IGCSE/GCSE/O & A Level/IB/University Student Forum
Qualification => Subject Doubts => GCE AS & A2 Level => Sciences => Topic started by: cs on September 14, 2010, 11:08:42 am
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Nov 2002
Q6) why is the arrow in upward direction ( as what i noticed, its from higher potential to lower), any one care to explain, here's the question attached with answer
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electron has a negative charge. Direction of force on electron will be towards the positive plate.
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electron has a negative charge. Direction of force on electron will be towards the positive plate.
+rep
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Unlike charges attract and like charges repel :)
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Nov 2002
Q6) why is the arrow in upward direction ( as what i noticed, its from higher potential to lower), any one care to explain, here's the question attached with answer
Actually the electric field is downwards from higher potential to lower potential. When the electron penetrates the electric field perpendicularly an electric force acts on it. This electric force is always opposite to the direction of the electric field for an electron.
Therefore the electron proceeds through the electric field between the two plates with a slight parabolic path towards the positively charged metal plate. Then it exits the field in a a straight line tangential to the point it exits the field.
But its easier to understand the way Garfield explained, i.e being negatively charge it is attracted to the positively charged metal plate.
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+rep Deadly_king :D Very informative mA :)
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+rep Deadly_king :D Very informative mA :)
Thanks.......though you said almost the same thing in only one line :P
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Thanks.......though you said almost the same thing in only one line :P
LOL! Now Cs has more ways to learn it..the easy and the hard way :P ;)
mA! Keep up this amazing work :D
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LOL! Now Cs has more ways to learn it..the easy and the hard way :P ;)
mA! Keep up this amazing work :D
Hahaha........yeah i guess so ::)
Sure........i actually love helping and it also helps me revise :)
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I really am proud that this forum has very helping members like you :D
And insA Allah will help you as you are helping others..and yea trueeee...when you explain and help someone..the info sticks to ur mind and reminds you if u know the topic well or not insA :D
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You know what??
The way you express yourself actually makes me want to help even more.........its very encouraging :)
Its a pleasure to be acquainted to such a nice person............you'll certainly be a good mod :)
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Thanks to all of you.. I am really glad that i found you guys in this forum.. :)
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Thanks to all of you.. I am really glad that i found you guys in this forum.. :)
although you guys made me feel that i am so weak in physics.. ;D
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Nan.......that was not our intention! Far from that we'll help you in any way we can :)
There's no shame in not knowing something ;)
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You know what??
The way you express yourself actually makes me want to help even more.........its very encouraging :)
Its a pleasure to be acquainted to such a nice person............you'll certainly be a good mod :)
Thank you..I am flattered :D :-[ :P I like that I encourage to help :D
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Thank you..I am flattered :D :-[ :P I like that I encourage to help :D
Hehe........i hope that 1 day i may be as encouraging as you are to others :)
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can someone help explain what is phase angle 60 degree as in the OCT/NOV 2002 Q5
it would be great if someone can show me how to do the Q5b(i) too :)
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can someone help explain what is phase angle 60 degree as in the OCT/NOV 2002 Q5
it would be great if someone can show me how to do the Q5b(i) too :)
By a phase angle of 60 degrees.......it meant that the wave T2 would be moving 60 degrees faster.
In other words, when wave T1 is at maximum amplitude at time t=0, wave T2 will be at maximum amplitude at time t=0.5
But both waves will be having same frequency, wavelength and period.
60 degrees is the phase angle. To be able to do Q5b(i) you should convert the phase angle to the time difference sine the graph is one of X against time.
Phase angle = (delta T)/T * 360
From the graph it is noted that period T= 3s.
Hence delta T will be 0.5s.
You just need to draw a wave with same amplitude, same frequency and wavelength........but which is just 0.5s faster.
Hope you understood........if not let me know and i'll try to be clearer :)
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Keep it up DK :D :D
+rep
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By a phase angle of 60 degrees.......it meant that the wave T2 would be moving 60 degrees faster.
In other words, when wave T1 is at maximum amplitude at time t=0, wave T2 will be at maximum amplitude at time t=0.5
But both waves will be having same frequency, wavelength and period.
60 degrees is the phase angle. To be able to do Q5b(i) you should convert the phase angle to the time difference sine the graph is one of X against time.
Phase angle = (delta T)/T * 360
From the graph it is noted that period T= 3s.
Hence delta T will be 0.5s.
You just need to draw a wave with same amplitude, same frequency and wavelength........but which is just 0.5s faster.
Hope you understood........if not let me know and i'll try to be clearer :)
Thank you DK ^^ your explanations are crystal clear! :D
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You are welcome :)
Glad to have been able to clear your doubts ;)
Thanks Asif :D
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i have a problem on O/N 2004 paper 2
its on Q6b (ii) & c
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b ii)Potential difference acrooss C&R are same as emf(because of parallel connection)
From the graph:
when v=2V
current in R=1.3A
current in C=0.75A
therefore,current in the battery+1.3+0.75=2.05A
c)The current is same in both R&C as they are connected in series.But the potential difference across C is higher than R.Since power=VI,therefore component c will dissipate thermal energy at a greater rate
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many thanks asif! :D
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U WELCOME ;D