afterwards are contradictory! OCR were probably still trying to learn their own specification at this point, it was the first exam after-all and they are usually poor being the first in a series haha.

Posted from TSR Mobile
Mate OCR are *******s...

I don't care if you tell me what's happening in the question, I need to understand the physics if I'm going to answer the question properly.

I'll ask my physics teacher tomorrow to see if he can make sense of it.
0
#518
(Original post by Elcor)
Mate OCR are *******s...

I don't care if you tell me what's happening in the question, I need to understand the physics if I'm going to answer the question properly.

I'll ask my physics teacher tomorrow to see if he can make sense of it.
Well for 2 marks and a 1 mark beforehand it was obviously just a quick Q = CV sub in

0
3 years ago
#519
(Original post by L'Evil Fish)
Well for 2 marks and a 1 mark beforehand it was obviously just a quick Q = CV sub in

(Original post by Hilton184)
Yeah I've noticed same problem and I think they are in series too. I realised the examiners wanted parallel however as there was only one voltage in the last question hence same voltage must have been across both.

Examiners report is patronising and says 'some candidates thought the capacitors were in series despite the question clearly stating they are in parallel'. The problem is however that statement and the description afterwards are contradictory! OCR were probably still trying to learn their own specification at this point, it was the first exam after-all and they are usually poor being the first in a series haha.

Posted from TSR Mobile
I think I might have worked it out. It might just be a matter of semantics. They say 'when the switch is closed', which implies you're working with the circuit at that very moment, when in actuality they mean some length of time has passed since that moment, and thus the current would have stopped flowing and the charges stored on either capacitor would be the same by then.

Do you think that's it?
0
3 years ago
#520
(Original post by Elcor)
I think I might have worked it out. It might just be a matter of semantics. They say 'when the switch is closed', which implies you're working with the circuit at that very moment, when in actuality they mean some length of time has passed since that moment, and thus the current would have stopped flowing and the charges stored on either capacitor would be the same by then.

Do you think that's it?
Sorry, I might be able to help, what question is this?
0
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