I need a simple formula to calculate the total coulombs

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Re: I need a simple formula to calculate the total coulombs

I'll give you a hand:

F = { (V^2) (e0) (er) (A) } / { (2) (X^2) },

where

F = Force, in Newtons
V = Voltage between the plates, in Volts
e0 = Dielectric constant of vacuum (8.854 E-12 farads per meter)
er = Relative dielectric constant (of the medium between the plates)
A = Area of one of the plates, in square meters
X = Distance between the plates, in meters

I'll leave the rest as a homework assignment.
 
Re: I need a simple formula to calculate the total coulombs

C. B., you are making this too hard.

F = 9*10^9*q1*q2*/d^2 (in a vacuum)

Where F is in newtons, q1 and q2 in coulombs, and d is in meters.
 
Re: I need a simple formula to calculate the total coulombs

Originally posted by rattus: C. B., you are making this too hard.
Click to expand...
And you are trying to make it too easy. That formula only applies to the force between two point charges. The question has to do with two charged, parallel, planes.
 
Re: I need a simple formula to calculate the total coulombs

Charlie, I am responding to a post by Sam who quotes Bryan, and he is asking for the force between two electrons. No mention of planes.
 
Re: I need a simple formula to calculate the total coulombs

Sorry. You're right. I was still thinking about capacitors. Mea culpa. :(
 
Re: I need a simple formula to calculate the total coulombs

I looked it up later the other night and I realized that Charlie had already posted Coulomb's law, or a form of it.

I wasn't even close. :D

F = kQQ'/d^2

k = 8.93 x 10^9 in air
k = 8.987 x 10^9 in space

an electron (or both Q and Q' in this case) = 1.6 x 10^-19

d = 10^-3

F = (8.93 x 10^9) x (1.6 x 10^-19)^2 / 10^-6 in Newtons, in air.

I don't have my scientific handy so I'll avoid embarrasing myself further be screwing up the exponents and leave the final computation omitted.

And now I've just read your posts Charlie and Rattus. I kind of do things backwards. :D Rattus is right, if he's allowed to round.

Edit: I have to pick on you a little Rattus, If you're gonna use k = 9 x 10^9 then you hardly need to specify whether it's in air or free space. The difference in those two values is less than the error from rounding. :p

[ October 21, 2005, 05:24 PM: Message edited by: Physis 2 ]
 
Re: I need a simple formula to calculate the total coulombs

Sam,

Now tell us about Milliken's oil drop experiment.
 
Re: I need a simple formula to calculate the total coulombs

I'd rather discuss Michaelson Morely (I hope I haven't butchered their names too badly).

A real land mark experiment. :cool:
 
Re: I need a simple formula to calculate the total coulombs

They used an observatory in LA.

Wwhhaatt'ss the name of that place??? :confused:
 
Re: I need a simple formula to calculate the total coulombs

I thought it was Griffith observatory?
 
Re: I need a simple formula to calculate the total coulombs

Palomar is associated with Caltech. You are the native son, you ought to know which is which. I am just an ignernt Texan.
 
Re: I need a simple formula to calculate the total coulombs

Not Gabrial's horn now. :D

There's no such thing as one dimentional paint! :p
 
Re: I need a simple formula to calculate the total coulombs

Bradlee -

You almost have all you need.

P2 is right, Q=CV. And Steve and P2 are dead on about the actual capacitance value.

But you can get a good measurement of the actual C value using:

i = C(dV/dT)

Set up a constant current source, say 50 to 100ma. You should be able to do this with a couple of small transistors, a zener, a couple of resistors and a 9V battery. Look up a circuit called a "current mirror".

Hook up the current source to the cap, put an ma meter is series, connect a voltmeter across the cap. Turn on the current source. The cap voltage should climb smoothly (linear). Time (in seconds)the change between say 1V to 6V, or 5V Delta V.

So C = I X (5V)/Time
should give a pretty accurate picture of the capacitance, depending on your meters. Measuring I to +/- 0.1ma, V to +/- 0.01V, and T to +/- 1.0 Sec ($50 DVMs and a $10 digital watch) you should be within 1% of the actual value.

Assuming a nominal 3f and 50ma, the time would be about 300 seconds - 6 minutes.

That number for C, and Q = CV will get you the numbers you asked for.

The current mirror project should be interesting. I build one using a 10 turn pot with a 10 turn dial for the variable, and a 9v battery, for a 4-20ma source or sink. For $6 worth of parts it worked great - as good as a $1000 dollar Transmation (if you didn't need better than 1% :)

carl
 
Re: I need a simple formula to calculate the total coulombs

Carl, it is even simpler than that. With your current source, one can compute charge by,

Q = I*T

Of course, the voltage must not exceed the working range of the current source, and you had better not short a 3Farad cap either.
 
Re: I need a simple formula to calculate the total coulombs

Originally posted by rattus:
Carl, it is even simpler than that. With your current source, one can compute charge by,

Q = I*T
Click to expand...
Well, not exactly. Their test circuit likely does not allow them to easily integrate the current over time. However, having an accurate measure of the capacitance and checking the cap voltage before and after their test will give an accurate number for the change in charge.

Originally posted by rattus:
...Of course, the voltage must not exceed the working range of the current source ...
Click to expand...
Yes, that's why I suggested a 9V battery.

Originally posted by rattus:
.. and you had better not short a 3Farad cap either.
Click to expand...
I don't know. Can't say I ever looked at what would happen. Let's see, stored energy is 1/2CV^2

So, at 6V and 3f, E = 108W-sec. How much energy is that? A flashlight battery, or a car battery?

carl

edited to get rid of some editing fluff

[ October 25, 2005, 07:20 PM: Message edited by: coulter ]
 
Re: I need a simple formula to calculate the total coulombs

Carl,

Your current source provides a constant current. You don't need any special equipment--just an ammeter and a timer. Of course you would monitor the voltage as well--with a hi-Z voltmeter.

Just charge the cap to the desired voltage then stop the timer and disconnect the charging current.

My point is that the answer to the original question is simply,

Q = I*T

where I is a constant amplitude pulse and T is the width of the pulse.
 
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