Physics: Post your doubts here!

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A small charge q is placed in the electric field of a large charge Q.
Both charges experience a force F.
What is the electric field strength of the charge Q at the position of the charge q?

A) F/(Qq)

B) F/Q

C) FqQ

D) F/q

Answer is D, why isn't it B? :O HELP
 
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A small charge q is placed in the electric field of a large charge Q.
Both charges experience a force F.
What is the electric field strength of the charge Q at the position of the charge q?

A) F/(Qq)

B) F/Q

C) FqQ

D) F/q

Answer is D, why isn't it B? :O HELP


Solution 596
 
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[QUOTE =“ kandelbiz,帖子:131649,成员:4642”]
回复:物理在这里帮助!卡在某个地方?在这里问!:)

在8分钟的时间内,3.6 * 10 ^ 16的氯(Cl-)离子在阳极被中和并释放,而1.8 * 10 ^ 16的铜(Cu2 +)离子被中和并沉积在阴极上。
1.计算此时通过电解质的总电荷
[/引用]
 
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IMG_20210223_023324.jpgcan anyone explain Q11 (b) (i)? I read an explanation somehwere and it said that the two forces have to be subtracted because they are opposite in direction. Why subtract when both forces are in same direction? Shouldn't we add? Both springs will exert a force to the left to bring it to the equilibrium position right?
 
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Hi everyone, AsSalamoAlaikum Wr Wb...

To get things organized in a better way, I am making this thread. As othewise, some queries remain unanswered!

So post your PHYSICS doubts in this thread. InshaAllah other people here will help me and you all. :D ;)

NOTE: If any doubts in the pastpapers, please post the link! You can find links here!

Any Physics related notes and links will be added here in this post. Feel free to provide the links to your notes around the forum, or any other websites! :)
Thanks!
Jazak Allah Khair!


Physics Notes:

Some links & Notes - by destined007

As physics p1 MCQS YEARLY Solved [explaination]

Physics Practical Tips - by arlery

Notes for A2 Direct Sensing (Applications) - shared by sweetiepie

Physics Summarised Notes (Click to download)

AS and A-Level Physics Definitions

A2: Physics Revision notes - by smzimran

Paper:5 Finding uncerainty in log - by XPFMember

Physics Paper 5 tips - by arlery


Physics Compiled Pastpapers: <Credits to CaptainDanger for sharing this..>

Here are the compiled A level topical Physics questions in PDF form...

Paper 1 : http://www.mediafire.com/?tocg6ha6ihkwd

Paper 2 & Paper 4 : http://www.mediafire.com/?g65j51stacmy33c

(Source : http://www.alevelforum.com/viewtopic.php?f=15&t=14)
482aebd9-bd89-4f23-a39f-64aa8aeae733.jpg
in the last part why cant we find the work done using change in potential=workdone/charge. and since workdone is the change in energy it should be valid right? it doesnt give the right answer.
 
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View attachment 65745
in the last part why cant we find the work done using change in potential=workdone/charge. and since workdone is the change in energy it should be valid right? it doesnt give the right answer.
It is because the charge is constant but the potential is changing which means you've got to integrate the dE = QdV expression over the given limits of V1 and V2. Since the 1/2 factor is only introduced due to integration, therefore this gives the right answer. The formula (change in V) = E/Q doesn't have the 1/2 factor in it so you can't get the correct answer because of it.
 

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View attachment 65741can anyone explain Q11 (b) (i)? I read an explanation somehwere and it said that the two forces have to be subtracted because they are opposite in direction. Why subtract when both forces are in same direction? Shouldn't we add? Both springs will exert a force to the left to bring it to the equilibrium position right?
To answer this question, you need to consider the free body diagrams (ignoring trolley weight of course) before and after the displacement of trolley. The right side can be considered as positive for force and displacement direction and vice versa. So initially the forces from both springs act in opposite directions in equal magnitude, thus keeping the trolley in equilibrium, i.e. F = ke.
After the trolley is displaced to the right, the force of the right spring becomes F= k(e-x) and for the the spring on left, it becomes F = k(e+x). The key is to keep the directions same as during the equilibrium, BUT changing the values of spring extension part of the Hooke's law formula. Since the forces are still in opposite direction, simple force balance calculation yields F= - 2kx as ke parts cancel out. The negative sign indicates that the restoring force acts in the opposite direction of the convention we set earlier ( right is +ve) so essentially the magnitude of restoring force is 2kx. I hope I explained it thoroughly so you could understand. 😃
 

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It is because the charge is constant but the potential is changing which means you've got to integrate the dE = QdV expression over the given limits of V1 and V2. Since the 1/2 factor is only introduced due to integration, therefore this gives the right answer. The formula (change in V) = E/Q doesn't have the 1/2 factor in it so you can't get the correct answer because of it.
but the potential needs to be changing; we need to have two values of potential to get potential difference which is exactly what was given. What i have figured is that the charge is what's changing cuz in one of the earlier parts of question it says charge required to raise sphere to a potential, so when the potential drops, sphere loses some of its charge in the spark so we cant use potential difference=workdone/charge cuz its the charge that doesnt stay constant. for potential difference we need to two different values. so I found the energy using different potential and charge values at both points and substracted them to get the loss in the energy and it was correct.
 
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but the potential needs to be changing; we need to have two values of potential to get potential difference which is exactly what was given. What i have figured is that the charge is what's changing cuz in one of the earlier parts of question it says charge required to raise sphere to a potential, so when the potential drops, sphere loses some of its charge in the spark so we cant use potential difference=workdone/charge cuz its the charge that doesnt stay constant. for potential difference we need to two different values. so I found the energy using different potential and charge values at both points and substracted them to get the loss in the energy and it was correct.
yeah that is understandable...totally makes sense.
 
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Re: Physics Help here! Stuck somewhere?? Ask here! :)

Problems in june 2011 paper 11

No 25, 27, 29, 32, 35,, 36,37..

And also can u send me some notes on potential divider, internal resistance along with some working examples which cover the whole chapter...
Thnks in adv..
Q25/. The distance between a node and an antinode is wavelength/4 meters. the frequency of this note is 75 hertz. v= frequency x wavelength. So, the frequency is equal to 75 = v/4L where L is the distance between a node and an adjacent antinode. L=wavelength /4. Wavelength = 4L. The v of the stationary wave is the same. Thus, the length of the tube in the second higher note will be 3 x wavelength / 4. 4L=3 x wavelength. wavelength = 4L/3. So, the frequency is equal to = 3v/4L = 3 x. (v/4L) = 3x 75 = 225 Hz. DO the same for the third higher note. And the answer will be D.
 
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Bro the 2021 papers will come out exactly when the Indian student's AS and A level receives their grades from Cambridge. Idk what date is that on tho. SO,...
Actually they get leaked after they take their exam since sometimes leftover exam copies are kept by their teachers and they give it to their students. I came across the math papers for feb/march 2021 already somewhere.
 
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