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4hv.org :: Forums :: General Science and Electronics
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Capacitor theory

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IamSmooth
Mon Mar 13 2006, 07:43PM Print
IamSmooth Registered Member #190 Joined: Fri Feb 17 2006, 12:00AM
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Posts: 1567
Is the relationship V*C = Q an emperically derived relationship or is there a theoretical basis? Does anyone know the theory, proof or a link-reference?
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HV Enthusiast
Mon Mar 13 2006, 08:27PM
HV Enthusiast Registered Member #15 Joined: Thu Feb 02 2006, 01:11PM
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Posts: 3068
There are several approaches to take to derive this equation but it usually involves solving some nasty differential equations in the end. Usually Laplace's equation (or similar) to calculate the voltage potential in a region surrounding two plates. From the potential, you can then derive e-field and from there charge. Finally, you can solve for capacitance.
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IamSmooth
Mon Mar 13 2006, 08:38PM
IamSmooth Registered Member #190 Joined: Fri Feb 17 2006, 12:00AM
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Posts: 1567
My question is not how to solve for the capacitance of an arbitary geometry. My question is how was the basic relationship C*V=Q discovered?
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Madgyver
Mon Mar 13 2006, 10:02PM
Madgyver Registered Member #177 Joined: Wed Feb 15 2006, 02:16PM
Location: Munich, Germany
Posts: 214
Nothing nasty.

Its the basic definiton for Farads. Look for SI unit conversion tables.
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Sulaiman
Mon Mar 13 2006, 10:06PM
Sulaiman Registered Member #162 Joined: Mon Feb 13 2006, 10:25AM
Location: United Kingdom
Posts: 3141
As far as I'm aware, the equation should be C=Q/V
Capacitance is defined by this formula (empirically)
Q -charge, is a fundamental unit
V -voltage is a derived unit (one volt requires one Joule to move one Coulomb)
All other capacitor equations come from this empirical definition.
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Simon
Tue Mar 14 2006, 12:21AM
Simon Registered Member #32 Joined: Sat Feb 04 2006, 08:58AM
Location: Australia
Posts: 549
V=Q/C and Q=VC are completely equivalent ways of writing the same thing. Some people find Q=VC more elegant and others V=Q/C more meaningful.

If you have two unlike charges you have an electric field between them, hence voltage, since voltage is just e-field times distance. It so happens that adding more charge increases the voltage, in proportion. So you think, hey, let's call this proportionality constant C (capacitance) for any system.

Working out C analytically from the geometry is where calculus comes in.
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Sulaiman
Tue Mar 14 2006, 02:08AM
Sulaiman Registered Member #162 Joined: Mon Feb 13 2006, 10:25AM
Location: United Kingdom
Posts: 3141
Since the original question was whether or not the equation was empirical,
I presented the equation in the form that shows it's derivation,
Capacitance in Farads can be defined in terms of Volts and Coulombs;
Coulombs and Volts are never defined in terms of Capacitance/Farads

Once you have a derived formula you can manipulate it in all sorts of mathemetical ways to determine a variable of interrest, depending on circumstances/application/information etc.
e.g. Volts are defined in terms of current and resistance.
(One Ohm = kg. m2 · s-3 · A-2 )
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