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    Two questions about capacitors which I'm not sureo n how to solve

    1)A capacitor with a capacitance of 1.77 x 10^-10 F is charged to 10 V. It is then disconnected from the power supply and the air gap is increased to 1mm. Determine the new potential across the plates.

    (it is a parallel plate capacitor)

    previous air gap was 0.5mm and ε0 = (8.85 x 10^-12 Fm^-1) and the area is 10cm x 10cm

    2) In the circuit below, the switch is thrown upwards to the the 1 μF capacitor and is then thrown downwards. Determine the final potential difference between A and B.

    (note I forgot to draw on A and B, these two points are on the bottom 2 corners of the circuit diagram, either side of the 5 μF capacitor)

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    (Original post by luke90)
    2) In the circuit below, the switch is thrown upwards to the the 1 μF capacitor and is then thrown downwards. Determine the final potential difference between A and B.
    The p.d. would be 10V. However, the charge wouldn't be stored evenly between the capacitors.

    (Original post by luke90)
    1)A capacitor with a capacitance of 1.77 x 10^-10 F is charged to 10 V. It is then disconnected from the power supply and the air gap is increased to 1mm. Determine the new potential across the plates.

    (it is a parallel plate capacitor)

    previous air gap was 0.5mm and ε0 = (8.85 x 10^-12 Fm^-1) and the area is 10cm x 10cm
    Well I'm not sure about this one, but as the separation has doubled, the p.d. might have doubled (it would definately increase).
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    Starofale is wrong (sorry mate)

    the 1st capacitor gats a certain amount of charge Q = C V

    when the switch is closed this charge is shared between the two capacitors in such a way that the p.d. across each of them is the same.

    so V = Q1 / C1 = Q2 / C2 so Q1/Q2 = C1/C2

    the bigger capacitor therefore gets 5 times more charge

    so it gets 5/6 of the original charge. Knowing this you can now do V = Q/C to find V
 
 
 
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