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    Afternoon all,

    I was thinking about this yesterday, when two capacitors are in parallel in a closed circuit they have the same voltage being applied to them both. But why is it when you have a fully charged capacitor at voltage V being passed through another capacitor in parralell they must have (lower) combined voltage?

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    Because the charge on the first capacitor (Q=CV) is shared between the two when they are joined. Some change moves from it to the other capacitor because the other one is at a lower pd. (zero to start with) This results is less charge on the original and therefore a lower voltage. V=Q/C
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    (Original post by Stonebridge)
    Because the charge on the first capacitor (Q=CV) is shared between the two when they are joined. Some change moves from it to the other capacitor because the other one is at a lower pd. (zero to start with) This results is less charge on the original and therefore a lower voltage. V=Q/C
    Thanks, makes sense. Is it always the electrons moving opposite to the direction of the electrons when the capacitor was first charged up, after the capacitor is put in parallel with the other capacitor? So no electrons actually pass the dielectric?

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    The electrons shouldn't pass through the dielectric unless it is "leaky". (This won't happen at A Level)
    They just flow along the connecting wires from the original capacitor to the second one until the pd is the same across both.
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    (Original post by Stonebridge)
    The electrons shouldn't pass through the dielectric unless it is "leaky". (This won't happen at A Level)
    They just flow along the connecting wires from the original capacitor to the second one until the pd is the same across both.
    Righto, thanks mate.
 
 
 
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