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Capacitor and Force

J

Jon Slaughter

Jan 1, 1970
0
I'm trying out a little experiment but its not working ;/

Since

Q = C*V

and

F = k*Q1*Q2/r^2,

The force one parallel plate capacitor due to the charge is

F = k*(CV)^2/r^2

So for a ceramic capacitor of 20nF with V = 20V, F ~= 3.6N. (assuming r ~=
1mm)

I've tried this with two pennies and a piece of paper(as the
dielectric/insulator) which gives about 20pF and about 1nF(theoretically and
to small for me to measure).

I grinded down one side of 40nF ceramic plate capacitor and a penny and it
has a total of 40nF but I experience no force when moving the plates close
together(which is surely < 1mm). But why? Surely since they act as a 40nF
capacitor and there is 20V across is then there should be a significant
force between the two. A large enough force to feel at when trying to
seperate the two?

But this isn't the case so I must be wrong either in theory or application.
The theory is pretty straight forward and even if my calulations are off by
1000 I probably should feel some force between the two but as far as I can
tell there is nothing. Not sure about the application either as it also is
pretty straight forward(the main problem is getting a large enough
capacitance in a simple way).

Any ideas where I went wrong?

Thanks,
Jon
 
J

Jim Thompson

Jan 1, 1970
0
I'm trying out a little experiment but its not working ;/

Since

Q = C*V

and

F = k*Q1*Q2/r^2,

The force one parallel plate capacitor due to the charge is

F = k*(CV)^2/r^2

So for a ceramic capacitor of 20nF with V = 20V, F ~= 3.6N. (assuming r ~=
1mm)

I've tried this with two pennies and a piece of paper(as the
dielectric/insulator) which gives about 20pF and about 1nF(theoretically and
to small for me to measure).

I grinded down one side of 40nF ceramic plate capacitor and a penny and it
has a total of 40nF but I experience no force when moving the plates close
together(which is surely < 1mm). But why? Surely since they act as a 40nF
capacitor and there is 20V across is then there should be a significant
force between the two. A large enough force to feel at when trying to
seperate the two?

But this isn't the case so I must be wrong either in theory or application.
The theory is pretty straight forward and even if my calulations are off by
1000 I probably should feel some force between the two but as far as I can
tell there is nothing. Not sure about the application either as it also is
pretty straight forward(the main problem is getting a large enough
capacitance in a simple way).

Any ideas where I went wrong?

Thanks,
Jon

Trying to remember some fundamentals...

Isn't Force = dE/dx

change of energy with spacing?

...Jim Thompson
 
B

Bruce Varley

Jan 1, 1970
0
Jon Slaughter said:
I'm trying out a little experiment but its not working ;/

Since

Q = C*V

and

F = k*Q1*Q2/r^2,

The force one parallel plate capacitor due to the charge is

F = k*(CV)^2/r^2

So for a ceramic capacitor of 20nF with V = 20V, F ~= 3.6N. (assuming r ~=
1mm)

I've tried this with two pennies and a piece of paper(as the
dielectric/insulator) which gives about 20pF and about 1nF(theoretically
and to small for me to measure).

I grinded down one side of 40nF ceramic plate capacitor and a penny and it
has a total of 40nF but I experience no force when moving the plates close
together(which is surely < 1mm). But why? Surely since they act as a 40nF
capacitor and there is 20V across is then there should be a significant
force between the two. A large enough force to feel at when trying to
seperate the two?

But this isn't the case so I must be wrong either in theory or
application. The theory is pretty straight forward and even if my
calulations are off by 1000 I probably should feel some force between the
two but as far as I can tell there is nothing. Not sure about the
application either as it also is pretty straight forward(the main problem
is getting a large enough capacitance in a simple way).

Any ideas where I went wrong?

Thanks,
Jon
What's your value of K? I can't recall exactly what it would be, but it
would include 'epsilon-0', an electrostatic constant, which is very small,
somewhere around 1.0E-07 IIRC. Electrostatic forces are weak. You can get a
feel for the situation by doing one of the classic electrostatic experiments
like picking up tiny pieces of lint with a charged plastic or glass rod. The
voltages involved are kilovolts.
 
I'm trying out a little experiment but its not working ;/

Since

Q = C*V

and

F = k*Q1*Q2/r^2,

The force one parallel plate capacitor due to the charge is

F = k*(CV)^2/r^2

So for a ceramic capacitor of 20nF with V = 20V, F ~= 3.6N. (assuming r ~=
1mm)

I've tried this with two pennies and a piece of paper(as the
dielectric/insulator) which gives about 20pF and about 1nF(theoretically and
to small for me to measure).

I grinded down one side of 40nF ceramic plate capacitor and a penny and it
has a total of 40nF but I experience no force when moving the plates close
together(which is surely < 1mm). But why? Surely since they act as a 40nF
capacitor and there is 20V across is then there should be a significant
force between the two. A large enough force to feel at when trying to
seperate the two?

But this isn't the case so I must be wrong either in theory or application.
The theory is pretty straight forward and even if my calulations are off by
1000 I probably should feel some force between the two but as far as I can
tell there is nothing. Not sure about the application either as it also is
pretty straight forward(the main problem is getting a large enough
capacitance in a simple way).

I don't think that the force between two point charges is of much
relevance to the force between the two parallel plates of a capacitor.

An easier way of getting there is to note that the energy stored in a
capacitor is 0,5*C*V^2, or 8 microjoules for a 40nF capacitor.

40nF is quite high for an air-gapped capacitor with circular plates
1cm in diameter - that is only 8*10^-5 square metres. The permitivity
of free space is about 9*10^-12 F/m, so your spacing would be about
0.1 micron.

Energy is force times distance, and doubling the gap to 0.2 micron
would halve the capacitance and halve the stored energy, which gives
an attractive force of the order of ten newtons - about one kilogram
weight - but only over about 0.1 micron. Obviously the force is going
to decrease very rapidly as the gap gets larger ...

Hope this helps.
 
T

The Phantom

Jan 1, 1970
0
I'm trying out a little experiment but its not working ;/

Since

Q = C*V

and

F = k*Q1*Q2/r^2,

The force one parallel plate capacitor due to the charge is

F = k*(CV)^2/r^2

So for a ceramic capacitor of 20nF with V = 20V, F ~= 3.6N.

The (CV)^2 term alone has a value of 16*10^(-14), so the force can't be
anywhere near 3.6N
 
M

Martin Griffith

Jan 1, 1970
0
I'm trying out a little experiment but its not working ;/

Since

Q = C*V

and

F = k*Q1*Q2/r^2,

The force one parallel plate capacitor due to the charge is

F = k*(CV)^2/r^2

So for a ceramic capacitor of 20nF with V = 20V, F ~= 3.6N. (assuming r ~=
1mm)

I've tried this with two pennies and a piece of paper(as the
dielectric/insulator) which gives about 20pF and about 1nF(theoretically and
to small for me to measure).

I grinded down one side of 40nF ceramic plate capacitor and a penny and it
has a total of 40nF but I experience no force when moving the plates close
together(which is surely < 1mm). But why? Surely since they act as a 40nF
capacitor and there is 20V across is then there should be a significant
force between the two. A large enough force to feel at when trying to
seperate the two?

But this isn't the case so I must be wrong either in theory or application.
The theory is pretty straight forward and even if my calulations are off by
1000 I probably should feel some force between the two but as far as I can
tell there is nothing. Not sure about the application either as it also is
pretty straight forward(the main problem is getting a large enough
capacitance in a simple way).

Any ideas where I went wrong?

Thanks,
Jon
I listen to my capacitors

http://www.audioreview.com/cat/speakers/floorstanding-speakers/quad/PRD_120487_1594crx.aspx


Martin
 
T

Tim Wescott

Jan 1, 1970
0
I'm trying out a little experiment but its not working ;/

Since

Q = C*V

and

F = k*Q1*Q2/r^2,

The force one parallel plate capacitor due to the charge is

F = k*(CV)^2/r^2

So for a ceramic capacitor of 20nF with V = 20V, F ~= 3.6N. (assuming r ~=
1mm)

I've tried this with two pennies and a piece of paper(as the
dielectric/insulator) which gives about 20pF and about 1nF(theoretically and
to small for me to measure).

I grinded down one side of 40nF ceramic plate capacitor and a penny and it
has a total of 40nF but I experience no force when moving the plates close
together(which is surely < 1mm). But why? Surely since they act as a 40nF
capacitor and there is 20V across is then there should be a significant
force between the two. A large enough force to feel at when trying to
seperate the two?

But this isn't the case so I must be wrong either in theory or application.
The theory is pretty straight forward and even if my calulations are off by
1000 I probably should feel some force between the two but as far as I can
tell there is nothing. Not sure about the application either as it also is
pretty straight forward(the main problem is getting a large enough
capacitance in a simple way).

Any ideas where I went wrong?

Thanks,
Jon

Look at Jim Thompson's comment about calculating force.

For a reality check, consider that when you do electrostatics experiments
you're happy to lift little bitty styrofoam balls with multi-kV of static
electricity -- what in heck are you going to manage to do with a measly
20V?

--
Tim Wescott
Control systems and communications consulting
http://www.wescottdesign.com

Need to learn how to apply control theory in your embedded system?
"Applied Control Theory for Embedded Systems" by Tim Wescott
Elsevier/Newnes, http://www.wescottdesign.com/actfes/actfes.html
 
J

Jon Slaughter

Jan 1, 1970
0
Jim Thompson said:
Trying to remember some fundamentals...

Isn't Force = dE/dx

change of energy with spacing?

lol. No. F = q*E. E = grad(V) which is probably what you were thinking.
 
J

Jon Slaughter

Jan 1, 1970
0
Bruce Varley said:
What's your value of K? I can't recall exactly what it would be, but it
would include 'epsilon-0', an electrostatic constant, which is very small,
somewhere around 1.0E-07 IIRC. Electrostatic forces are weak. You can get
a feel for the situation by doing one of the classic electrostatic
experiments like picking up tiny pieces of lint with a charged plastic or
glass rod. The voltages involved are kilovolts.

k = 9*10^9 and is actually very large(it does have e in it but in the
denominator.. I think its 1/(4pie_0) or something).

k is what makes the electrostatic force so strong compared with the
gravitational force and the gravitational constant G.

Electrostatic forces in your example are no weak... unless you are comparing
it to the nuclear forces or something like that. You have to realize that in
your example the amount of charge is extremely small.

http://musr.physics.ubc.ca/~jess/hr/skept/E_M/node2.html

"the gravitational attraction between the two protons is roughly a trillion
trillion trillion times weaker than the electrostatic repulsion."
 
J

Jon Slaughter

Jan 1, 1970
0
I don't think that the force between two point charges is of much
relevance to the force between the two parallel plates of a capacitor.

Its relevant for an order approximation. Each plate has a charge Q on it.
Its a fact of like that columbs law works so since we have two charges,
equal and opposite in this case, we have -kQ^2/r^2 for the force. That force
must exist. No way around it unless somehow it is getting neuralized but I
don't see how.
An easier way of getting there is to note that the energy stored in a
capacitor is 0,5*C*V^2, or 8 microjoules for a 40nF capacitor.

If I do the same calculations I will arrive at the same answer...

C*V^2/
40nF is quite high for an air-gapped capacitor with circular plates
1cm in diameter - that is only 8*10^-5 square metres. The permitivity
of free space is about 9*10^-12 F/m, so your spacing would be about
0.1 micron.

Energy is force times distance, and doubling the gap to 0.2 micron
would halve the capacitance and halve the stored energy, which gives
an attractive force of the order of ten newtons - about one kilogram
weight - but only over about 0.1 micron. Obviously the force is going
to decrease very rapidly as the gap gets larger ...

Well, I was using the dielectric in the ceramic cap. It doesn't quite matter
what the capacitance is cause I measured it and its quite strong. So even if
I cannot measure r I can still measure the capacitance. Sure my F might be
off but I'm pretty sure its within < 1mm.
 
J

Jon Slaughter

Jan 1, 1970
0
Tim Wescott said:
Look at Jim Thompson's comment about calculating force.

For a reality check, consider that when you do electrostatics experiments
you're happy to lift little bitty styrofoam balls with multi-kV of static
electricity -- what in heck are you going to manage to do with a measly
20V?

Jesus christ... I thought you guys were like MIT grads or something?

Do you realize that the charges involved in "electrostatics" are on the
order of nC's if your lucky?

"Two balloons are charged with an identical quantity and type of
charge: -6.25 nC"

"The force of repulsion of two +1.00 Coulomb charges held 1.00 meter apart
is 9 billion Newtons."

Seeing that the parallel plate capacitors have a charge of ~1000 times
implies the force is 1M times larger(since it depends on Q^2).
 
J

Jon Slaughter

Jan 1, 1970
0
Jon Slaughter said:
Its relevant for an order approximation. Each plate has a charge Q on it.
Its a fact of like that columbs law works so since we have two charges,
equal and opposite in this case, we have -kQ^2/r^2 for the force. That
force must exist. No way around it unless somehow it is getting neuralized
but I don't see how.

Well, I should get the same answer either way but I don't... I'll have to
see whats going on later.



Its not air gapped as its a ceramic cap. The dielectric has a width of about
1mm as I broke one open and looked.

Of course cause its 1/r^2.

Even with simple electrostatics with charges much smaller one can see large
forces. The voltages are much higher than this but voltage has nothing to
do with it since we know the charge(of course the voltage is related to the
force)

I don't know. Maybe my logic is flawed somewhere. (obviously something is
flawed cause I'm not getting the proper results)
 
A

Adrian Jansen

Jan 1, 1970
0
Jon said:
Jesus christ... I thought you guys were like MIT grads or something?

Do you realize that the charges involved in "electrostatics" are on the
order of nC's if your lucky?

"Two balloons are charged with an identical quantity and type of
charge: -6.25 nC"

"The force of repulsion of two +1.00 Coulomb charges held 1.00 meter apart
is 9 billion Newtons."

Seeing that the parallel plate capacitors have a charge of ~1000 times
implies the force is 1M times larger(since it depends on Q^2).
If you are putting external electrodes around a piece of hi K dielectric
material, it seems to me the charge is stored inside the material, not
in the airgaps. So when you move the electrodes, you are not changing (
much ) the charge stored in the dielectric, so no force. You would have
to expand the dielectric itself to get a measurable force.

--
Regards,

Adrian Jansen adrianjansen at internode dot on dot net
Design Engineer J & K Micro Systems
Microcomputer solutions for industrial control
Note reply address is invalid, convert address above to machine form.
 
J

Jon Slaughter

Jan 1, 1970
0
Adrian Jansen said:
If you are putting external electrodes around a piece of hi K dielectric
material, it seems to me the charge is stored inside the material, not in
the airgaps. So when you move the electrodes, you are not changing (
much ) the charge stored in the dielectric, so no force. You would have
to expand the dielectric itself to get a measurable force.

That might be true. My calculations are assuming that the dielectric is
constant for all distances and thats not true.

But, charge is charge and that its what matters in calculating the force.
There is no question about that since F = k*q1*q2/r^2.

If I were to charge the plates up when the distance is minimal(i.e., when
the capacitance is actually high) then I will get the proper charge that I'm
using. Now when I move the plates apart the capacitance rapidly decreases
decreasing the charge and hence decreasing the force.

BUT, If the charge is fixed on the plates(so the power source is removed)
then there is no way it can change for any distance and the force must
therefor be constant too.

So potentially what is happening when I disconnect the power supply is that
all the charge is leaking away to fast for me to measure any force. This was
my initial idea but I don't know if its correct or not.

e.g., take a very large plate capacitor with capacitance C. That is, you fix
some distance r and dielectric and measure C. Charge the plates and
disconnect the power source. F = k*(C*V)^2/r and that is a fact. Now if Q
is drastically decreasing then F is too and that is the only real problem
because if Q is fixed then F depends on r which I have control of.

Chances are its cause the charge is leaking off though but again, this is
somewhat of a guess. I guess I would have to make much larger plates and
somehow reduce the leakage.

Jon
 
J

john jardine

Jan 1, 1970
0
Jon Slaughter said:
I'm trying out a little experiment but its not working ;/

Since

Q = C*V

and

F = k*Q1*Q2/r^2,

The force one parallel plate capacitor due to the charge is

F = k*(CV)^2/r^2

So for a ceramic capacitor of 20nF with V = 20V, F ~= 3.6N. (assuming r ~=
1mm)

I've tried this with two pennies and a piece of paper(as the
dielectric/insulator) which gives about 20pF and about 1nF(theoretically and
to small for me to measure).

I grinded down one side of 40nF ceramic plate capacitor and a penny and it
has a total of 40nF but I experience no force when moving the plates close
together(which is surely < 1mm). But why? Surely since they act as a 40nF
capacitor and there is 20V across is then there should be a significant
force between the two. A large enough force to feel at when trying to
seperate the two?

But this isn't the case so I must be wrong either in theory or application.
The theory is pretty straight forward and even if my calulations are off by
1000 I probably should feel some force between the two but as far as I can
tell there is nothing. Not sure about the application either as it also is
pretty straight forward(the main problem is getting a large enough
capacitance in a simple way).

Any ideas where I went wrong?

Thanks,
Jon
The formula I have is,
Newtons attract = [8.856e-12 x relative permeability x Area in mtr^2 x
Voltage^2] / [2 x gap in mtrs]
Would seem my 10n 400V multilayer poly cap has 1/4 ounce squeezing it
together at max voltage.
Maybe this is the source of dielectric absorbsion (sp?). E.g. the more rigid
dielectrics will 'spring' together under a voltage strain and relax at
leisure.
Suggests a dielectric with similar cold flow properties to putty would be
ideal. I.e once the plates have been squeezed together a few times then no
more compaction can take place.
 
J

Jon Slaughter

Jan 1, 1970
0
The formula I have is,
Newtons attract = [8.856e-12 x relative permeability x Area in mtr^2 x
Voltage^2] / [2 x gap in mtrs]
Would seem my 10n 400V multilayer

huh? e*A*V^2/2/d?

e*A/d is the capacitance and so you have C*V^2/2 which is the energy stored
in the electric field. This is not the same as the force but W = F*d so one
should theoretically be able to calculate the force due to the energy but
I'm not sure if this is the same as coulumbs force as it is the force
required to move one charge against the electric field.


Coulumbs law is quite simple and is analogous to the gravitational field. F
= k*q1*q2/r^2. The charge q1 and q2 are the same in a parallel plate
capacitor and can be found if the capacitance and applied voltage is known
since Q = C*V.

Its actually quite simple

Q = CV
F = k*Q^2/r^2

C and V are measured and hence F can be found.
 
T

Tim Wescott

Jan 1, 1970
0
Jesus christ... I thought you guys were like MIT grads or something?

By that logic _you_ are an MIT grad -- so what was it like? I graduated
from WPI where we actually did things.
Do you realize that the charges involved in "electrostatics" are on the
order of nC's if your lucky?

"Two balloons are charged with an identical quantity and type of charge:
-6.25 nC"

"The force of repulsion of two +1.00 Coulomb charges held 1.00 meter
apart is 9 billion Newtons."

Since you're obviously quoting, why aren't you putting in your sources?

By your own math, with C = 20pF and V = 20V, you get q = 400pC. That's
less than a nano-Coulomb.

Assuming your paper was 5 mils thick you get (9e9)(400e-12)^2/(127e-6)^3 =
144e-5 = 89 mN, or 1/3 of an ounce. Frankly, I don't believe this given
my reality check; I'm going to have to think of a way to test it.
Seeing that the parallel plate capacitors have a charge of ~1000 times
implies the force is 1M times larger(since it depends on Q^2).



--
Tim Wescott
Control systems and communications consulting
http://www.wescottdesign.com

Need to learn how to apply control theory in your embedded system?
"Applied Control Theory for Embedded Systems" by Tim Wescott
Elsevier/Newnes, http://www.wescottdesign.com/actfes/actfes.html
 
J

Jon Slaughter

Jan 1, 1970
0
Tim Wescott said:
By that logic _you_ are an MIT grad -- so what was it like? I graduated
from WPI where we actually did things.

huh? Are you serious?
Since you're obviously quoting, why aren't you putting in your sources?

Find them yourself... You don't seem to believe the logic of coulumbs law so
want can I say?
By your own math, with C = 20pF and V = 20V, you get q = 400pC. That's
less than a nano-Coulomb.


"So for a ceramic capacitor of 20nF"

Jesus christ... your off by a factor of 1000.

Remember, I did two experiements. One with pennies which as about 20pF and
one with a ceramic cap which was 40nF. Theres a huge difference and this is
what I said the first time but you obviously didn't read it.

To be clear just so you don't keep going down the wrong path

C = 40nF!!!!!!!!!!!!!!!!!!!!!!!
V = 20V!!!!!!!!!!!!!!!!!!!!!!!


NOW COMPUTE Q AND TELL ME WHAT YOU GET!!!

NOW COMPUTE F AND TELL ME WHAT YOU GET!!!

I bet you will be supprised.

Assuming your paper was 5 mils thick you get (9e9)(400e-12)^2/(127e-6)^3 =
144e-5 = 89 mN, or 1/3 of an ounce. Frankly, I don't believe this given
my reality check; I'm going to have to think of a way to test it.

Yes, you are correct... for C = 20pF. This is similar to the calculation of
got with two pennies and a piece of paper. It is insiginificant. That is why
I used half of a ceramic capacitor which has C = 40nF which is over 1000
times more. Now do the the calculation and see what you get. I bet you will
be astonished by the force... its not weak is it? Your first post is total
BS about how weak the force is.

Remember, that force is for V = 20 and not those voltages normally
associated with static electricity. Even if you take the distance much
larger, since I have a dielectric in the way, say its 2mm, then the force is
still > 1

Seriously. Compute the force for C = 20pF and C = 40nF for r = 1mm and note
the HUGE difference.(or 10pF and 10nF) Since F depends on Q^2 the
difference in C is magnified by the square.

So if we have C1 = 20pF and C2 = 20nF = 1000*C1, that factor of 1000 becomes
1000000. A huge difference. That is why I used the ceramic caps in the the
first place so I could get a much stronger force.

I suppose maybe you should go back to WPI and ask for a refund?

It amazes me how much you guys bitch about my grammar and spelling but don't
understand basic physics ;/

I mean, the first comment you made about how weak the electric force is just
pure idiocity. I'm sure 8th grade physics mentions just how strong it is.
Hell, most people only know 2 forces and so either you have gravitational
force or you have electric force and and the electric force wins hands down.
Of course maybe you had some other force in mind? Maybe the strong nuclear
force?


In any case the force for the 40nF cap is large enough to observe so there
is another problem. I imagine that the problem is actually keeping the
charge on the caps. Either its leaking(but it must be doing it at the speed
of light) or the charge is flowing back into the power supply when I
disconnect it from the cap. (this is probably understandable since even
diodes probably won't stop the flow of this small amount of charge)

I suppose I could try to electrostatically charge the caps and maybe that
will work but I don't know since the plates are conductors.
 
T

Tim Wescott

Jan 1, 1970
0
huh? Are you serious?


Find them yourself... You don't seem to believe the logic of coulumbs law so
want can I say?



"So for a ceramic capacitor of 20nF"

Jesus christ... your off by a factor of 1000.

Remember, I did two experiements. One with pennies which as about 20pF and
one with a ceramic cap which was 40nF. Theres a huge difference and this is
what I said the first time but you obviously didn't read it.

To be clear just so you don't keep going down the wrong path

C = 40nF!!!!!!!!!!!!!!!!!!!!!!!
V = 20V!!!!!!!!!!!!!!!!!!!!!!!


NOW COMPUTE Q AND TELL ME WHAT YOU GET!!!

NOW COMPUTE F AND TELL ME WHAT YOU GET!!!

I bet you will be supprised.



Yes, you are correct... for C = 20pF. This is similar to the calculation of
got with two pennies and a piece of paper. It is insiginificant. That is why
I used half of a ceramic capacitor which has C = 40nF which is over 1000
times more. Now do the the calculation and see what you get. I bet you will
be astonished by the force... its not weak is it? Your first post is total
BS about how weak the force is.

Remember, that force is for V = 20 and not those voltages normally
associated with static electricity. Even if you take the distance much
larger, since I have a dielectric in the way, say its 2mm, then the force is
still > 1

Seriously. Compute the force for C = 20pF and C = 40nF for r = 1mm and note
the HUGE difference.(or 10pF and 10nF) Since F depends on Q^2 the
difference in C is magnified by the square.

So if we have C1 = 20pF and C2 = 20nF = 1000*C1, that factor of 1000 becomes
1000000. A huge difference. That is why I used the ceramic caps in the the
first place so I could get a much stronger force.

I suppose maybe you should go back to WPI and ask for a refund?

It amazes me how much you guys bitch about my grammar and spelling but don't
understand basic physics ;/

I mean, the first comment you made about how weak the electric force is just
pure idiocity. I'm sure 8th grade physics mentions just how strong it is.
Hell, most people only know 2 forces and so either you have gravitational
force or you have electric force and and the electric force wins hands down.
Of course maybe you had some other force in mind? Maybe the strong nuclear
force?


In any case the force for the 40nF cap is large enough to observe so there
is another problem. I imagine that the problem is actually keeping the
charge on the caps. Either its leaking(but it must be doing it at the speed
of light) or the charge is flowing back into the power supply when I
disconnect it from the cap. (this is probably understandable since even
diodes probably won't stop the flow of this small amount of charge)

I suppose I could try to electrostatically charge the caps and maybe that
will work but I don't know since the plates are conductors.

I'm sorry, you clearly live so much closer to God than I do than nothing I
could say would be acceptable.

Plonk.

--
Tim Wescott
Control systems and communications consulting
http://www.wescottdesign.com

Need to learn how to apply control theory in your embedded system?
"Applied Control Theory for Embedded Systems" by Tim Wescott
Elsevier/Newnes, http://www.wescottdesign.com/actfes/actfes.html
 
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