Physics: Post your doubts here!

[qwertypoiu]

View attachment 53884
plzzzzzzzzzz fulll explanation??

Electric field lines show the direction in which any positive charge placed in it would move. Since electrons are negatively charged, they'll move in the opposite direction to the direction of electric field line arrows, ie. radially inwards...

Remember that strength of any field is dependent on number of field lines per unit area. Like how concentrated the lines are. In a uniform electric field, the electric field lines are parallel and equidistant, so the electric field strength is same throughout the field. Thus any two objects with the same charge would experience the same force no matter where it is placed. This is because:
F=EQ where E is electric field strength.

However in this diagram, the field lines are not equidistant. The electric field is not uniform. The closer you get to the middle, the more closer the individual lines are, and thus more lines exist per unit area, or another way of saying it is, the lines are more concentrated together.
Thus the charge nearer to the center, charge X, experiences a greater force, simply because the electric field strength is stronger there compared to further away.
F=EQ, the Q is same for both X and Y but E is greater for X.

 

[DESTROYER1198]

can someone give me an example of how to solve this question "Justify the number of significant figures that you have given for your values of k"

 

[The Sarcastic Retard]

can someone give me an example of how to solve this question "Justify the number of significant figures that you have given for your values of k"

https://uk.answers.yahoo.com/question/index?qid=20111113142756AAVpRMu

 

[The Chill Master]

7 a (ii) part I got the answer correct but I first did the 7 a (iii) part then I did the (ii) part so is that ok?
And if not then is their any method to do it directly?

The B part i know it's gonna be a F is directly proportional to 1/L graph but can anyone draw the graph and show ?

Thanks

 

[The Chill Master]

PArt b

 

[Turki AbdulAziz]

For determining whether a collision is elastic or not? Can i look at the relative speed of approach and separation even though the masses are different? or do the masses have to be the same?

 

[ashcull14]

Electric field lines show the direction in which any positive charge placed in it would move. Since electrons are negatively charged, they'll move in the opposite direction to the direction of electric field line arrows, ie. radially inwards...

Remember that strength of any field is dependent on number of field lines per unit area. Like how concentrated the lines are. In a uniform electric field, the electric field lines are parallel and equidistant, so the electric field strength is same throughout the field. Thus any two objects with the same charge would experience the same force no matter where it is placed. This is because:
F=EQ where E is electric field strength.

However in this diagram, the field lines are not equidistant. The electric field is not uniform. The closer you get to the middle, the more closer the individual lines are, and thus more lines exist per unit area, or another way of saying it is, the lines are more concentrated together.
Thus the charge nearer to the center, charge X, experiences a greater force, simply because the electric field strength is stronger there compared to further away.
F=EQ, the Q is same for both X and Y but E is greater for X.

omg thnk yoU

 

[ashcull14]

Electric field lines show the direction in which any positive charge placed in it would move. Since electrons are negatively charged, they'll move in the opposite direction to the direction of electric field line arrows, ie. radially inwards...

Remember that strength of any field is dependent on number of field lines per unit area. Like how concentrated the lines are. In a uniform electric field, the electric field lines are parallel and equidistant, so the electric field strength is same throughout the field. Thus any two objects with the same charge would experience the same force no matter where it is placed. This is because:
F=EQ where E is electric field strength.

However in this diagram, the field lines are not equidistant. The electric field is not uniform. The closer you get to the middle, the more closer the individual lines are, and thus more lines exist per unit area, or another way of saying it is, the lines are more concentrated together.
Thus the charge nearer to the center, charge X, experiences a greater force, simply because the electric field strength is stronger there compared to further away.
F=EQ, the Q is same for both X and Y but E is greater for X.

but the ans to this que is A?...u said tht X experieces greater force than Y? right?? cuz its closer to the center.....bt the ans says it experiences lesser force than Y...that's the point where im confused...can u help me....or is the ans wrong?

 

[The Chill Master]

The Sarcastic Retard

 

[qwertypoiu]

but the ans to this que is A?...u said tht X experieces greater force than Y? right?? cuz its closer to the center.....bt the ans says it experiences lesser force than Y...that's the point where im confused...can u help me....or is the ans wrong?

I really don't see how A could be the answer.... I won't be quick to say the answer is wrong because now I'm confused... Did u check the examiner's report? I hope someone else here can answer...

If you find out the reason to the answer please tell me

 

[The Sarcastic Retard]

http://physics-ref.blogspot.in/2015/01/physics-9702-doubts-help-page-43.html
Q 625

The Sarcastic Retard
View attachment 53936

 

[Maayee]

how was paper 31 practical?

 

[The Chill Master]

http://physics-ref.blogspot.in/2015/01/physics-9702-doubts-help-page-43.html
Q 625

Thanks
So I guess there's a problem with Mark Scheme of the book because it says 0.54ms^-2.

 

[ParulS]

how to calculate the absolute uncertainties ?

 

[Lyfroker]

http://freeexampapers.com/A-Level/Physics/CIE/2007-Jun/9702_s07_qp_5.pdf
Can somebody tell me how to calculate the absolute uncertainties in Q#2
I'm not getting them right

 

[princessnoor]

http://freeexampapers.com/A-Level/Physics/CIE/2007-Jun/9702_s07_qp_5.pdf
Can somebody tell me how to calculate the absolute uncertainties in Q#2
I'm not getting them right

Uncertainity in l^2 is
(Delta L/ L x 2) x L^2

 

[farhan141]

Q.38. Help needed with explanation thanks. Answer is C

 

[ashcull14]

I really don't see how A could be the answer.... I won't be quick to say the answer is wrong because now I'm confused... Did u check the examiner's report? I hope someone else here can answer...

If you find out the reason to the answer please tell me

Actually ds que had been posted by a friend who told me the ans A bt i found the Examiner report and unfortunately it says the ans B..deres no further detail.its november 2008...but i seriously dont undrstnd how cud the force b radially inwards ...thnks for the help btw

 

[qwertypoiu]

Actually ds que had been posted by a friend who told me the ans A bt i found the Examiner report and unfortunately it says the ans B..deres no further detail.its november 2008...but i seriously dont undrstnd how cud the force b radially inwards ...thnks for the help btw

If the answer is B it's perfect! It's supposed to be B!
Read my explanation again. I said that the force will be inwards because electrons are negatively charged, whereas field lines show the direction in which a positive charge would move.
As I said previously the lines are more dense near the centre so X experiences a larger force.
So I was expecting the answer to be B, but you said it's A, but now you found out its B, and yet you seem to be confused?
Finding that the answer was B should've been something to be happy about!

 

[ashcull14]

see solution 428 at
http://physics-ref.blogspot.com/2015/03/physics-9702-doubts-help-page-81.html

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