I have a 2 mm radius steel ball and I want to charge it in a 0.025 Coulomb charge. Can I do that domestically or the charge is too high? And how can I do that assuming I don't have another conductive material that I know it's charge?
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Not if you value your life!. The capacitance of a sphere is given by :-
\$C=4\pi\epsilon_0R\$
The voltage on a capacitor is given by :-
\$V=\dfrac{Q}{C}\$
Putting your figures into those equations we find that the capacitance of your sphere is about 0.22pF and the voltage required for that much charge would be 110GV!
Edit ...
If you had another conductive object, you could make a two-plate capacitor in which case the required voltage could be much lower, so in theory you could charge a two plate capacitor using a much lower voltage, then move the other plate away and this would leave your original sphere charged, but this charge would soon leak away.
MikeJ-UK
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so if I put 110GV through that ball it's charge would be 0.025 Colon? I soppose that won't be easy... and there is no other practical way? – Someonation Jul 25 '12 at 12:57
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If the steel ball can be close to another conductive object iit can be done. – MikeJ-UK Jul 25 '12 at 13:00
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but that condactive object has to have charge... how do I charge that object? – Someonation Jul 25 '12 at 13:06
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The Van de Graaf generator is the best kind of experiment to show how this works. – MikeJ-UK Jul 25 '12 at 13:10
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but how could I know that I reached the wanted charge? – Someonation Jul 25 '12 at 13:20
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In theory, you can separate the plates of the capacitor, but it would take 900 billion pounds of force (using F=Q^2/eps*A) – Justin Jul 25 '12 at 14:06
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1@Someonation: probably because the steel ball fires off in a random direction at a few dozen kilometers per second. – Bryan Boettcher Jul 25 '12 at 14:26
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1Rule of thumb for having a spark (=discharge) in normal air is about 1kV/mm. So if your experiment is in the open air then you're talking about 110km separation between the two conductors ;o) – jippie Jul 25 '12 at 20:44