Re: Capacitors, charge, and electrons

On 3 Apr 2005 21:28:02 -0700, grug2005@xxxxxxxxxxx wrote:

>Hi guys,
>
>I'm a CS major, but an EE major at heart :)
>
>I know how to build circuits, I use SPICE, but I am now trying to
>figure out the more physical properties of devices.
>
>As for capacitors, I know how to charge them, discharge them, use them
>for timing circuits, etc.
>
>However, I'd like to know what it actually means at the physical level.
>
>For example:
>
>Let's take a Cap rated at 35V, 100uf.
>
>Let's apply a voltage source to it, say 10V - in other words, charge it
>up :)
>
>Taking another Cap that is exactly the same, and charge it with 20V.
>
>Now, let's take a Cap rated at 35V, but has 1000uf capacity.
>
>We charge with 10V as before to fully charge.
>
>What I would like to know now is what is the difference between all
>three Caps at the physical property level?
>
>For example, the two 100uf caps are charged with different voltage
>levels. What does that mean exactly? Does the one charged with 20V
>have more electrons stored than the one charged with 10V? Or are they
>at different energy levels? What gives?
>
>The one with 1000uf certainly has more 'charge', but again, does the
>mean more electrons?

---
No analogy is perfect, but let's say that instead of capacitors we
have three pitchers which are right circular cylinders: two with
volumes (capacitances) of 100cm^3, one with a volume of 1000cm^3, and
all three with heights (voltage ratings) of 35cm. Let's also say
we've attached identical pressure gauges (voltmeters) which read gage
pressures from zero g/cm² to 35g/cm² to the bottoms of the pitchers,
and that the pitchers are all empty (discharged) and resting on their
bottoms with the gauges reading zero.

Now let's say that we pour a fluid (charge) with a specific gravity of
1g/cm^3 into the first pitcher until its pressure gauge reads 10g/cm²
(10V). Since the density of the fluid is 1g/cm^3, the height of the
fluid column will be 10cm, and the pitcher will be about 28.57% full.

If we now pour enough fluid into the second pitcher to cause its
pressure gauge to read 20g/cm², the height of the fluid column will be
20cm and the pitcher will be about 57.1% full, so the quantity of
fluid (charge) in the pitcher with the gauge reading twice the
pressure (charged to twice the voltage) will be double that of the
other pitcher.

Now let's pour enough fluid into the third pitcher to make its
pressure gauge read 10g/cm². As with the first pitcher, since the
density of the fluid is the same, the fluid column will be 10cm high,
but because the pitcher has a larger capacity it will contain more
fluid. How much more? Since the first pitcher had a capacity of
100cm^3 and the third pitcher a capacity ten times greater than that,
for the same height of column the third will contain ten times as much
fluid as the first.

So, since the quantity (Q)of fluid in the pitcher is going to depend
on the capacity (C) of the pitcher and the pressure (Voltage) exerted
by the fluid, we can write:


Q = CV


Where, for a capacitor,

Q = the quantity of charge in the capacitor in coulombs,
C = the capacitance of the capacitor in farads, and
V = the voltage across the capacitor in volts.

For the first capacitor, then:


Q = C * V = 100E-6F * 10V = 1E-3coul = 1 millicoulomb


for the second:


Q = C * V = 100E-6F * 20V = 2E-3coul = 2 millicoulomb


and for the third:


Q = C * V = 1000E-6F * 10V = 1E-2coul = 10 millicoulombs


Just like the fluid being poured into the pitchers was made out of
molecules, the fluid charging the capacitors will be made out of
electrons, with one coulomb of charge containing about 6.02E18
electrons.

--
John Fields
Professional Circuit Designer
.

Quantcast