Capacitor experiments?

[Capiert]

Hi, I need some help (diagnosing

 & defining

 (some of) my experiments,

 to be explained correctly).

 

Here is the 1st 1.

 

I have taken a 10 cm diameter styropor ball,

 wrapped it with Al foil (to be like a Gaussian sphere, e.g. the center is always zero (charge?, E_field?),

 so it's a self_discharging single electrode capacitor, being charged externally from the environment (surroundings).)

 & connected it to a J-FET input oscilloscope's probe.

 

The (oscillo)scope indicated RF noise (voltage on the ball, exponentially)

 when that ball was brought near the ground

 (or walls, ceiling).

(Much larger) AC voltage also showed (on the display) when the ball neared AC cables,

 & increased as distance decreased.

 

I made 2 large capacitor plates (each 1 m x 0.5 m x 0.1 m,

 from styropor plates covered with Al foil),

 separated them (e.g. d~0.5 m) as a parallel plate capacitor

 & connected them (in parallel) to the AC (house) power (outlet, via wires & alligator clips).

 

I moved the ball_probe between the 2 plates (back & forth).

As the Al_ball probe neared an inside plate surface

 the AC voltage ((that) displayed on the oscilloscope) increased (exponentially);

 & the polarity was opposite for the other plate;

 but (ruffly) zero (voltage) in the middle between the plates.

That looks to me like (a single probe) plus+minus (=adding) cancelation effect

 so please identify what I measured.

(=What was I measuring? Electric potential?)

 

Thanks

 

 

Edited December 11, 2017 by Capiert

[studiot]

40 minutes ago, Capiert said:

(=What was I measuring? Electric potential?)

 

Yes.

A question for you to think about.

You describe the scope display as increasing when you move the probe towards the room walls or the capacitor plates.

Did you try placing the probe statically at decreasing distances and recording the scope readings at various distances after the display had settled?

That is did the movement make any difference?

Do you know anything about the effect of capacitance on scope inputs?

[swansont]

2 hours ago, Capiert said:

 

(Much larger) AC voltage also showed (on the display) when the ball neared AC cables,

 & increased as distance decreased.

This may just be acting like an antenna. You're picking up the mains frequency; the field is driving a current on the ball.

Which may be happening with the capacitor plates, too.

[studiot]

1 hour ago, swansont said:

This may just be acting like an antenna. You're picking up the mains frequency; the field is driving a current on the ball.

Which may be happening with the capacitor plates, too.

Indeed it may be.

It is the electric field that antenna respond to.

[Capiert]

3 hours ago, studiot said:

Yes (you were measuring electric potential).

What does that electric potential

 look like mathematically.

I familiar with voltages

 which are said to be potential difference.

But I'm missing something there.

I'm also familiar with Gauss's (surface) charge_density Q/A, A=4*Pi*(r^2),

 but I don't know how they are related (to electric potential)?

Quote

A question for you to think about.

You describe the scope display as increasing when you move the probe towards the room walls or the capacitor plates.

Yes.

Quote

Did you try placing the probe statically at decreasing distances and recording the scope readings at various distances after the display had settled?

Yes, that is what I meant.

I measuured the voltage

 at specific distances

 but the voltage values

 did not have a linear relation

 wrt distance;

so I tried to develop a formula

to predict those (voltages)

wrt (also) probe area.

Quote

That is did the movement make any difference?

If you mean vibration,

 no not really,

The real time response

 was fast enough

 (for me) to ignore wiggling.

(I was only interested in the static DC values anyway;

 even though I used AC instead

 as a statistical method, =average of ruff values, thumb rule.)

There is no sense in high accuracy measurements

 for low quality apparatus.

I just wanted an idea of how it worked,

 not how well.

 

The experiment measurements

 were very ruff approximations

 (to give me an idea of what to expect,

 so I could calculate the exact values).

E.g. I doubled distances

 (to determine the tendancies).

Quote

Do you know anything about the effect of capacitance on scope inputs?

Maybe a bit, I would have to guess the rest & figure it out.

(But I don't think that (swaying or vibration) was important for 50..60 Hz. 110..230 V AC.)

 

I guess the probe capacitance C would act like a low pass filter

 as it charged & discharged

 thru the (series) wire resistance R.

 

(The capacitor's charging voltage would be

vc=Vmax*(1-(0.5^(-(t/(R*C)*(2^0.5)))))

&

the capacitor's discharge voltage

vc=Vmax*0.5^((t/(R*C))*(2^0.5))

where Vmax is the max voltage applied

& t is the half_(voltage)_time (in seconds)

 e.g. the time t it takes for half the capacitor's voltage.)

 

Thus it would round off a noisey waveshape, e.g.

 making a square shaped wave

 look (a little) like a rounded sine form.

At least (it would) tend to do that

 some amount.

2 hours ago, swansont said:

This may just be acting like an antenna. You're picking up the mains frequency; the field is driving a current on the ball.

I can not imagine a D'Arsenval analog voltage measurement

 without a current.

Quote

Which may be happening with the capacitor plates, too.

Where is the return path (current flow)

 i.e. 2nd electrode (for the probe's measurement)?

Edited December 11, 2017 by Capiert

[swansont]

If you have charges on a metal surface, they will respond to changes in the electric field at any given point, and you will measure a voltage on your oscilloscope. Since it's AC, the signal should go through zero.

Also, the probe itself will affect the charge on the plates, especially one that's 10 cm across.

10 minutes ago, Capiert said:

 Where is the return path (current flow)

 i.e. 2nd electrode (for the probe's measurement).

You have a 10 cm ball. There can be currents flowing on the surface of it.

[Capiert]

1 hour ago, studiot said:

Indeed it may be.

It is the electric field that antenna respond to.

(Yes) But we have a current (from the antenna)?

I.e. Current flow.

What is the motivating (=motion) mechanism? Abracadabra?

(Newton's 2nd law: (accelerating force), requires the 3rd law: recoil.)

51 minutes ago, swansont said:

If you have charges on a metal surface,

When do we not (have charges on metal & in metal)?

The atoms are full of them (electrons).

I suppose you mean free(d, =unbound) electrons.

But doesn't the presence

 of (free) charges

 influence materials

 to also have (freed) charge(s)?

Naturally different (amounts)

 for the various materials

 (insulators & conductors).

Quote

they will respond to "changes" in the electric field at any given point, and you will measure a voltage on your oscilloscope.

That (changes, of E_field) means DC will not NOT produce a current flow

 to produce a voltage (measurement). ?

(Back to the Millikan oil drop experiment (thread);

 that means

 a horizonal setup

 (instead of vertical)

 would not accelerate e.g. move,

 a charge(d oil drop) sideways.)

?

Sorry you're not making (enough) sense to me.

Quote

Since it's AC, the signal should go through zero.

Yes, it does.

Quote

Also, the probe itself will affect the charge on the plates, especially one that's 10 cm across.

Quite probably (true). Loading,

e.g. like a resistance

 (is a small conductor, that)

 (allows a curent flow),

i.e. an impedance.

 

That means

 the probe's presence

 acts to produce

 (something like)

 less (insulation) resistance

 than the air's (resistance).

Edited December 11, 2017 by Capiert

[swansont]

25 minutes ago, Capiert said:

(Yes) But we have a current (from the antenna)?

I.e. Current flow.

What is the motivating (=motion) mechanism? Abracadabra?

Electric field.

From the wires carrying the AC.

As already mentioned.

25 minutes ago, Capiert said:

(Newton's 2nd law: (accelerating force), requires the 3rd law: recoil.)

That (changes, of E_field) means DC will not NOT produce a current flow

 to produce a voltage (measurement). ?

You said you were doing this with AC

The word changes does not have quotes around it in my post. Why are they there in the quote of my post? Did you edit them in?

25 minutes ago, Capiert said:

(Back to the Millikan oil drop experiment (thread);

 that means

 a horizonal setup

 (instead of vertical)

 would not accelerate e.g. move,

 a charge(d oil drop) sideways.)

?

A static E field from a DC will accelerate charges, since F = qE

But you should really bring it up in the Millikan thread. Unless you're talking about AC, in which case it's not a Millikan experiment.

25 minutes ago, Capiert said:

Sorry you're not making (enough) sense to me.

You rarely make sense to me. 

 

[Capiert]

1 hour ago, swansont said:

Electric field (is moving the charges).

Yes, but can you elaborate

 how the E_field moves things like charge(d matter).

I'm still looking for the mechanism.

Quote

From the wires carrying the AC.

I think it's obvious

 from where we got the electric power

 but maybe that should be stated

 for security

 that I don't go astray.

Quote

As already mentioned.

Yes.

Quote

You said you were doing this with AC.

Yes, but as stated (above?)

 the purpose is to interpret

 how (so_called) static electricity

 is (responding).

The AC is a statistical method (technique)

 for acquiring (behaviour) info.

I want to find out how Electrostatic behaves,

 but don't have the equipment,

 so the most I can do

 is lower the frequency (to near DC)

 & observe the tendencies.

Quote

The word changes does not have quotes around it in my post. Why are they there in the quote of my post? Did you edit them in?

Yes, absolutely.

I wanted to accentuate it

 (draw it to your attention)

 for my following comments.

How should I do that?

I don't use quote marks

 like you do.

My technique stems from the days before I had Winword:

Different editors & PC software

 were unstable,

 did not last long & crashed.

I needed (my own) standards

 for recovery

 & used ASCIi

 (in an encrypted way, you will say)

 instead of loose valuable info & files.

I also experimented with language

 searched for a ideal, never found.

I can't stand my own style

 so it's no wonder you can hardly understand me

 or like it either.

Quote

A static E field from a DC will accelerate charges, since F = qE

That might be useful (for what I'm looking for)?

Is there any more to it

 that can be broken down. E.g.

What does

 E=F/q

 the force F

 per charge Q mean?

 

It looks like

 the electric field

 E=(m/e)*a

 is a mass to charge ratio being accelerated,

 but it seems to me corrupt(ed)=peculiar

 if that ratio is not inverted

 for a passive (inefficient) mass (factor) m.

That equation is failing symmetry for me.

E.g.

We're dealing with an "electric" field E (on the left side)

 & the right side shows NOT enough corelation

 with the charge "inverse".

If both E & e were in nummerators respectively,

 then there would be a direct correlation

 between both,

 for what(ever) electric (parameter) is (defined as).

I (also) get the idea Newton's 2nd law is poorly defined

 for a harmony with electromagnetism.

It would have been better (sometimes)

 to have used an inverse mass symbol

 because I tend to notice 2 types of mass:

 (storable) active; & (inefficiency) passive (coupling)

 (masses).

Sorry for thinking too loud.

 

How do I get nearer to my quest for potential('s definition, & relations)?,

 since that is what I have measured;

 or has that (potential, as idea)

 bean dropped (=kicked out)

 due to your antenna interpretation.

Quote

But you should really bring it up in the Millikan thread.

Why should I be hopping back a forth

 between the different threads

 like a rabbit

 when this thread is capacitor experiments (plural).

I evaluate based on comparisons.

I'll do it (for you) if you suggest a good reason.

I'm trying to unify 

 my scattered concepts,

 (to universal formula, by narrowing in),

 not diverge them

 although I often need your divergent, isolating method.

Quote

Unless you're talking about AC, in which case it's not a Millikan experiment.

This thread is for both AC & DC,

 however we are concentrating on AC.

Quote

You rarely make sense to me. 

It's interesting

 that some people have the talent

 (to understand,

 even when things aren't always logical).

I don't know what that gift is,

 special software (for de_encryption)

 that they possess

 or have aquired?

Some are born as natural teachers,

 others are not.

You cannot convince me that conventional flow

 is not (utter) non_sense.

Edited December 11, 2017 by Capiert

[swansont]

16 hours ago, Capiert said:

Yes, but can you elaborate

 how the E_field moves things like charge(d matter).

I'm still looking for the mechanism.

An electric field exerts forces on charged particles. E if the force per unit charge.

16 hours ago, Capiert said:

 Yes, but as stated (above?)

 the purpose is to interpret

 how (so_called) static electricity

 is (responding).

It's not static if you use an AC source.

16 hours ago, Capiert said:

The AC is a statistical method (technique)

 for acquiring (behaviour) info.

I want to find out how Electrostatic behaves,

 but don't have the equipment,

 so the most I can do

 is lower the frequency (to near DC)

 & observe the tendencies.

It's not DC if it's AC. 

16 hours ago, Capiert said:

Yes, absolutely.

I wanted to accentuate it

You don't get to do that unless you specifically state you have done so. People usually use bold or italics and say "emphasis added"

Failure to do that is dishonest.

 

16 hours ago, Capiert said:

 

That might be useful (for what I'm looking for)?

Is there any more to it

 that can be broken down. E.g.

What does

 E=F/q

 the force F

 per charge Q mean?

It's the amount of force per unit charge. The equation explains it. If you double the field, or double the charge, the force will be twice as big.

 

16 hours ago, Capiert said:

It looks like

 the electric field

 E=(m/e)*a

 is a mass to charge ratio being accelerated,

The electric field, and thus the force, does not depend on mass. The acceleration will, because F = ma

16 hours ago, Capiert said:

  We're dealing with an "electric" field E (on the left side)

 & the right side shows NOT enough corelation

 with the charge "inverse".

If both E & e were in nummerators respectively,

 then there would be a direct correlation

 between both,

 for what(ever) electric (parameter) is (defined as).

This makes no sense to me. "enough correlation"? Correlation of what with what?

E = F/q, or rearrange to F = qE

More charge or stronger field means a larger force. That's about as straightforward as it gets.

 

16 hours ago, Capiert said:

I (also) get the idea Newton's 2nd law is poorly defined

 for a harmony with electromagnetism.

It would have been better (sometimes)

 to have used an inverse mass symbol

 because I tend to notice 2 types of mass:

 (storable) active; & (inefficiency) passive (coupling)

 (masses).

Sorry for thinking too loud.

Electrostatics fields do not depend on mass.

16 hours ago, Capiert said:

You cannot convince me that conventional flow

 is not (utter) non_sense.

Then what's the point of discussion, if you out-and-out reject the science that has been shown to work?

[Capiert]

On 12 December 2017 at 1:04 PM, swansont said:

An electric field exerts forces on charged particles. E if the force per unit charge.

I still have a big gorge (=lack of info)

 in between the words

 "exerts force"s

(Surely you mean pressure too?)

That's not telling me how things are happening

 (as a mechanism);

 it's only telling that they happen

 & by how much.

Quote

It's not static if you use an AC source.

Are not static charges

 moving more than 100s of m/s

 on this rotating earth?

I only see that you are speeding them up or slowing them down.

Thus static is an illusion,

 a lie, deceipt.

Quote

It's not DC if it's AC.

Perhaps you mean

 the non_linear acceleration

 of AC,

 which is (too) difficult (=complicated) to follow

 linearly?

Quote

You don't get to do that unless you specifically state you have done so. People usually use bold or italics and say "emphasis added"

Failure to do that is dishonest.

We don't want that. Sorry, I'll try to be careful.

Quote

It's the amount of force per unit charge. The equation

E=F/q

Quote

explains it

(=itsself)

Quote

. If you double the field, or double the charge, the force will be twice as big.

If you double the (E) field

 for constant charge (q)

 then the force F will double (agreed);

 but if you double the charge q

 then the E field will half

 (for a constant force F)

 (= something is (really) screwy about that);

 or (else) the E field must stay the same

 & the force F will double (as you said).

E.g.

Isn't more charge

 more electric field

E=Q/(4*Pi*Epsilon*(r^2))?

Quote

The electric field, and thus the force, does not depend on mass.

But (isn't it so, that)

 you can NOT have charge

 without mass.

It (=charge e)

 is always (found) in (nature as) an e/m ratio.

Charge does NOT exist alone without mass.

Charge is (always) a property of matter (=mass);

 & NOT mass is a property of charge

 (although I am not sure about that last 1 (phrase)?).

Charge with zero mass

 does NOT exist in nature.

The (charged) electron has mass.

Quote

The acceleration will,

(Will) what?

Quote

because F = ma

.

Please explain, you lost me (on that sentence).

Quote

This makes no sense to me. "enough correlation"? Correlation of what with what?

Electric "co"_relation="direct"_proportionality;

 instead of inverse proportionality.

E.g. Working together for something;

 instead of (counter productively) against.

If I'm talking about "electric" charge

 & "electric" field,

 then I expect

 they both have something (electric)

 in common;

 instead of behaving oppositely.

Quote

E = F/q, or rearrange to F = qE

More charge or stronger field means a larger force. That's about as straightforward as it gets.

Not for me.

(That's not what I see.

Something there is NOT working right.)

In that equation E=F/q

 says

 more charge q

 would produce "less" field E.

Which makes no sense at all.

At least that's what I can see there.

Quote

Electrostatics fields do not depend on mass.

I can't quite completely agree with you there.

Quote

Then what's the point of discussion, if you out-and-out reject the science that has been shown to work?

But conventional flow does NOT work right. (Thus it's non_sense.)

Electron flow does (work right).

(They are opposites: 1 makes sense because it works;

 the other does not make sense, because it does not work.)

I think the point is to clear the misunderstandings.

What's the purpose of perpetuating non_sense? (To avoid booby traps? as cautions?)

(Please don't get offended when I say

 you have been brainwashed

 to accept something that does NOT work

 because your brain is not ticking

 on that point.

It has shut off any questioning

 about it.

That's not an insult,

 it is a fact.

It happens to me (often) & everyone (else) also.

(It's a short_circuit, bypass program.)

Descartes said "the greatest thing you can do is to doubt",

 perhaps because you will search thru everything

 till you find the (correct) answer.

That's the purpose of discussion,

 to find the right answer(s).

So I will repeat, (more mildly in other words):

 I doubt you can convince me

 that conventional flow is correct,

 as not (confusing) non_sense.)

Sorry, appollogies in advance.

Edited December 14, 2017 by Capiert

[John Cuthber]

41 minutes ago, Capiert said:

(Surely you mean pressure too?)

No

Pressure is force/ area and there's no area defined.

 

This goes to show the depth of your ignorance which puts the rest of your comments in context

[Capiert]

On 11 December 2017 at 4:07 PM, swansont said:

This may just be acting like an antenna. You're picking up the mains frequency; the field is driving a current on the ball.

Which may be happening with the capacitor plates, too.

1st I'd like to thank both of you (Studiot & Swansont)

 for your careful handling (hands on) approach

 to get me thinking.

My reactions might sound harsh (e.g. Abracadabra)

 but they are only reflexes,

 (which accelerate me to the point of importance)

 I'm not angry.

Thanks Studiot for asking about capacitance (my depth)

 before proceeding;

 & for those few helpful

 reassuring small phrases

 (to steer me along).

(E.g. That led up to Abracadabra.)

(Although I'm ambivalent) I've got mixed feelings

 about Swansonts post above:

 I cringed at (1.) the "guess" of (2) an "antenna";

 but was intrigued by the idea

 that a current was forced (thru the probe).

It's very (importantly) interesting.

I can't quite tell (=figure out)

 why I don't like the idea of an antenna.

I suspect I wanted to hear

 (an exacter (=pin pointed) answer like):

 the probe

 is acting like a capacitor plate.

Maybe bidirectional:

 "sender"; & receiver=sensor?

But I'm still curious (though)

 why the electric potential

 measures zero

 in the middle

 between

 the 2 plates.

That (zero voltage, on the probe)

 means zero current?

How can that be?

(You might (probably) say

 because there is zero (net) force?

Such an answer as that

 leaves a vast pillar like column, hole tunnel

 in my mind, an abis (=bottomless pit).)

 

1 hour ago, John Cuthber said:

No

Pressure is force/ area and there's no area defined.

This goes to show the depth of your ignorance which puts the rest of your comments in context

Matter is 3D, so a 2D cross_section(al area)

 is expected by me.

Pressure was my suggestion to him.

I think I would call it (my) lack of knowledge

 (=not knowing),

 but I have to admit

 I must sort things out (into the right sequence) first:

 I can not accept anything but the key sequence (to the major problems) I need.

Everything else is a distraction.

Edited December 14, 2017 by Capiert

[Capiert]

On 11 December 2017 at 6:05 PM, swansont said:

you have charges on a metal surface, they will respond to changes in the electric field at any given point, and you will measure a voltage on your oscilloscope. Since it's AC, the signal should go through zero.

Also, the probe itself will affect the charge on the plates, especially one that's 10 cm across.

Please explain a bit (how or how much, an example)?

What will happen

 to the charge on the plate?

On 11 December 2017 at 6:05 PM, swansont said:

On 11 December 2017 at 5:56 PM, Capiert said:

 Where is the return path (current flow)

 i.e. 2nd electrode (for the probe's measurement)?

You have a 10 cm ball. There can be currents flowing on the surface of it.

What happens if we use a flat_probe

 (instead of the round ball)

 e.g. Square Al(uminum) 10 cm x 10 cm x 1 mm,

 parallel to the 2 plates.

(I chose 10 cm to make the math easy;

 but that could be smaller,

 e.g. of only a few cm square foil, instead,

 if needed.)

 

The problem with electric potential (voltage measurement)

 is, there doesn't seem to be

 a 2nd electrode

 (=return wire, or ground)

 for the (forced) current

 flowing thru the probe

 into the Oscillscope.

Edited December 14, 2017 by Capiert

[studiot]

4 hours ago, Capiert said:

But conventional flow does NOT work right. (Thus it's non_sense.)

Electron flow does (work right).

I don't know if you are talking about the ill fated attempt to replace 'conventional current' with 'electron current' in electric threory.

That was a really bad idea and lead to much confusion in the 1990s.

 

The plain fact is that currents can be the flow of either positive or negative charges or both.

It is difficult to say which is more common.

 

The issue was resolved by choosing a convention many years ago, now called 'conventional current'.

This was neither the right nor wrong way round, but it did allow the whole of electric theory to develop to what we have today.

Whatever convention you choose you have to actually choose not one but two sign conventions and you cannot avoid the situation where some quantity seems  'the wrong way round'.

Therefore there is no point changing current direction conventions and I strongly recommend you stick with conventional current.

 

Thanks for the vote of confidence expressed about my posts.

 

Edited December 14, 2017 by studiot

[swansont]

1 hour ago, Capiert said:

Please explain a bit (how or how much, an example)?

What will happen

 to the charge on the plate?

If the ball is charged, he charges on the plate will feel a force. Like charges attract, opposites repel.

 

Quote

What happens if we use a flat_probe

 (instead of the round ball)

 e.g. Square Al(uminum) 10 cm x 10 cm x 1 mm,

Why use such a large probe? Every probe I've seen has been small.

Quote

  

The problem with electric potential (voltage measurement)

 is, there doesn't seem to be

 a 2nd electrode

 (=return wire, or ground)

 for the (forced) current

 flowing thru the probe

 into the Oscillscope.

Your oscilloscope doesn't have a ground input?

You are measuring a voltage, not a current. The current through the probe is supposed to be small. And if it's AC, the electrons don't move very far anyway. 

[John Cuthber]

12 hours ago, Capiert said:

I must sort things out (into the right sequence) first:

Yes.

The right sequence is to start by learning some science.

If you

can also learn

how to stop

your text appearing

like

this it will

be a step

forward too

[Capiert]

16 hours ago, studiot said:

I don't know if you are talking about the ill fated attempt to replace 'conventional current' with 'electron current' in electric threory.

That was a really bad idea and lead to much confusion in the 1990s.

 

The plain fact is that currents can be the flow of either positive or negative charges or both.

It is difficult to say which is more common.

 

The issue was resolved by choosing a convention many years ago, now called 'conventional current'.

This was neither the right nor wrong way round, but it did allow the whole of electric theory to develop to what we have today.

Whatever convention you choose you have to actually choose not one but two sign conventions and you cannot avoid the situation where some quantity seems  'the wrong way round'.

Therefore there is no point changing current direction conventions and I strongly recommend you stick with conventional current.

 

Thanks for the vote of confidence expressed about my posts.

 

Thanks too for the good reply.

Do you have a link or info

 to reinforce getting me on track there.

E.g. both polarities moving in a wire?

I though only electrons (if they can be called that) hopped around

 because the heavy nuclei were bound (=bonded)

 as solid.

I suspect the 1990s confusion was the vast amount

 of older literature with conventional flow

 sabbotaging progress.

All literature (that made any reference to electricity) needed review.

For then mission impossible.

But I don't know the issues.

Now (it looks to me like)

 we close an eye

 (& make a compromise)

 to get around the (polarity confusion) problem. (?)

[Capiert]

9 hours ago, John Cuthber said:

Yes.

The right sequence is to start by learning some science.

How do you define learning?

(Memorizing?)

I'd prefer to digest

 by sorting things into the right boxes.

There is too much confusion,

 to proceed otherwise.

9 hours ago, John Cuthber said:

If you

can also learn

how to stop

your text appearing

like

this it will

be a step

forward too

Hi John

I can't tell you how delightful it was to read that.

It looked almost like mine.

[Capiert]

20 hours ago, swansont said:

If the ball is charged, (t)he charges on the plate will feel a force. Like charges attract, opposites repel.

Very good, that's what I'm looking for. (Attraction & repulsion.)

((But now let's back up.)

How is the ball charged?

From the (voltage) on the plate. ?

So the plates feel their own forceful effects?

(The electrified (=charged) plates cause the ball to be electrified (=charged)

 so both ball & plate will feel forces. ? (So that's:)

Action at a distance.

Only air in between.

Air is also electrified.

Electrical chains (=series)?

(Electrification) also happens without air (=vacuum). Why?

Electrons are shot out (by the acceleration),

 but that acceleration

 is caused by (electric) repulsion.

I.e. Too many electrons in the volume

 of metal (plate).

 

Seems ok,

 but I can NOT explain attraction;

 only (higher) pressure (repulsion).

Quote

Why use such a large probe? Every probe I've seen has been small.

Maximum accuracy, simple math.

My original intent

 was to predict the probe's measured voltage

 based on (any) area.

Then scale down

 (smaller).

I.e. Calibration.

Quote

Your oscilloscope doesn't have a ground input?

No, my oscioscope also has a ground,

 like any other;

 but it's NOT used for the potential measurements

 (except to calibrate zero, by shorting the input (manually)

 for a few seconds).

Quote

You are measuring a voltage,

Yes.

Quote

not a current.

Yes. The ground (wire) is not used.

Quote

The current through the probe is supposed to be small.

Yes, 10 M Ohm impedance.

Quote

And if it's AC, the electrons don't move very far anyway. 

(..from what?)

I don't understand that 1 (sentence).

Do you mean, the electrons redistribute in the (probe) wire (metal)

 as charges per volume?

I assume

 the number

 of free electron( charge)s Q

 per vol(ume)

 (=volume charge density rho=Q/vol)

 is changing

 in the (probe) conductor.

Those repelling electrons

 affect the J-FET's (gate G (e.g. pinch off by affecting

 the silicon's charge(_density) nearby G, & thus the)

 drain to source (electron_)flow (decreases).

 

More electrons per volume means more (electric) potential.

Subtracting potentials is voltage (difference).

 

I enjoyed your questions.

 

Edited December 15, 2017 by Capiert

[swansont]

3 hours ago, Capiert said:

Very good, that's what I'm looking for. (Attraction & repulsion.)

((But now let's back up.)

How is the ball charged?

You were the one who claimed it was charged.

"I have taken a 10 cm diameter styropor ball,

 wrapped it with Al foil (to be like a Gaussian sphere, e.g. the center is always zero (charge?, E_field?),

 so it's a self_discharging single electrode capacitor, being charged externally from the environment (surroundings).)"

(emphasis added)

Quote

 

Seems ok,

 but I can NOT explain attraction;

Opposite charges attract.

Quote

  Maximum accuracy, simple math.

Since the voltage changes with position, a large probe does not give you maximum accuracy.

Quote

 

No, my oscioscope also has a ground,

 like any other;

 but it's NOT used for the potential measurements

Then your measurement is going to be crap. Potentials have to be measured with respect to a reference. The voltage is with respect to two measurements. Your unused reference could have a lot of noise on it, and that will be superimposed on any signal you measure. 

Quote

  

(..from what?)

I don't understand that 1 (sentence).

Do you mean, the electrons redistribute in the (probe) wire (metal)

 as charges per volume?

Electrons in current flow from 50 Hz or 60 Hz have a very slow drift velocity (of order 1mm/sec), and only move for ~20 ms before the direction changes. So lack of a return path does not have the same impact as it does in DC. You can just be getting eddy currents in the conducting ball.

 

[studiot]

8 hours ago, Capiert said:

Thanks too for the good reply.

Do you have a link or info

 to reinforce getting me on track there.

E.g. both polarities moving in a wire?

I though only electrons (if they can be called that) hopped around

 because the heavy nuclei were bound (=bonded)

 as solid.

I suspect the 1990s confusion was the vast amount

 of older literature with conventional flow

 sabbotaging progress.

All literature (that made any reference to electricity) needed review.

For then mission impossible.

But I don't know the issues.

Now (it looks to me like)

 we close an eye

 (& make a compromise)

 to get around the (polarity confusion) problem. (?)

 

If you really want to look into electric sign conventions, then we should have a new thread.

It would be a diversion here.

So I will just put up one diagram and ask you to consider the terminals of a battery, particularly the negative one.

 

 

When the switch is closed conventional current is  said to flow around the circuit in the direction of the red arrow.

No actual discussion of charge carriers is needed.

 

The Disney fairy tale about electrons runs like Nellie the Elephant who packed her bags and said goodbye to the circus.

So the tale goes that the electrons pack their bags and say goodbye to the negative terminal and go on their holidays around the circuit, arriving at the positive terminal.

Question one

As each electron leaves the negative terminal does that terminal become slightly less negative ?

If not that must mean new electrons arrive frome somewhere to take their place.

 

So Question two

Where do all these electrons come from (that is how do they get to the negative terminal?) ?

 

Question three

Why is the positive terminal positive?

[Capiert]

On 15 December 2017 at 12:08 PM, swansont said:

You were the one who claimed it was charged.

..by the nearby (things) of the environment,

 e.g. plate's voltages.

Quote

"I have taken a 10 cm diameter styropor ball,

 wrapped it with Al foil (to be like a Gaussian sphere, e.g. the center is always zero (charge?, E_field?),

 so it's a self_discharging single electrode capacitor, being charged externally from the environment (surroundings).)"

(emphasis added)

Yes, just (trying to) narrow in

 on trying to understand the mechanism (cause details, for the effects).

Quote

Opposite charges attract.

Yes, (very) good tip. Thanks (for making it simpler).

Maybe it's (= the cause details are) not as difficult as I expect,

 but I doubt my optimism.

Quote

Since the voltage changes with position, a large probe does not give you maximum accuracy.

For my crude measuring instruments it does

 (help accuracy to deal with large approximate voltages).

I can't measure small values very accurately.

I don't have a big buget for best instruments

 like gov officials do.

(Good) Instruments are rare for me;

 & too expensive.

No place to store everything either.

Quote

Then your measurement is going to be crap.

I thought that too at 1st;

 but got reproduceable values.

& they were stable.

I tried similar measurements

 with other (company's better) equipment

 & got (only) crap.

So I guess I had luck

 finding that (Russian) Oscilloscope.

It caught my eye;

 contemplating

 what I could do with it

 to answer my questions.

(I was curious about the emotional effects

 radiation produced. E.g. stress.)

It wasn't enough for all my needs,

 (an extra channel would have been helpful)

 but I'll never regret buying it.

I had lived in an old house

 where the (electrical wiring's) cloth insulation

 had burn marks, all over.

 I was curious why I felt certain sensations

 that reoccured

 & wanted to find out why.

A lot of strange things happened.

The strangest is,

 all that happened

(but mostly) only in that house.

It had peculiar problems

 that I could not reproduce elsewhere.

(So that house('s problems) & its circumstances

 was a rare artifact (for me)

 or (my lab) reference.

Something like Marie Curie's pitch blend.

Something strange, noticeable, repeats, not explained. What is it?

How could it be explained? Build theories, & test them with the observations.

It's subjective testhing (for an egoist);

 that does not interest (objective) physicists.

Quote

Potentials have to be measured with respect to a reference.

That reference seems to be taken care of (automatically)

 by (some) FET's wiring circuits.

Quote

The voltage is with respect to two measurements.

Correct;

 & a single potential is not. ?

Quote

Your unused reference could have a lot of noise on it,

Yes.

Quote

and that will be superimposed on any signal you measure.

Yes, perhaps that's the tip.

The Oscilloscope was grounded

 (by its power cable, 3 wires)

 although I didn't use the ground input,

 that might have been internally grounded

 (at the input).

Expensive scopes leave the ground input open (=not connected)

 so the user can & must connect that (everytime).

Quote

Electrons in current flow from 50 Hz or 60 Hz have a very slow drift velocity (of order 1mm/sec),

That's interesting,

 & might explain

 why I could almost see arcing

 in clumps

 (when a boy while playing around with DC motors & batteries).

Quote

and only move for ~20 ms before the direction changes.

Seems reasonable (& good description).

Quote

So lack of a return path does not have the same impact as it does in DC.

uhmm, (ruffly) yuk! (for the approximating);

 but its a good=excellent description.

(=I don't like that AC & DC are treated as so different;

 but I have to face the facts (even though I don't like guesses (too well),

 because I have enough of my own (guesses, & assumptions)

 trying to get to the bottom of things).

Quote

You can just be getting eddy currents in the conducting ball.

What does that mean then?

I had eventually thought

 the house had RF on the power cables

 as a (wireless) carrier for LF (low frequency) noise

 because the radio transmitter was 5 Km away

 in the next villiage.

Sometimes the stress was so (high=bad),

 it hurt more to cry=weep,

 than not.

All I wanted was peace (& quiet)

 but could NOT get it

 (without further R&D).

Edited December 17, 2017 by Capiert

[Capiert]

On 15 December 2017 at 12:53 PM, studiot said:

 

If you really want to look into electric sign conventions, then we should have a new thread.

It would be a diversion here.

So I will just put up one diagram and ask you to consider the terminals of a battery, particularly the negative one.

 

 

When the switch is closed conventional current is  said to flow around the circuit in the direction of the red arrow.

No actual discussion of charge carriers is needed.

 

The Disney fairy tale about electrons runs like Nellie the Elephant who packed her bags and said goodbye to the circus.

So the tale goes that the electrons pack their bags and say goodbye to the negative terminal and go on their holidays around the circuit, arriving at the positive terminal.

Question one

As each electron leaves the negative terminal does that terminal become slightly less negative ?

(Very) good question.

Maybe yes.

My (virtual, working theory) guess (=hypothesis) is

 (molecules or atoms, ions of) bipole chains build

 along a voltage drop V=E*d

 (e.g. in a wire).

Each bipole has a neutral center.

d is the bipole's length.

The wire's resistance & voltage

 would help determine d.

Any voltage drop will produce a string of bipoles.

 

You probably explain things (=neutrality, zero middle charge)

 with dipole moment('s center) & (an insulator's) dielectric polarization. ?

Quote

If not that must means new electrons arrive from somewhere to take their place.

Yes, that's probably true.

Quote

So Question two

Where do all these electrons come from (that is how do they get to the negative terminal?) ?

I think from nearest neighbours

 (along the voltage drop, direction).

Quote

Question three

Why is the positive terminal positive?

Maybe a (severe) "lack of negative" charges

 is why it's (called (the)) positive (terminal).

 

That's a mighty good question.

Charge might be spin direction

 (e.g. left & right hand rules, curled fingers vs upright thumb)

 for + & -.

If charged particles

 have a spin or rotation

 then they have magnetism (you say a magnetic field;

 but I have great difficulty trying to comprehend

 what a field is (in detail);

 so I tend to avoid using the word (field).

Occationally I use that word (like everyone else)

 when I'm careless

 & casual

 (taking it for granted).

It's a big problem.

 

Anyway, in reverse: a magnetic field

 indicates a moving charge(d material, matter).

As we've seen voltage affects matter

 & charges (that matter).

(But (charged), voltage in a capacitor

 only indicates

 the amount of charges

 (per volume?).)

I suppose it's the difference

 of nearby materials

 (e.g. electronegativity?)

 that charges them initially

 to make a battery (differential cell).

Considering (a generator, & forceably) turning

 copper wire

 thru a magnetic field (careless again, I should say magnetism instead)

 at 90 degrees

 we get electricity (=moving charged electrons=electron_charges?).

I assume an electron always has a charge

 & that it is negative.

I don't know if a neutral electron can exist?;

 or a positron electron pair. ?

Most atomic orbitals can pair opposite electrons,

 but nuclear physics distinguishes (positive) positrons

 from (negative) beta electrons.

It's a wonder why a positive (electron?) orbital is ignored.

Maybe that has to do with our instrumentation

 is grounded wrt negative

 for measurements.

What would the (new) world be like

 with positive grounding=earthing.

Who knows?

Maybe we would find a few new missing (non_heavy) isotopes.

E.g. Charging the environment

 affects the measurement.

The common ground earth

 is made 1 of the noisest conductors.

I don't know why positive is positive;

 but I enjoy

 the lack of electrons;

 especially when it's sunny

 & the light drives them away

 (with the photoelectric effect)

 like in our photocopiers.

Electrons (scattered (all over)) are the culprit

 for our (modern) stress.

Positive (electric) potential calms, maybe

 because our eyes are an exposed nerve,

 a (electric) zero_crossing detector.

Subconsciously maybe we can notice

 what is NOT there;

 but it's the (increasing negativity) negative voltage spikes

 (=when the noise voltage goes downwards from zero,

 seen as a falling flank because it's negative going)

 that bother us, I think?

(When our body has a similar potential.

Mine was ~0.14 V AC (based on body size, capacitively) when touching a bare probe.

Any noise approx. my body's potential (voltage)

 made me ill,

 probably because random cancellation confused

 my automatic vegetative nervous system signals (mostly visually).

I concluded that's what made people sick from Electric noise.

Large or smaller values

 did NOT affect illness as much.

But larger was worse.

Thanks for the question,

 I didn't know, maybe

 the eye (subconsciously) reacts exclusively to only electrons' (increasing) negative charge?

That's screwy to explain positively

 if electrons are the major charge carrier

 & positive charge

 is the lack of electrons.

I guess, it's the atomic kick_out "sequence"

 that might count

 (maybe due to the coreolis direction?

 earth, sun, galactic rotation direction?

 which has priority, spin up vs down).

1 is easier to do (kick out)

 because of a (rotational) momentum assist (against the whole?);

 or the other spin direction

 sits better, as

 more stable, in harmony,

 rotating (or spinning)

 in a similar direction

 to the larger cosmic scale's

 rotational direction.

 

Positive is a lack of electrons.

Is it possible

 electrons spinning direction

 is in the opposite direction

 than most of the mass

 in the universe rotates?

 Or galaxy rotation.

I often wondered why galaxies rotated.

 

 

 

Edited December 17, 2017 by Capiert

[swansont]

11 hours ago, Capiert said:

 For my crude measuring instruments it does

 (help accuracy to deal with large approximate voltages).

I can't measure small values very accurately.

 

Measuring a volt or more should be trivial even with inexpensive equipment.

 

8 hours ago, Capiert said:

  That's a mighty good question.

Charge might be spin direction

No, it's not.

8 hours ago, Capiert said:

If charged particles

 have a spin or rotation

 then they have magnetism (you say a magnetic field;

Yes, they do.

8 hours ago, Capiert said:

  I don't know if a neutral electron can exist?;

No, it can't.

8 hours ago, Capiert said:

  It's a wonder why a positive (electron?) orbital is ignored.

They aren't, but systems with positrons are rare. The solutions to the QM equations with positrons and antiprotons look exactly like those with normal matter.

8 hours ago, Capiert said:

  (maybe due to the coreolis direction?

 earth, sun, galactic rotation direction?

No. If you could do the physics, it would be trivial to show this: the Coriolis force is way too small.

8 hours ago, Capiert said:

 which has priority, spin up vs down).

1 is easier to do (kick out)

 because of a (rotational) momentum assist (against the whole?);

 or the other spin direction

 sits better, as

 more stable, in harmony,

 rotating (or spinning)

 in a similar direction

 to the larger cosmic scale's

 rotational direction.

See above.

8 hours ago, Capiert said:

Positive is a lack of electrons.

Is it possible

 electrons spinning direction

 is in the opposite direction

 than most of the mass

 in the universe rotates?

Irrelevant.