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    http://filestore.aqa.org.uk/subjects...2-QP-JUN15.PDF

    Question 2 (c) i

    I haven't a clue where they've got the formula used in the markscheme from, so any help would be greatly appreciated.


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    (Original post by iMacJack)
    http://filestore.aqa.org.uk/subjects...2-QP-JUN15.PDF

    Question 2 (c) i

    I haven't a clue where they've got the formula used in the markscheme from, so any help would be greatly appreciated.


    Cheers
    Two ways:

    1) Standard formula way:

    Electric potential is defined to be

     V = \frac{Q}{4 \pi \varepsilon_0 r} \Rightarrow Q = 4\pi\varepsilon_0 V r by rearranging. Select a value of V and r from the graph and substitute.

    2) Notice from the formula above, if you have a graph of V against 1/r, then the gradient will be  \frac{Q}{4 \pi \varepsilon_0 } . Work out the gradient and multiply by  4 \pi \varepsilon_0 .
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    (Original post by kingaaran)
    Two ways:

    1) Standard formula way:

    Electric potential is defined to be

     V = \frac{Q}{4 \pi \varepsilon_0 r} \Rightarrow Q = 4\pi\varepsilon_0 V r by rearranging. Select a value of V and r from the graph and substitute.

    2) Notice from the formula above, if you have a graph of V against 1/r, then the gradient will be  \frac{Q}{4 \pi \varepsilon_0 } . Work out the gradient and multiply by  4 \pi \varepsilon_0 .
    Ahhh! A simple re-arragement, how have I missed that! Oops!

    Thanks a bunch, also the second method is interesting and good for understanding it more too.

    Cheers
 
 
 
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