Pls someone help me understand why this gives the initial reading on a voltmeter..?

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BrainJuice
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Paper: http://pmt.physicsandmathstutor.com/...Capacitors.pdf

Mark scheme: http://pmt.physicsandmathstutor.com/...itors%20MS.pdf

It is question 2cii, I don't understand why the calculation is such. It might be very obvious though I know. Please can you explain what is being found and how the equation helps it.
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lizardlizard
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The defining equation for a capcitor shows that V=Q/C. C is a constant, so you know that voltage is directly proportional to charge. If charge is divided by 1000, voltage will also be divided by 1000 and so the voltage drops from 5000V to 5V.
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Joinedup
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You've got to pay a lot of attention to the text of the question concerning the order in which things happen... tbh if you see a diagram with capacitors and flyleads or switches you should see a warning flag waving.

the sphere alone is charged to 5000V - then it;s disconnected from the PSU and then connected to a capacitor.

The fixed amount of charge the sphere had when it was at 5000V is now distributed to the much larger capacitance of the sphere plus capacitor and you work out the new voltage as shown by Lizard.
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BrainJuice
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Thanks guys/girls.
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