IGCSE/GCSE/O & A Level/IB/University Student Forum
Qualification => Subject Doubts => GCE AS & A2 Level => Sciences => Topic started by: Xam_Vampire on December 12, 2010, 07:18:17 pm
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Hey peeps help me out:
june 07 ques 3 b
nov 07 ques 5 b ii
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Hey peeps help me out:
june 07 ques 3 b
nov 07 ques 5 b ii
I apologize for the late reply. :-[
Jun 07 p4
3.(b) You need to find the area under the graph shown in Fig 3.2 as said in the question itself. But you need to be careful because the units have to be converted to metres and volt per metre respectively. ;)
Potential difference = 530 V
Nov 07 p4
5.(b)(ii) Again you need to find the area under graph as shown. Divide into different tirangles and trapeziums to improve accuracy.
Area under graph = (1/2 x 4 x 1) + (1/2 x 2 x 1) + 10 + 1/2(4 + 1.6)(2) + (1/2 x 4 x 1.4) + (1/2 x 4 x 0.2) = 21.8 cm2
1 cm2 = 0.125 mA x 1.25s = 1/6400 C
Hence initial charge = 1/6400 x 21.8 = 3400 uC
Hope it helps :)