Milikan's oil drop experiment?

[Capiert]

You assume the (Coulomb's) electric force

is a "constant" wrt the distance d

between

the 2 capacitor plates

(e.g. for linac acceleration).

 

But how can that be?

 

I.e. When the equipotential (e.g. the (surface_)charge_density (sigma=charge/area))

in the middle between the plates

is zero!

 

Due to a "bipolar" Gauss (surface_charge) law sum,

each plate receives half the voltage V/2,

but has opposite polarity (i.e. +V/2 & -V/2)

so the sum is the total voltage V

between the 2 plates,

but zero (voltage) in the middle

of the air dielectric.

 

Surely a "non_linear" acceleration

must exist (instead)

wrt distance

between the 2 plates

because "zero potential" difference (at the dielectrric middle)

has no driving force.

(=Net (+/-) force equals zero!)

 

Or does it?

 

E.g. The force near the plates is maximum;

but is minimum (=zero) in the middle

(between the 2 plates).

Edited July 7, 2017 by Capiert

[Sensei]

but how can that be

 

Oil drop is levitating. This is what you can see on your own eyes. This device should be presented to the all primary school students. Prior even learning about Coulomb's force.

 

So, what are you asking.. ?

Get device, and check results.

This is not "theoretical physics".

Edited July 6, 2017 by Sensei

[Capiert]

I'm asking about the oil drops(' accelerated) speed. (Is that speed linearly or non_linearly accelerated?)

Are they (=those oil drops) non_linear(ly accelerated, wrt position, e.g. height)?

 

If the(ir accelerating) force (F=m*a) is NOT linear wrt distance,

then the(ir) acceleration (a=F/m) can NOT be (linear wrt distance) either.

 

I apologize

for the spelling typo,

the title should read "Millikan's.." (not Milikan).

Edited July 6, 2017 by Capiert

[swansont]

You assume the (Coulomb's) electric force

is a "constant" wrt the distance d

between

the 2 capacitor plates

(e.g. for linac acceleration).

 

But how can that be?

 

I.e. When the equipotential (e.g. the (surface_)charge_density (sigma=charge/area))

in the middle between the plates

is zero!

 

Due to a "bipolar" Gauss (surface_charge) law sum,

each plate receives half the voltage V/2,

but has opposite polarity (i.e. +V/2 & -V/2)

so the sum is the total voltage V

between the 2 plates,

but zero in the middle

of the air dielectric.

 

Surely a "non_linear" acceleration

must exist (instead)

wrt distance

between the 2 plates

because "zero potential" difference (at the dielectrric middle)

has no driving force.

(=Net (+/-) force equals zero!)

 

Or does it?

 

E.g. The force near the plates is maximum;

but is minimum (=zero) in the middle

(between the 2 plates).

How does the potential difference relate to force?

[Capiert]

The force (applied to the (oil drop's) charge q)

is directly proportional

to the voltage V (=potential "difference").

 

& the middle (=in between the 2 plates)

of any (functionable) capacitor,

is always 0 V(olts, potential difference)

no matter what voltage (below its limit(ing voltage (rating)), to avoid damage)

is applied.

Charged, or uncharged!

(Thus is there NO (electrical) acceleration there?)

 

The potential difference (=voltage) is

V=E*d.

The electric field

E=V/d

is the (potential difference=) voltage V

divided by distance d,

& the Coulomb's electric force

F=q*E

is the charge q

multiplied by the electric field E (=Q/(Epsilon_o*Spherical_Area)=Q//(Epsilon_o*4*Pi*(r^2))).

So the force

F=q*V/d

is the charge q

multiplied by the (potential difference=) voltage V

divided by the (charge's) distance d

(e.g. away from the electric field's (source) charge Q.

In that case (((it is) the distance) away from) a capacitor plate).

Edited July 7, 2017 by Capiert

[Strange]

What connection is there between capacitors and Millican's experiment?

[Capiert]

Millikan used a voltage

applied to 2 plates

that are spaced

with air distance d.

That is a simple capacitor (to me)

C=Epsilon*A/d

(with Area A

& having an applied voltage).

i.e. The (=his) capacitor was charged.

 

Thus he used the electric field

of a capacitor

with the dielectric, air.

Edited July 7, 2017 by Capiert

[swansont]

On 7/6/2017 at 7:11 PM, Capiert said:

The force (applied to the (oil drop's) charge q)

is directly proportional

to the voltage V (=potential "difference").

 

& the middle (=in between the 2 plates)

of any (functionable) capacitor,

is always 0 V(olts, potential difference)

no matter what voltage (below its limit(ing voltage (rating)), to avoid damage)

is applied.

Charged, or uncharged!

(Thus is there NO (electrical) acceleration there?)

 

The potential difference (=voltage) is

V=E*d.

The electric field

E=V/d

is the (potential difference=) voltage V

divided by distance d,

& the Coulomb's electric force

F=q*E

is the charge q

multiplied by the electric field E (=Q/(Epsilon_o*Spherical_Area)=Q//(Epsilon_o*4*Pi*(r^2))).

So the force

F=q*V/d

is the charge q

multiplied by the (potential difference=) voltage V

divided by the (charge's) distance d

(e.g. away from the electric field's (source) charge Q.

In that case (((it is) the distance) away from) a capacitor plate).

 

 

You have a d in your equation for V. That should be a clue. Also the name: potential difference

 

IOW, while V has a value at some position, it is always in reference to some other point. This is like potential energy, or speed — you can always define the place (or frame) where the value is zero. It's only the difference in value between points that matters. We could just as easily choose some other reference point such that 0V is not at the center between the plates, but the potential difference will not change. The potential difference between the plates is still V.

 

There is no valid application of this formula that says that E = 0 anywhere between the plates. And F = qE. As long as there is a field, there is a force.

[Capiert]

My question was aimed

at the (net) force intensity

(onto the charged oil drop q)

wrt position

in the middle between the 2 plates,

based on the surface_charge_density (sigma=Q/A)

of the 2 opposite polarity plates (e.g. +Q/A, & -Q/A).

From what I see,

that force is zero.

The (net) surface_charge_density

is zero

in the middle

between both plates.

Thus little (=zero) affect.

 

(Otherwise, apparently uncharged objects

would appear to attract (& repel) other objects,

without electrostatic( law)s.

A form of gravitation, & anti_gravitation (e.g. levitation)?)

 

I suppose you will maintain

that the negative plate

would repel a negative charged oil drop;

& the positive plate would continue

to attract that same drop

(thus ignoring a ((volume)_charge_density rho=Q/vol) cancelation effect),

so that a (constant) continuum

of force exists

(wrt position

between the plates).

Edited July 7, 2017 by Capiert

[swansont]

My question was aimed

at the (net) force intensity

(onto the charged oil drop q)

wrt position

in the middle between the 2 plates,

based on the surface_charge_density (sigma=Q/A)

of the 2 opposite polarity plates (e.g. +Q/A, & -Q/A).

From what I see,

that force is zero.

The (net) surface_charge_density

is zero

in the middle

between both plates.

Thus little (=zero) affect.

What you see is wrong. You cannot equate force with a voltage at a particular point.

 

(Otherwise, apparently uncharged objects

would appear to attract (& repel) other objects,

without electrostatic( law)s.

I have no idea how you come to that conclusion.

 

I suppose you will maintain

that the negative plate

would repel a negative charged oil drop;

& the positive plate would continue

to attract that same drop

(thus ignoring a ((volume)_charge_density rho=Q/vol) cancelation effect),

so that a (constant) continuum

of force exists

(wrt position

between the plates).

That would be a valid way of looking at it (other than whatever you mean by a volume charge density cancellation effect)

[Capiert]

[Ref Swansont's "no idea..". PS: How do I get "quote" "copy & paste" to work in Windows explorer too on this website, instead of iPad?]

 

I equate the force

is proportional

to a (volume) charge_density "difference"

(between (the charged oil_drop) q & the plates total (+ & -) affect,

considering q has its own charge_density=cd).

 

If the volume_charge_densities

are different

(between q & its position),

then they (both q & its position's cd)

will attempt

to equalize

causing charge_flow

(i.e. the oil_drop will (be forced to) move).

That is the motivating force,

(that moves the (bound) mass (of the charge)),

the mechanism why it works, =moves (the charged oil drop mass).

 

(The charges attempt

to equally space

(& distance) from 1 another

because they all repell.)

 

 

The volume_charge_density's

"cancelation effect"

is just another way

to express

the (Gauss's law) surface_charge_density's

cancelation effect,

when positive

& negative

(surface_charge_densities sigma=Q/A)

overlap

(onto the same position).

"Surface (charge_density)" is the (simplified) math theory (trick=technique, model),

that (2D, charge per 2D area)

means (=implies)

the real (=physical, =3D)

charge per volume (rho).

Edited July 7, 2017 by Capiert

[swansont]

[Ref Swansont's "no idea..". PS: How do I get "quote" "copy & paste" to work in Windows explorer too on this website, instead of iPad?]

 

I equate the force

is proportional

to a (volume) charge_density "difference"

(between (the charged oil_drop) q & the plates total (+ & -) affect,

considering q has its own charge_density=cd).

 

If the volume_charge_densities

are different

(between q & its position),

then they (both q & its position's cd)

will attempt

to equalize

causing charge_flow

(i.e. the oil_drop will (be forced to) move).

That is the motivating force,

(that moves the (bound) mass (of the charge)),

the mechanism why it works, =moves (the charged oil drop mass).

 

 

The volume_charge_density's

"cancelation effect"

is just another way

to express

the (Gauss's law) surface_charge_density's

cancelation effect,

when positive

& negative

overlap

(onto the same position).

"Surface (charge_density)" is the (simplified) math theory (trick=technique, model),

that (2D) means (=implies) the real (=physical, =3D)

charge per volume (rho).

 

 

There is no overlap of charge, there is no cancellation. There are + charges on one plate, and - charges on the other. They are physically separated.

[Capiert]

The (capacitor's) plates

are physically separated (naturally, no argument there);

but the(ir) (positive & negative, as) "sum"

of their surface_charges (sigma=+/- Q/A)

at the "position" of the oil drop

has a cancelation effect

(=resulting total, which is less (than a single plate)).

Edited July 7, 2017 by Capiert

[swansont]

The (capacitor's) plates

are physically separated (naturally, no argument there);

but the(ir) (positive & negative, as) "sum"

of their surface_charges (sigma=+/- Q/A)

at the "position" of the oil drop

has a cancelation effect

(=resulting total, which is less (than a single plate)).

 

 

 

No, it absolutely does not have a cancellation effect. You cannot sum the charges in this situation.

 

The only way you could look at them as cancelling is if you are very far away from the capacitor plates. But that is decidedly not true for an oil drop that is between the plates.

[Manticore]

You need to look at the charge gradient - not the actual charge relative to the external world.

[Capiert]

I assume,

a charge gradient

can be the difference

wrt distance (position)

of (either)

the surface_charge_density sigma=Q/A

(or

the volume_charge_density rho=Q/vol).

 

Isn't the charge gradient

maximum

near the plates;

& minimum

in the middle

between the 2 plates?

 

Is not the (electric) potential

maximum

at each plate;

but minimum

in the middle

(between both plates),

because of opposite polarities?

 

Is not the electric_potential

zero

in the middle

as an equipotential?

I.e. (both potentials are the)

same amount,

but opposite polarity.

 

Does not positive potential

cancel negative (potential)

at the same position

(of overlap)?

 

(Otherwise, zero (=neutral) charge

would never exist, anywhere.

A discharged capacitor

would NOT be possible!

But we don't see that (impossible neutrality).

Instead we see that

things can be charged (up),

& (can be) discharged.

E.g. A thing

is either charged,

or not (=neutral).

 

The (dielectric's) polarization

in the middle of the capacitor

is minimum (e.g. zero);

but maximum nearest the plates

(having the opposite polarity

of a nearby plate).

Edited July 9, 2017 by Capiert

[John Cuthber]

I assume,

a charge gradient

can be the difference

wrt distance (position)

of (either)

the surface_charge_density sigma=Q/A

(or

the volume_charge_density rho=Q/vol).

 

Isn't the charge gradient

maximum

near the plates;

& minimum

in the middle

between the 2 plates?

 

No

That's why, when it comes down to it, the experiment works.

[swansont]

I assume,

a charge gradient

can be the difference

wrt distance (position)

of (either)

the surface_charge_density sigma=Q/A

(or

the volume_charge_density rho=Q/vol).

There is no charge gradient, as such. There is charge at two positions. It is not distributed along the path. A charge gradient anywhere along that path is zero.

 

Stop making up physics. Read a text book.

[Capiert]

Well then I'm sorry

but I do NOT know

what Manticore meant

by "charge gradient".

 

I could find nothing on the topic.

[swansont]

I think "difference" or "differential" would have been the appropriate term.

[Capiert]

Then may we continue there?

I.e. The difference of charge density.

[Sensei]

Well then I'm sorry

but I do NOT know

what Manticore meant

by "charge gradient".

 

I could find nothing on the topic.

 

Charged particle, charged object, charged plate, creates electric field:

 

[math]F=q_1 E[/math]

 

[math]E=k_e \frac{q_2}{r^2}[/math]

 

Change r-distance, and you have gradient.

It's the best to visualize in some 3D software, as 3D array of arrows, which have different length, or color, to indicate their strengths (scalar),

and pointing in some direction (vector) (up or down, mostly, in the case of this experiment).

 

[math]\vec{F}=q_1 \vec{E}[/math]

 

[math]\vec{E}=k_e \frac{q_2}{\vec{r}^2}[/math]

 

What you're interested in oil-drop experiment, is when Coulomb's force is cancelled by gravitation force (both force scalars at some point, have the same magnitude, but reverse vectors).

 

Edited July 9, 2017 by Sensei

[Mordred]

(both force scalars at some point, have the same magnitude, but reverse vectors).

 

 

Your wording is incorrect here. It should be force vectors with same magnitude but reverse directions

[Sensei]

Your wording is incorrect here. It should be force vectors with same magnitude but reverse directions

 

I made shortcut "reverse vectors" = "reverse directions of vectors"

[Mordred]

Your first term force scalar is magnitude only. Its a vector magnitude+direction.