# Could you please explain why Q=C*V?

Discussion in 'Electronic Basics' started by Boki, Jul 24, 2003.

1. ### BokiGuest

Hi, All:

Could you please explain why Q=C*V?

Thanks.

Boki

3. ### grahamkGuest

The larger the Capacitor is, and the higher the Voltage is, then the larger
the charge( Q) is.

4. ### Jonathan KirwanGuest

That's a dimensional analysis way of looking at it. For a given
capacitor, the higher the voltage the more charge you expect,
so:

q proportional to V

and for a given voltage, the bigger the capacitor is the more
charge you expect, so:

q proportional to c

There don't appear to be other obvious factors at this point,
so:

q proportional to c*v

or

q = k1 * c * v

with 'k1' being yet to be determined.

Or, rearranged, C = k2*Q/V.

Turns out, C is *defined* so that k2 is just 1.000 when Q is in
Coulombs and V is in Joules per Coulomb. Capacitance is in
Coulumb^2/Joule.

Add to this, an Ohm is a Joule-sec/Coulomb^2 (a Joule-sec is a
measure of angular momentum, for example, and is the units of
Planck's constant -- it's also just kg*(m/s)*m.)

Multiply units of R*C:

(Joule-sec/Coulomb^2) * (Coulomb^2/Joule)

and you get seconds! Good thing.

Since a Henry is just a Joule-sec^2/Coulomb^2 (a kind of
Ohm-second or Joule per amp^2), the SQRT(L*C) also works out to
seconds. Also a good thing. And L/R is also in seconds.
Another good thing.

Jon  