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# capacitors watch

1. can somebody explain the difference between these two formulas?

(check the attatchment)
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2. Don't take my word for it, but i think that the work done by the supply is QV, where as the energy stored on the capacitor is only half of this; 1/2QV

The other half of the energy is 'lost' due to heating effect etc.
3. If you had a constant charge, rather than an increasing one, the graph would be a horizontal line, and so the area underneath it would be a rectangle. This would then have the area of QV rather than half QV.

They are both the same formula, just talking about different situations. I think the only one you'd need is the 1/2QV one, because the charge on a capactitor always increases as the energy stored increases.
4. thanks, but that definatly isnt the answer. you have any other clues?

edit:at hanratarsed, i think you got it thanks
5. i just think of it graphically: since area underneath the graph is energy, if the area is of a triangle shape, then it must be 1/2 QV to give work done.

btw, i'ave got the same book as u have, page 17 lol
6. (Original post by jez06)
Don't take my word for it, but i think that the work done by the supply is QV, where as the energy stored on the capacitor is only half of this; 1/2QV

The other half of the energy is 'lost' due to heating effect etc.
Can someone tell me if this is 100% correct??
7. No. Some energy will be lost, but its unlikeley to be half of the total energy stored - that would be a very inefficient capacitor.
8. (Original post by popop12345)
Can someone tell me if this is 100% correct??
yes it is.

It turns out that the amount of energy disippated as heat in the wires is independent of the reisstance of the wires.

Its fun to use the expeonential equation for the current and integrate I sqaured R wrt to time. Almost magically the energy is 0.5QV.

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Updated: June 16, 2006
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