The Student Room Group

OCR Physics A - G485: Fields, Particles & Frontiers of Physics - June 2012

Scroll to see replies

Reply 240
does anyone know where i can find good cosmology notes? thank you in advaance :smile:
Original post by Lindizya
does anyone know where i can find good cosmology notes? thank you in advaance :smile:


The Robbie Peck notes in the attachments are pretty good.
If you're looking for something a little more in depth, you could try the CyberPhysics Notes. :smile:
Reply 242
Original post by magdaplaysbass
The Robbie Peck notes in the attachments are pretty good.
If you're looking for something a little more in depth, you could try the CyberPhysics Notes. :smile:


life saver thank you!
hi does anyone have the g484 jan 12 paper?
Havnt seen a thread for the up and coming exam and thought that we could have a discussion on what we think is difficult and what might come up.

I find the MRI machine really hard and it hasn't came up for a few papers so it will be sods law! Its the wording that the mark scheme wants that gets me.
http://www.thestudentroom.co.uk/showthread.php?t=1949108 - I think this may be what you're looking for
Yes, and exam discussion threads have their own section. Not here in the homework help section.
Reply 247
Does anyone have model answers (e.g. from a markscheme!) about how x-rays, PET scans, CAT scans, and contrast mediums work. In fact, any of the medical physics stuff - I know the basics but I'm really struggling to write what the mark schemes actually want.
Reply 248
Original post by Toshiya
Does anyone have model answers (e.g. from a markscheme!) about how x-rays, PET scans, CAT scans, and contrast mediums work. In fact, any of the medical physics stuff - I know the basics but I'm really struggling to write what the mark schemes actually want.


This! i think i'm just going to have to learn the mark scheme for these questions it's the only way
I'm having trouble with this question:

Ionic solids consist of a regular arrangement of positive and negative ions. The figure
below shows two neighbouring ions in a particular ionic solid. The ions A and B may be
considered as two point charges of equal magnitude, 1.6 × 10–19 C, and opposite sign,
with a separation of 2.0 × 10–10 m. The ion A is positive.

(ii) Calculate the magnitude of the electric field strength at the mid point between the
charges.


I'm using E=Q4πϵ0r2E={Q\over{4\pi\epsilon_0 r^2}}

As this is asking for the magnitude between the midpoint, I am using r=1×1010r = 1\times10^{-10}. This is not correct in the markscheme. Why does r need to be the whole distance if this is only for 1 particle?

EDIT: Markscheme:

(ii) Use of E = (1/4πε0)Q/r^2

Sum of two equal terms

E = 2 × 9 × 109 × 1.6 × 10–19 / (2.0 × 10–10)^2

E = 7.2 × 1010 N C–1 or V m–1

(edited 11 years ago)
Original post by ROFLWaffle
I'm having trouble with this question:



I'm using E=Q4πϵ0r2E={Q\over{4\pi\epsilon_0 r^2}}

As this is asking for the magnitude between the midpoint, I am using r=1×1010r = 1\times10^{-10}. This is not correct in the markscheme. Why does r need to be the whole distance if this is only for 1 particle?

EDIT: Markscheme:


r is defined as the distance between the charges; so although the question is asking you to find E at the midpoint between the charges, the charges which create E are 2×10102\times10^{-10}m apart.
The mark scheme has worked out the magnitude of E provided by one of the ions; and then doubles the answer because the other ion provides an equal magnitude of E. :smile:
Reply 251
guys i was wondering what is the total ums needed to get and an A? and what is the total need for a B?
like thorughout the whole course?
Original post by stunnaman
guys i was wondering what is the total ums needed to get and an A? and what is the total need for a B?
like thorughout the whole course?


480/600 ums for an A (80%) and i think it's 420/600 ums for a B (70%)

This was posted from The Student Room's Android App on my GT-S5830
Original post by magdaplaysbass
r is defined as the distance between the charges; so although the question is asking you to find E at the midpoint between the charges, the charges which create E are 2×10102\times10^{-10}m apart.
The mark scheme has worked out the magnitude of E provided by one of the ions; and then doubles the answer because the other ion provides an equal magnitude of E. :smile:


The mark schemes wrong on this question. You should indeed use r=1E=10 not r=2E=10.
Original post by teachercol
The mark schemes wrong on this question. You should indeed use r=1E=10 not r=2E=10.


Thank you.

Original post by stunnaman
guys i was wondering what is the total ums needed to get and an A? and what is the total need for a B?
like thorughout the whole course?


120 UMS in G485 is needed for an A on the Paper, 105 for a B. The total is 150 UMS with 100 marks on the paper. This paper counts for 1/2 the A2 grade and 1/4 of the total A-Level.

For an A in the A2 modules, 240/300 is needed and for the whole course 480/600 is needed. (80% UMS).

The grade boundaries for the paper are as follows (the A*, U and full marks and are estimates. This paper is out of 100). The A* is equivalent to 90% and assumes you have 90% in the other paper and coursework. Depending on your coursework grades and the grade in G484, this will vary.


Paper 100% A* A B C D E

UMS 150 135 120 105 90 75 60

Jun 2010 82 74 66 58 50 43 36
Jan 2011 81 73 65 57 50 43 36
Jun 2011 78 71 64 57 50 44 38
Jan 2012 88 82 76 70 64 58 52
(edited 11 years ago)
Original post by teachercol
The mark schemes wrong on this question. You should indeed use r=1E=10 not r=2E=10.


i'm confused :confused: is r not meant to be the distance between the two charges? thanks :smile:

This was posted from The Student Room's Android App on my GT-S5830

EDIT: Never mind; I get it now :smile:
(edited 11 years ago)
Original post by magdaplaysbass
i'm confused :confused: is r not meant to be the distance between the two charges? thanks :smile:

This was posted from The Student Room's Android App on my GT-S5830


From my understanding, if you're using F=Q1Q04πϵ0r2F={Q_1Q_0\over4\pi\epsilon_0r^2} then you're measuring the force between the two point charges. Therefore you need to use the whole distance.

The question before asked for the field at the midpoint. It wants you to find the electric field from both of the particles at that point, then add them up.

When you use E=Q4πϵ0r2E={Q\over4\pi\epsilon_0r^2}, you single out one particle and completely ignore the other particle when working out the electric field strength. Therefore the distance between the point charge and the midpoint is half the distance between the two point charges.

After you work out the effect of both of the particles' fields on the midpoint, you add or take away the field strengths to get the total (add if the charges are opposite, subtract if they're the same).
Original post by ROFLWaffle
From my understanding, if you're using F=Q1Q04πϵ0r2F={Q_1Q_0\over4\pi\epsilon_0r^2} then you're measuring the force between the two point charges. Therefore you need to use the whole distance.

The question before asked for the field at the midpoint. It wants you to find the electric field from both of the particles at that point, then add them up.

When you use E=Q4πϵ0r2E={Q\over4\pi\epsilon_0r^2}, you single out one particle and completely ignore the other particle when working out the electric field strength. Therefore the distance between the point charge and the midpoint is half the distance between the two point charges.

After you work out the effect of both of the particles' fields on the midpoint, you add or take away the field strengths to get the total (add if the charges are opposite, subtract if they're the same).


thanks :h: yeah, I remember learning that when working out the force, the distance is between the two charges; and just assumed that this applied to the Electric field strength as well. Thanks for clearing that up :smile: hopefully, i won't be making silly mistakes like that in the exam :crossedf:
I'd like some clarification about Question 21 b ii (pictured):

VDJ3f.png

The left hand rule should be used as a force is being induced. (Left hand rule for a force, right hand for a current).

The magnetic field is going upwards to the left (to the centre of the imaginary circle that the electron moves on).

From my understanding, the charge would be in the opposite direction to the motion of the electron as it has a negative charge.

Therefore the Force would be going into the page.

The mark scheme seems to assume a conventional current, even though I can't see any references to this in the paper.

Is my reasoning correct? How should I know to use a conventional current if a similar question comes up on the exam?
Reply 259
Original post by ROFLWaffle
I'd like some clarification about Question 21 b ii (pictured):

VDJ3f.png

The left hand rule should be used as a force is being induced. (Left hand rule for a force, right hand for a current).

The magnetic field is going upwards to the left (to the centre of the imaginary circle that the electron moves on).

From my understanding, the charge would be in the opposite direction to the motion of the electron as it has a negative charge.

Therefore the Force would be going into the page.

The mark scheme seems to assume a conventional current, even though I can't see any references to this in the paper.

Is my reasoning correct? How should I know to use a conventional current if a similar question comes up on the exam?


You're a bit mixed up with your terms.

Force acts up the page, this is why the electons accelerate upwards in the magnetic field region.
The conventional current is flowing left. Remember, electron flow is opposite to conventional current.

Flemming's LHR ALWAYS uses conventional current.

[Edit: to answer your last question, conventional current flows opposite to the direction of negative charge flow, and in the same direction as positive charge flow.]
(edited 11 years ago)

Quick Reply

Latest

Trending

Trending