---> Physics Paper 4 Past Paper Discussion <---

[1357913579]

Yes, that is right, in the middle an electron would experience equal attraction from the two positive charges and a proton would experience equal repulsion from the same two charges and thus the net force (and the net field strength) would be equal to zero.

If they ask the question with a proton, I assume you're talking about part (c), then the work done should be the positive value of the work done there, because the potential at a point is the work done by an external force in moving unit test charge from infinity to that point; an electron would accelerate towards the protons, and to ensure that it goes at a constant velocity a negative amount of work has to be done. However, a proton will not move there due to the repulsion so an external applied force would have to do positive work in moving the proton to any point in the diagram.

The question says the work done by the electron, not the external force, so the work done is positive. If a proton were involved, I would guess that the answer would be the negative version of the same value since both the proton and the electron have the same charge magnitudes.

Hope this helped!
Good Luck for all your exams!

thanks alot bro for your cooperation
this is whats written in the examiner report:
Very few candidates were able to cope with the two charges and determine the potential change
between the two points. A large number did not realise that the potential at each point was the
sum of the potentials for each charge. A number of candidates assumed the midpoint was at zero
potential or that there was only one charge. A significant number tried to determine the work done
using W = Fd or Vq. Very few were able to determine the change in potential between the two
points for both charges and then determine the work done on an electron.
Bro can you please solve this question to get the value of work done as i have tried for hours and hours still getting the wrong answer.

 

[1357913579]

But since they're both in opposite directions, wouldn't that mean that they cancel each other out? Because that's what the top answer on this Yahoo Answers page says:

http://answers.yahoo.com/question/index?qid=20120322031104AAaEyQJ

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w11_qp_41.pdf
number-4 please can u help iam totally confused about this question
thanks alot
iam only able to a(i) part nothing other than this iam totally 0 about how to solve the rest of the parts of this question

 

[MaxStudentALevel]

sagar65265 thanks once again!!!
Could you please take a look at Q10 for ON 2009 41??

Im getting the opposite answer of the MS?
Wouldnt the red turn on for 18? And be green for 16? As V- would be greater an V+ here
Because the thermistor adds overall more resistance thus more voltage to that side?

 

[MaxStudentALevel]

Also another VERY small query, how to find Freq of information signal? On ON 2009 42 Q11 b (ii) ????

 

[sagar65265]

thanks alot bro for your cooperation
this is whats written in the examiner report:
Very few candidates were able to cope with the two charges and determine the potential change
between the two points. A large number did not realise that the potential at each point was the
sum of the potentials for each charge. A number of candidates assumed the midpoint was at zero
potential or that there was only one charge. A significant number tried to determine the work done
using W = Fd or Vq. Very few were able to determine the change in potential between the two
points for both charges and then determine the work done on an electron.
Bro can you please solve this question to get the value of work done as i have tried for hours and hours still getting the wrong answer.

Man, man , i'm sorry, there may not be any force on a stationary electron at that point and the net field strength may be zero, but the potential definitely isn't zero, man, i'm really sorry about that, Ashique was absolutely right - the potential energy of an electron in between the two positive charges will not be zero, it will be the sum of two systems, the electron and one of the charges and the electron and the other charge!

Just try your calculation by including the potential in the beginning, see if it works!

Good Luck for all your exams!

 

[sagar65265]

sagar65265 thanks once again!!!
Could you please take a look at Q10 for ON 2009 41??

Im getting the opposite answer of the MS?
Wouldnt the red turn on for 18? And be green for 16? As V- would be greater an V+ here
Because the thermistor adds overall more resistance thus more voltage to that side?

The input potential to the inverting input of the op amp is the potential across resistor P, not across the thermistor T - the potential across T is the number of Volts lost when unit charge travels through T, and the potential into the inverting input is the remaining potential, which is the potential across the resistor R, not the thermistor T. So the calculation becomes:

16 Degrees Celcius:
Inverting Input: V- = (2000/4100) * 2 = 0.97 V
Non Inverting Input: V+ = (2000/4000) * 2 = 1 V

Since the potential applied at the non inverting input is greater than that applied at the inverting input, the output is positive (the op amp acts as a comparator) and the diode R is on while the diode G is off. The opposite process occurs for 18 degrees celcius, where the inverting input has a greater potential than the non inverting input and thus G is on while R is off.

Hope this helped!
Good Luck for all your exams!

 

[1357913579]

The input potential to the inverting input of the op amp is the potential across resistor P, not across the thermistor T - the potential across T is the number of Volts lost when unit charge travels through T, and the potential into the inverting input is the remaining potential, which is the potential across the resistor R, not the thermistor T. So the calculation becomes:

16 Degrees Celcius:
Inverting Input: V- = (2000/4100) * 2 = 0.97 V
Non Inverting Input: V+ = (2000/4000) * 2 = 1 V

Since the potential applied at the non inverting input is greater than that applied at the inverting input, the output is positive (the op amp acts as a comparator) and the diode R is on while the diode G is off. The opposite process occurs for 18 degrees celcius, where the inverting input has a greater potential than the non inverting input and thus G is on while R is off.

Hope this helped!
Good Luck for all your exams!

bro help required at this as wel please
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w11_qp_41.pdf
number-4 please can u help iam totally confused about this question
thanks alot
iam only able to a(i) part nothing other than this iam totally 0 about how to solve the rest of the parts of this ques

 

[1357913579]

Man, man , i'm sorry, there may not be any force on a stationary electron at that point and the net field strength may be zero, but the potential definitely isn't zero, man, i'm really sorry about that, Ashique was absolutely right - the potential energy of an electron in between the two positive charges will not be zero, it will be the sum of two systems, the electron and one of the charges and the electron and the other charge!

Just try your calculation by including the potential in the beginning, see if it works!

Good Luck for all your exams!

thanks bro but iam still confused like if electric field is 0 isnt potential 0 can you please elaborate

 

[sagar65265]

Also another VERY small query, how to find Freq of information signal? On ON 2009 42 Q11 b (ii) ????

Since the carrier wave is amplitude modulated with the information signal, the carrier wave will have an "envelope" that joins the crests of each wavelet on the carrier signal. The wavelength of the carrier wave is from one crest on this envelope to the next.
On the graph you can see that the first peak shown is the maximum height for that modulated wave, and that peak appears at t = 2.2 seconds or something (that isn't so important, you can take the part at time t = 0 and use the same zero displacement position for the next maximum). The next maximum amplitude of the carrier (and information) signal comes at t = 102.2 seconds or something of that sort. In other words, from one point on the envelope to the next point on the same envelope that is in phase with the first point is a time difference of 100 microseconds, i.e. 100 * 10^-6 seconds.

Since f = 1/T,

f = 1/(100 8 10^-6) = 10,000 Hz = 10 kHz.

Hope this helped!
Good Luck for all your exams!

 

[sagar65265]

thanks bro but iam still confused like if electric field is 0 isnt potential 0 can you please elaborate

Sorry, man, but i'll try to answer both as soon as possible, have to go and revise for other exams too.
(BTW even i'm confused with question 4 w_11, I think the following link should be helpful:
http://www.physicsclassroom.com/Class/estatics/u8l4d.cfm)

 

[MaxStudentALevel]

sagar65265 THANKYOUSOMUCH!!
Honestly, i do hope you get the grades you wish for!!

 

[1357913579]

sagar65265 THANKYOUSOMUCH!!
Honestly, i do hope you get the grades you wish for!!

bro can u please help me solve thse 2 problems that i have posted please

 

[1357913579]

Sorry, man, but i'll try to answer both as soon as possible, have to go and revise for other exams too.
(BTW even i'm confused with question 4 w_11, I think the following link should be helpful:
http://www.physicsclassroom.com/Class/estatics/u8l4d.cfm)

so ur also confused about both the problems that i posted out?
thanks alot for trying and for the link

 

[sagar65265]

thanks bro but iam still confused like if electric field is 0 isnt potential 0 can you please elaborate

Yahoo Answers is such a savior - I was doubting that statement right after I made it, but see if this explanation helps out:

http://answers.yahoo.com/question/index?qid=20120131161711AAekKKO

Think of it in this way - suppose the electron was being brought towards the two particles on the perpendicular bisector of the line joining them, i.e. along the line through the mid - point. There would be no net sideways pull, but there would still be a net forward pull, right? So this forward pull would do work, and to make sure the electron moves at a constant velocity, the work done on it by the external force (and thus the electric potential) would be negative. Meanwhile, if a proton was being brought into place there, it would be repelled along that same line and a positive amount of work would have to be done on it to get it there, so the potential would be positive!

Hope this helped!
Good Luck for all your exams!

 

[sagar65265]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w11_qp_41.pdf
number-4 please can u help iam totally confused about this question
thanks alot
iam only able to a(i) part nothing other than this iam totally 0 about how to solve the rest of the parts of this question

Okay, so for a (ii), the basic statement is that both charges have the same sign, or the same polarity.
If both the charges have the same sign, they will both affect a charged particle in the same way, they will either both attract or both repel.
Taking a look at the graph, there is a point where the net field strength, and thus the net force (remember, E = Force per unit charge, so if the field strength at a point is zero, the force per unit charge is zero, and thus force on any charge at that point will be zero) will also be zero.

Imagine this - a positive charge is on the left, a negative charge is on the right. If you place a negative charge in the field, it will always be pushed away from the negative charge and towards the positive charge, so there will be no point where there is a zero net force! If you place a positive charge in the field at any point, the same conclusion can be arrived at - there is no point where the net force is zero!

So if the net force has to be zero at any poiny, the pull (or push) from the charge on the left has to be equal in magnitude and opposite in direction to the pull (or push) from the charge on the right, and in that case, the net force vectors add to zero. This can only happen if the two charges have the same polarity, or sign!

b) i) The relationship is that the Electric Field Strength at a point is the negative rate of change of the Electric Potential at a point with respect to time, i.e.

E = - Delta(Potential)/Delta(time)
The reason it is negative is that if the electric potential energy is increasing very fast, then the force on the particle will be a repulsive force, and since the field strength is a vector, it will be negative at this point, since the force is acting away from the charge causing the electric field.

ii) Since the rate of change of potential with respect to time is equal to -E (from the formula above), the rate of change of Electric Potential with respect to time is maximum when the field strength is minimum, which is at t = 11.4 cm on the graph, and the rate of change of potential is zero when the electric field strength is zero too, which is at around 8 cm or so on the graph (when the line touches the axis).

Hope this helped!
Good Luck for all your exams!

 

[sagar65265]

thanks bro but iam still confused like if electric field is 0 isnt potential 0 can you please elaborate

Also, this link:

http://answers.yahoo.com/question/i...Vz7yGIwjzKIX;_ylv=3?qid=20100409093730AAgc8FR

Good Luck for all your exams!

 

[blueberryyums]

http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_s04_qp_4.pdf

Number 4 c

Is this the correct drawing? Just a sketch

 

[seneria]

can anyone solve question 5 part b and c(11) of nov 12 paper 43 physics?

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w12_qp_43.pdf​

 

[seneria]

and also question 9part b(1) of nov 12 41
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w12_qp_41.pdf

please reply soon guys..

 

[seneria]

nov 2 question 5 (a)
http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_w02_qp_4.pdf