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    hi
    Can anybody pls explain the last part of Q2 on Unit 5 Jan 05
    how does the resultant electric field strength change 2 zero???

    im guessing dis sort of Q wont *** up tormorrow jus it came u p last tyme...? :confused:
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    my answer was that ther must be another particle on the other side of Y. but that was a guess. lol
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    (Original post by cheerfulsakib)
    hi
    Can anybody pls explain the last part of Q2 on Unit 5 Jan 05
    how does the resultant electric field strength change 2 zero???

    im guessing dis sort of Q wont *** up tormorrow jus it came u p last tyme...? :confused:
    :reddy:
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    wow....is there a Jan 05 paper? could someone email it to me? i would be erally grateful




    are there papers from january in otheryears? does anybody have copies of them?

    thanks
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    maybe wanna seperate the email address to prevent spam
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    (Original post by cheerfulsakib)
    hi
    Can anybody pls explain the last part of Q2 on Unit 5 Jan 05
    how does the resultant electric field strength change 2 zero???

    im guessing dis sort of Q wont *** up tormorrow jus it came u p last tyme...? :confused:
    At Y: the force due to the -1uC point charge is attractive, the force due to the +3uC is repulsive. So the forces are in opposite directions. At Y, the +3uC charge is further away but is a bigger charge, whilst the -1uC charge is closer but smaller. So the magnitude of the two forces at Y are the same, but their directions are opposite. So the resultant field strength is zero.
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    Can ne one tell me how do this qu form the jan 05 paper

    jan 05
    qu 4

    Explain,if the wire had twice the diameter, the current would have to be increased by a factor fo four to just lift the wire above the surface of the bench.

    ( the previous bit shows magnetic fields lines with a wire of lenght 0.4 m of mass 0.5g and the magnetic flux is 2 * 10 ^-2 T
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    (Original post by Imaad)
    Can ne one tell me how do this qu form the jan 05 paper

    jan 05
    qu 4

    Explain,if the wire had twice the diameter, the current would have to be increased by a factor fo four to just lift the wire above the surface of the bench.

    ( the previous bit shows magnetic fields lines with a wire of lenght 0.4 m of mass 0.5g and the magnetic flux is 2 * 10 ^-2 T
    If the diameter is 2 times more, the volume is 4 times more (since v=L(pi)r^2), so the mass is 4 times more, so the force needed to raise the wire is 4 times more, so the current needs to be 4 times more.
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    sweet thanks alot i knew it was something to do with somehting ^2. i was tryin to do it with the magnetic field strength and stuff and failin big time. thanks alot
 
 
 
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