C
ChrisGibboGibson
Guest
If we charge an air-spaced cap to 100 volts. Then increase the spacing between
the plates, what happens to the voltage?
Gibbo
the plates, what happens to the voltage?
Gibbo
J
John Popelish
Guest
ChrisGibboGibson wrote:
The opposite charges on the plates cause them to attract each other.
It takes work to pull them apart against this attractive force. The
increase in distance also lowers the capacitance. The electrical
energy stored in the cap is 1/2 * C * V^2. The only way the energy
can go up (because mechanical work was done on the cap, conservation
of energy and all that) in spite of the drop in capacitance is if the
voltage increases.
Another way to visualize this is to have two similar capacitors except
for the difference in spacing. Apply enough voltage across each to
force the same charge to be displaced through each. The one with the
bigger spacing (lower capacitance) will require more voltage to force
the same charge to move. If there were some way to get them so far
apart (or otherwise change the plates) that there was zero capacitance
between them it would take an infinite voltage to move that charge.
--
John Popelish
It goes up. There are several ways to imagine this without math.
The opposite charges on the plates cause them to attract each other.
It takes work to pull them apart against this attractive force. The
increase in distance also lowers the capacitance. The electrical
energy stored in the cap is 1/2 * C * V^2. The only way the energy
can go up (because mechanical work was done on the cap, conservation
of energy and all that) in spite of the drop in capacitance is if the
voltage increases.
Another way to visualize this is to have two similar capacitors except
for the difference in spacing. Apply enough voltage across each to
force the same charge to be displaced through each. The one with the
bigger spacing (lower capacitance) will require more voltage to force
the same charge to move. If there were some way to get them so far
apart (or otherwise change the plates) that there was zero capacitance
between them it would take an infinite voltage to move that charge.
--
John Popelish
P
PN2222A
Guest
"ChrisGibboGibson" <chrisgibbogibson@aol.com> wrote in message
news:20041108202846.06439.00000347@mb-m18.aol.com...
C = epsilon A / D (ignoring fringing effects)
V = D / epsilon A (where epsilon A is a constant)
so V rises directly as D increases
news:20041108202846.06439.00000347@mb-m18.aol.com...
Q = C V
C = epsilon A / D (ignoring fringing effects)
V = D / epsilon A (where epsilon A is a constant)
so V rises directly as D increases
C
ChrisGibboGibson
Guest
Thank you John and PN
Now what, exactly, causes the potential difference between the 2 plates?
Gibbo
Now what, exactly, causes the potential difference between the 2 plates?
Gibbo
J
John Popelish
Guest
ChrisGibboGibson wrote:
this field is spread across.
--
John Popelish
The electric field caused by the separated charges and the distance
this field is spread across.
--
John Popelish
P
PN2222A
Guest
"ChrisGibboGibson" <chrisgibbogibson@aol.com> wrote in message
news:20041108211326.06323.00000175@mb-m26.aol.com...
I'm not sure I understand the question.
When opposite electric charges are separated, the electric field exerts a
force on those charges. That force is the "electromotive force" -- e.g.
the voltage.
ft = 300MHZ minimum.
news:20041108211326.06323.00000175@mb-m26.aol.com...
Um, the electrons?
I'm not sure I understand the question.
When opposite electric charges are separated, the electric field exerts a
force on those charges. That force is the "electromotive force" -- e.g.
the voltage.
ft = 300MHZ minimum.
N
nospam
Guest
chrisgibbogibson@aol.com (ChrisGibboGibson) wrote:
the plates apart.
It goes up, and so does the stored energy to balance the work done pulling
the plates apart.
R
Rich The Philosophizer
Guest
On Tue, 09 Nov 2004 02:13:26 +0000, ChrisGibboGibson wrote:
Cheers!
Rich
Thank you John and PN
Now what, exactly, causes the potential difference between the 2 plates?
Whatever it was you used to charge the cap in the first place.
Cheers!
Rich
B
Ban
Guest
ChrisGibboGibson wrote:
The metal conductor of which the plates consist has free electrons, which
are vibrating between the grid of the nucleusses. In an uncharged state both
plates have the same density of free electrons. A voltage source will "pump"
some of those electrons from one plate to the other until the resulting
electric field is in equilibrium with the applied voltage.
Now one plate has a higher electron density than the other. If you remove
the voltage source, this distribution stays the same. We call this
"potential difference" and the amount of moved electrons "charge". The
situation creates an electrostatic field.
It is similar when you move a mass upwards, the energy is transformed into a
potential energy, which is released when you drop the mass. The weight of
the mass(or the charge Q) stays the same, but the energy(or voltage V)
augments, the higher the mass is moved.
Compare this to the formulas of PN, and you hopefully might have a few more
insights.
--
ciao Ban
Bordighera, Italy
this question would be more appropriate in s.e.b.
The metal conductor of which the plates consist has free electrons, which
are vibrating between the grid of the nucleusses. In an uncharged state both
plates have the same density of free electrons. A voltage source will "pump"
some of those electrons from one plate to the other until the resulting
electric field is in equilibrium with the applied voltage.
Now one plate has a higher electron density than the other. If you remove
the voltage source, this distribution stays the same. We call this
"potential difference" and the amount of moved electrons "charge". The
situation creates an electrostatic field.
It is similar when you move a mass upwards, the energy is transformed into a
potential energy, which is released when you drop the mass. The weight of
the mass(or the charge Q) stays the same, but the energy(or voltage V)
augments, the higher the mass is moved.
Compare this to the formulas of PN, and you hopefully might have a few more
insights.
--
ciao Ban
Bordighera, Italy
R
Robert Baer
Guest
ChrisGibboGibson wrote:
So, decrease C and V goes up.
Seems that has been done before in a rotary fashion to make rather
high voltages.
There may even be a patent somewhere on them thar new-fangled
sparkers...
Lemme see.. Q = C * V
So, decrease C and V goes up.
Seems that has been done before in a rotary fashion to make rather
high voltages.
There may even be a patent somewhere on them thar new-fangled
sparkers...
A
Active8
Guest
On Tue, 09 Nov 2004 02:44:17 GMT, PN2222A wrote:
Joules/Coulomb.
--
Best Regards,
Mike
needed to move the charge divided by the amount of charge,
Joules/Coulomb.
--
Best Regards,
Mike
C
CBarn24050
Guest
This is completely wrong Ban, it's hard to imagine a more fundamental
missunderstanding.
B
Ban
Guest
CBarn24050 wrote:
I do not know what exactly you refer to, but the gravitational field and the
electrostatic field have a lot in common. Look at Newtons Law of
Gravitation:
F = (-G* m* M)/r^2 F=force G=const. m= mass1 M= mass2
and Coulombs Law:
F = (k* q* Q)/r^2 F= force k= const. q= charge1 Q= charge2
Now charge has a sign, whereas gravity is alway attracting, but the fields
have similarities.
--
ciao Ban
Bordighera, Italy
I'm missing your correction here!
I do not know what exactly you refer to, but the gravitational field and the
electrostatic field have a lot in common. Look at Newtons Law of
Gravitation:
F = (-G* m* M)/r^2 F=force G=const. m= mass1 M= mass2
and Coulombs Law:
F = (k* q* Q)/r^2 F= force k= const. q= charge1 Q= charge2
Now charge has a sign, whereas gravity is alway attracting, but the fields
have similarities.
--
ciao Ban
Bordighera, Italy
M
Marlboro
Guest
Rich The Philosophizer <null@example.net> wrote in message news:<pan.2004.11.09.02.29.48.749713@neodruid.org>...
human called charge
Or.. becauses human calls that the voltage, the energy space for whatOn Tue, 09 Nov 2004 02:13:26 +0000, ChrisGibboGibson wrote:
Thank you John and PN
Now what, exactly, causes the potential difference between the 2 plates?
Whatever it was you used to charge the cap in the first place.
Cheers!
Rich
human called charge
C
CBarn24050
Guest
Yes quite so Ban, but that isn't what you said at the start of your post.
J
James Meyer
Guest
On Tue, 09 Nov 2004 16:47:06 GMT, "Ban" <bansuri@web.de> wroth:
matter?
Jim
Doesn't anti-matter fall UP in a gravity field produced by ordinary
matter?
Jim
J
John Popelish
Guest
James Meyer wrote:
http://math.ucr.edu/home/baez/physics/ParticleAndNuclear/antimatterFall.html
But it has been pretty well tested that antimatter has the same sort
of momentum as matter. It couldn't be contained in accelerators if
this was not the case.
--
John Popelish
I don't think the experiment has been definitively done, yet.
http://math.ucr.edu/home/baez/physics/ParticleAndNuclear/antimatterFall.html
But it has been pretty well tested that antimatter has the same sort
of momentum as matter. It couldn't be contained in accelerators if
this was not the case.
--
John Popelish
T
Tim Wescott
Guest
John Popelish wrote:
But they don't mention the difficulty of keeping one's lab notebooks
intact when it touches down...
--
Tim Wescott
Wescott Design Services
http://www.wescottdesign.com
we simply drop some antimatter in a lab, and see how fast it falls. "
But they don't mention the difficulty of keeping one's lab notebooks
intact when it touches down...
--
Tim Wescott
Wescott Design Services
http://www.wescottdesign.com
R
Rich The Philosophizer
Guest
On Tue, 09 Nov 2004 05:43:01 -0500, Active8 wrote:
;^j
Rich
Use The Force, Luke.
;^j
Rich
R
Rich The Philosophizer
Guest
On Tue, 09 Nov 2004 19:26:46 -0500, John Popelish wrote:
charge?
Actually, I'm trying to ask this seriously. But I speculate about
stuff like, can the equations that describe a proton be mapped onto
the equations that describe a black hole?
And I don't have the maths, which might be a good thing, since my
head isn't all clogged up with "what everybody already knows." ;-)
Thanks,
Rich
So, it's not really anti-"matter", but just matter with the opposite
charge?
Actually, I'm trying to ask this seriously. But I speculate about
stuff like, can the equations that describe a proton be mapped onto
the equations that describe a black hole?
And I don't have the maths, which might be a good thing, since my
head isn't all clogged up with "what everybody already knows." ;-)
Thanks,
Rich