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    (Original post by hajira)
    here's the graph
    check this out though :confused:
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    For measuring the current dont you just use a resistor of a known value and a voltmeter.

    For the earths electric field i drew a circle around the earth, with lines pointing towards the earth
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    (Original post by xstream)
    what does the earths B electric field look like and why is it caused??
    How many marks was that worth? I think I messed it up.

    I think its supposed to look like a circle(earth) with radial lines coming out (from the surface of the earth, not centre of the earth), stopping at the ionosphere.

    Not sure what possessed me but I did circles around the earth.

    How many marks was it? 2?

    Ok im lost again. B electric?

    B field- magnetic
    E field - electric no?

    Did it even have the B in the exam? If it did it would be a right angles to E, so maybe my picture is right? Im lost. Why cant they give us a bloody copy of the paper to take home. I always take my math question paper to check at home. Synoptic question answer paper used to be seperate -
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    Surely the lines would be towards the earth because positive always points to negative?
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    (Original post by sequence123)
    Surely the lines would be towards the earth because positive always points to negative?
    yep your right. it says ionosphere +ve. earth -ve.

    Did it say anything about B field anywhere in the question?

    How much was that diagram worth?
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    sorry that was just a typo, it didnt say B anywhere..just electric
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    The diagram was the second part of a question which was worth 5 marks, so id say the diag is worth 3 marks max
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    i dunno why you peeps are drawing radial fields for :confused:

    it was an electric field E

    straight lines, i.e like a capacitor locally.
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    http://baubiologie.us/learn/sample/c...Abb1elektr.jpg
    This picture clearly shows that a radial field around the earth
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    (Original post by mackin boi)
    ahh man im loosing marks all ovr the place for stupid things...

    LOL - Welcome to my world.

    I could really kick myself sometimes.

    But the main problem is leaving study and revision for the last minute.

    Im never going to do that again. I bought a tonn of past papers plus answers and never finished them all. Doing past papers makes all the difference.

    Just once in my life I dont want to be cramming a few hours before the exam - I just cant cope. I want to finish all revision at least a good 3weeks before the exam. Next year its going to be different I hope.
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    i drew the arrows down to the earth, then crossed the diagram out and drew it with the arrows the other way, ahh well, second qu i ahve chnaged the rtite answer to the rong 1
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    (Original post by nas7232)
    i dunno why you peeps are drawing radial fields for :confused:

    it was an electric field E

    straight lines, i.e like a capacitor locally.
    that what is know as radial lines.

    the other is know as circles.
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    well, it's easy paper
    Q1:
    -E = F/q.
    -The passage did tell you that ionosphere is +, the Earth is -. So I assumed E = constant = V/d ... then calculated something I forgot.
    -Diagram: 2 circle, inside is the Earth, out side is ionosphere, electric field lines are towards the centre of the Earth
    -I use V voltmeter to measure peak voltage across the resistor of 1Mohms, range of 0.1mV - 10mV.
    -The peak current graph is like inverse of UUU with period is 0.02s

    Q2: I said electron ionises atoms/molecules and bubble is form along these atoms. The light shines so that we can see the visible white track.
    r decreases as electron lost energy due to collisions, and because F = Bev = mv^2/r, so r = mv/Be.
    And measure rP, rQ, rR are 6.5, 4.6 and 3.0 cm respectively. Then use p = Ber.
    After that use m = p/v (which v = c) -> mass of electron is much bigger as it travel with high speed( nearly speed of light).

    Q3: I used Q = 60mgh -> dT = 60gh/c, got something like 3.6K
    The actually dT might be less cuz of loss of heat to surrounding....
    Water has much higher s.h.c than lead -> dT = 60gh/c is very small -> might not be detected.
    V(25) = 25*1.5/(47025)V = Vthermocouple.
    Thermocouple is tiny so it can be put in the tube.

    Q4: -The star is moving away as frequency decreases
    - v = c*20/440 (don't remember the number) I said it moving towards cuz I didn't remember blueshift or redshift.
    - Use Q = CV = n.e, I got n about 8.10^7
    The diagram contains Cathode which is made of the metal plate and is illuminated by nomochromatic light, and Anode connected to ammeter. Not so sure about this.

    Well, tomorrow Chemistry 6 and P6. Nearly finish
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    (Original post by xstream)
    http://baubiologie.us/learn/sample/c...Abb1elektr.jpg
    This picture clearly shows that a radial field around the earth
    locally it's a straight lines. Otherwise for the next question we can't use E=V/d.

    Hopefully, either way is right.
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    (Original post by sequence123)
    For measuring the current dont you just use a resistor of a known value and a voltmeter.

    For the earths electric field i drew a circle around the earth, with lines pointing towards the earth
    oh yeh what did u put the resistor value as... I think i put 100M ohms cos the current was pA(x10^-12).....otherwise there would be a very low reading in the voltmeter as V=IR
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    (Original post by mikeA1)
    check this out though :confused:
    I'm afraid the answer may look different. Please remember the charge equation: q=epsilon*E*A.

    As there are instances when q=0, there must be some 'backward' currents which carry charges from the more negatively-charged plate to the more positively-charged one. Hence the graph should look rather like a sine (although not that curved)
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    i did like a sin graph with period 0.02s...i think thats right...thats what my friends did too
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    (Original post by DmAsTeR)
    i did like a sin graph with period 0.02s...i think thats right...thats what my friends did too
    The period in my graph was 0.010s as in one revolution, there are two instances when the shapes of superimposition of the 2 plates are identical, hence 2 positive peaks (and 2 negative ones) in one rev, meaning 2 periods in 1/50 of a second.
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    that doesnt matter...in 1 period there's 1 max positive and 1 max negative...it doesnt matter if the current is negative max...its still a max!...only 1 period with 1 positive and 1 negative max i say...
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    (Original post by DmAsTeR)
    that doesnt matter...in 1 period there's 1 max positive and 1 max negative...it doesnt matter if the current is negative max...its still a max!...only 1 period with 1 positive and 1 negative max i say...
    Yes, it is the period of the current that has such charateristic. But 50Hz is the rotational period of the vane plate.

    I'm trying to convert the rotational period to the period of the current by saying 1 period of rotation contains 2 periods of the current.
 
 
 
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