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.