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Natural Science Forum / Physics / Relativity / June 2006Since k varies but not G suggests an Eather
F= kQq/r^2 F = GMm/r^2 k varies but never G: 1. k ***VARIES**** depending on the density(or structure) of the other charges in the medium (since medium's are made of electrons and protons) excluding Q&q. 2. but as opposed Newton's G remains ****CONSTANT**** regardless of the other masses(planets, atoms) and densities surrounding M&m. *************************************** Theoretical probabilistic Conclusion to the above: For this reason it leads me to believe that the Electric Field regulates the permittivity and not vice-versa, and for this reason that space is not empty but also has a weaker Electric Field (energy field) known as the EATHER. **************************************** It also leads me to believe that Bilge is Bald. > F= kQq/r^2 F = GMm/r^2 > [quoted text clipped - 14 lines] > space is not empty but also has a weaker Electric Field (energy field) > known as the EATHER. It only appears that way simply because of the scales involved. In the end both of these are derivable from Gauss's law, which is easier to use with variable charge and/or mass densities. > > F= kQq/r^2 F = GMm/r^2 > > [quoted text clipped - 18 lines] > end both of these are derivable from Gauss's law, which is easier to > use with variable charge and/or mass densities. Mass densities? How much would the weight of an industrial gas cylinder increase if you put 10^42 electrons in it ? Sue... > How much would the weight of an industrial > gas cylinder increase if you put 10^42 electrons in it ? Since the mass of an electron is 9.1 x 10^-31 kg, then the mass of 10^42 electrons is 9.1 x 10^11 kg. Why do you ask? - Randy > F= kQq/r^2 F = GMm/r^2 > > k varies No, the electrostatic CONSTANT does not vary. That's why they call it a constant. > but never G: > > 1. k ***VARIES**** depending on the density(or structure) of the other > charges in the medium (since medium's are made of electrons and > protons) excluding Q&q. No, it doesn't. - Randy > > F= kQq/r^2 F = GMm/r^2 > > [quoted text clipped - 10 lines] > > No, it doesn't. But k will certainly have different values in different media, since it depends on the total permittivity of the medium. > > > F= kQq/r^2 F = GMm/r^2 > > > [quoted text clipped - 13 lines] > But k will certainly have different values in different media, since it > depends on the total permittivity of the medium. Sorry... all media manufacturers get thier electrons from the same supplier. http://hyperphysics.phy-astr.gsu.edu/Hbase/electric/elefor.html Sue... > > > > F= kQq/r^2 F = GMm/r^2 > > > > [quoted text clipped - 19 lines] > > Sue... That's a poor explanation Sue, that link simply explains that "k" is constant in space THEREFORE FOR THE ELECTRIC FORCE EXCERTED BETWEEN TWO CHARGES IN ***AIR*** OR ***WATER*** IS *********NOT THE SAME********* AS THE ELECTRIC FORCE EXCERTED BETWEEN TWO CHARGES IN SPACE: F = KQq/R^2 (K VARIES WITH THE MEDIUM..... ....MEANING K VARIES WITH THE PRESENCE OF OTHER CHARGES IN THE SAME REGION WHERE AS "G" DOES NOT VARY WITH THE PRESENCE OF OTHER OBJECTS(MASSES) IN THE SAME REGION OF M&m (F= GMm/R^2) ). > > > > > F= kQq/r^2 F = GMm/r^2 > > > > > [quoted text clipped - 29 lines] > REGION WHERE AS "G" DOES NOT VARY WITH THE PRESENCE OF OTHER > OBJECTS(MASSES) IN THE SAME REGION OF M&m (F= GMm/R^2) ). You just answered your own question so I contend the URL is effective. ;-) k is sumed over all space because it is radiative. G is sumed over all the matter because it is induced. You have to apply forces to move a mass from moon to earth. You do not have to apply force to move a charge from one plate to another plate of a charged capacitor. So... You see some important difference in the behavior of fundamental and composite particles ? Sue... > > > > > > F= kQq/r^2 F = GMm/r^2 > > > > > > [quoted text clipped - 35 lines] > k is sumed over all space because it is radiative. > G is sumed over all the matter because it is induced. coils induce magnetism or current, G is neither > You have to apply forces to move a mass from moon to earth. not if there's no moon and only earth & mass > You do not have to apply force to move a charge from one > plate to another plate of a charged capacitor. WELL DONE! That's because the attractive charge force along the wire is stronger then the attractive charge force between the dielectric plate..... Therefore "k" indeed varies with the medium (dielectric medium versus copper medium) and the main difference between mediums is the electron density(& proton) that surrounds Q&q! Where as "G" remains constant regardless of the other planets(objects) or mass densities that surround M&m! F= k_medium (Qq/r^2) F = G_invariant (Mm/r^2) > So... You see some important difference in the behavior > of fundamental and composite particles ? > > Sue... > > > > > > > F= kQq/r^2 F = GMm/r^2 > > > > > > > [quoted text clipped - 37 lines] > > coils induce magnetism or current, G is neither The analysis of the Tajmar-de Matos experiment would indicate otherwise. > > You have to apply forces to move a mass from moon to earth. > > not if there's no moon and only earth & mass If you can remove the moon but leave the debris on its surface in place I'll accept that. Something tells me the debris is going to tag along. :o) > > You do not have to apply force to move a charge from one > > plate to another plate of a charged capacitor. [quoted text clipped - 15 lines] > > So... You see some important difference in the behavior > > of fundamental and composite particles ? k even varies with the person that is using it, which is why you have to be careful about the assumptions made in different unit systems. << Any discussion of magnetism is complicated because the physical notation for magnetic fields is not standardized. mks-SI system regards magnetism as an effect of electric currents - where as - cgs system begins with forces between poles >> http://www.geo.mtu.edu/~jdiehl/magnotes.html So it is imprudent to accept the assumptions of a particular metrology, especially when you have evidence the assumptions may not hold in practice. Sue... > > Sue... > > > > > > > > F= kQq/r^2 F = GMm/r^2 > > > > > > > > [quoted text clipped - 40 lines] > The analysis of the Tajmar-de Matos experiment would > indicate otherwise. cooling reduces motion such as freezing your food slows down the rotting process. supercooling reduces resistivity....whatever that means since it's only a terminology, I guess it helps the electron jump to it's neighjbor valences with less interference from other electrons of the neighbors atom? eitherway by INCREASING the MOTION gradually (spinning motion), you may also be gradually contradicting the FREEZING MOTION of the superconductor therefore it may still be an electromagnetic effect. (Brief from above: gradually increasing motion of spin may be gradualliy negating the freezing motion of supercooling effect). > > > You have to apply forces to move a mass from moon to earth. > > [quoted text clipped - 3 lines] > surface in place I'll accept that. Something tells me > the debris is going to tag along. :o) ?? I don't know what you were getting at since you started the debate but likewise you if you can remove a positive charge but leave it's pairing negative charge in place I'll accept that also???? > > > You do not have to apply force to move a charge from one > > > plate to another plate of a charged capacitor. [quoted text clipped - 19 lines] > have to be careful about the assumptions made in different > unit systems. Well according to Randy in another thread k in F= k Qq/r^2 remains the constant permittivity of space REGARDLESS of which medium it's in, I find that illogical (why would the velocity decrease due to permittivity but not the force of a charge.....especially since the main difference between mediums is the charge density)? > << Any discussion of magnetism is complicated because the > physical notation for magnetic fields is not standardized. [quoted text clipped - 10 lines] > > > > Sue... > > > > > > > > > F= kQq/r^2 F = GMm/r^2 > > > > > > > > > [quoted text clipped - 68 lines] > but likewise you if you can remove a positive charge but leave it's > pairing negative charge in place I'll accept that also???? The original point was the need to apply a force. Reconsidering, I should have offered something showing the homogenity of a sea of charges as opposed to the homogenity of a sea of masses. That occured to me in reading the Randy's post which you reference below. > > > > You do not have to apply force to move a charge from one > > > > plate to another plate of a charged capacitor. [quoted text clipped - 25 lines] > permittivity but not the force of a charge.....especially since the > main difference between mediums is the charge density)? Yes... that is a good example to disassemble because we have Fitzpatrick's rigourous formalism that isn't jumping between unit systems and it based in proven plasma theory. http://farside.ph.utexas.edu/teaching/em/lectures/node98.html Read his page carefully noting how he considers currents and magnetic effects in the dielectric. Many discussions don't do that and it contributes immensly to all the confusion that surrounds the issue. Sue... > > << Any discussion of magnetism is complicated because the > > physical notation for magnetic fields is not standardized. [quoted text clipped - 10 lines] > > > > > > Sue... > > > > > > > > > > F= kQq/r^2 F = GMm/r^2 > > > > > > > > > > [quoted text clipped - 115 lines] > that and it contributes immensly to all the confusion that > surrounds the issue. Quote from above link: "There are no free charges or free currents in the medium. There is also no bound charge density" 1. Is this more of an exception then a general case, meaning it's only related to inert gases since only they don't have any electron valences (free charge, free current) as well as perhaps bound charge density when atoms exchange electron valences to form molecules? ----------------------------- 2. The same link(node 99) IMMEDIATELLY AFTER IT actually specifies it's speaking of gases at: http://farside.ph.utexas.edu/teaching/em/lectures/node99.html Quote: "Dielectric constant of a gaseous medium " > Sue... > [quoted text clipped - 12 lines] > > > > > > > > Sue... > > > > > > > > > > > F= kQq/r^2 F = GMm/r^2 > > > > > > > > > > > [quoted text clipped - 125 lines] > (free charge, free current) as well as perhaps bound charge density > when atoms exchange electron valences to form molecules? Over the volume of the universe it is the most general case. It is only exceptional when we put equipment near it to try and study it. Then the equipment spoils the homogenity. 'Fitz' knows what will spoil our study of epsilon and cause us to drag out this: http://en.wikipedia.org/wiki/Multiple_integral ...so the theatre is 'sterilized' for us. Later we can introduce conductive and magnetic inhomogenities one at a time and see their effects on epsilon and mu and express epsilon and mu in any system of units. > ----------------------------- > [quoted text clipped - 5 lines] > Quote: > "Dielectric constant of a gaseous medium " Yes... Some low pressure gas similar to helium probably meets the specification. But he has specified it so we don't need know about those gasses but rather in terms that we can see the characteristics needed to avoid doing a balancing act with eps and mu while we are trying to study how they are derived. Jackson uses a similar trick with his E plane only, spherical antenna: http://arxiv.org/abs/physics/0506053 Sue... > > Sue... > > [quoted text clipped - 12 lines] > > > > > > > > > > Sue... > Well according to Randy in another thread k in F= k Qq/r^2 remains the > constant permittivity of space Yes. > REGARDLESS of which medium it's in, Perhaps you've forgotten that "medium" is just an approximation, and at the microscopic level a medium consists of isolated atoms with pure vacuum between. > I > find that illogical (why would the velocity decrease due to > permittivity Perhaps that's because you've forgotten that the idea of "velocity decreasing" is just an approximation, and that what happens at the microscopic level involves energy traveling at c, the vacuum speed, in vacuum between isolated atoms. > but not the force of a charge.....especially since the > main difference between mediums is the charge density)? No, that is not the main difference between mediums. Mostly our discussion of dielectric materials involves neutral media, i.e. charge density = 0. - Randy > > Well according to Randy in another thread k in F= k Qq/r^2 remains the > > constant permittivity of space > > Yes. Your words from another thread: " The electrostatic force felt by a particle q in a medium....using the standard, unchanging value of k." But the link below specifies that "k" does vary with the medium for ELECTROSTATIC FIELDS?: http://www.plus2physics.com/electrostatics/study_material.asp?chapter=2 Electrostatic field: (The electric field stength is E= k Qq/r^2) Quote: "Lines of force are a convenient way of visualizing an electric field .....The total number of lines of force is inversely proportional to the ***PERMITTIVITY(thus k) OF THE MEDIUM**** in which the charge is located." > > REGARDLESS of which medium it's in, > [quoted text clipped - 20 lines] > > - Randy > > > Well according to Randy in another thread k in F= k Qq/r^2 remains the > > > constant permittivity of space [quoted text clipped - 5 lines] > " The electrostatic force felt by a particle q in a medium....using the > standard, unchanging value of k." Yes. > But the link below specifies that "k" does vary with the medium for > ELECTROSTATIC FIELDS?: [quoted text clipped - 8 lines] > inversely proportional to the ***PERMITTIVITY(thus k) OF THE MEDIUM**** > in which the charge is located." The reason for your confusion is that there are two equivalent ways of looking at what happens with a medium. A "medium" is a convenient fiction, an approximation. It can be described in terms of such things as dielectric constant and index of refraction, and light "slowing down". However, all of these quantities are DERIVED from the vacuum equations. You calculate how the charges in the medium respond to electromagnetic fields (by polarizing for instance), and then what fields result from those actions of the charges in the medium. These calculations are all done using vacuum values, as those are the fundamental equations which are always true. When you do those calculations, you can summarize the effects due to charges in the medium as "medium properties". Thus, it is true that the force on a charge is always, always, ALWAYS the sum of kQq/r^2 where k is the vacuum value and the sum is over ALL OTHER CHARGES, including those in the medium. It is also true that you don't have to look at the individual charges in the medium, but can approximate their effects by introducing the idea of a "permittivity of a medium". When you do that, you are doing the process I am saying: summing up VACUUM kQq/r^2 over all the charges of the medium. Note that when people do this, they no longer consider the individual charges in the medium. That's just an approximation, albeit a good one. But things like "permittivity of a medium" can actually be calculated, by considering those individual charges and using the vacuum equations. I know. I had to do it for exams. - Randy > > > > Well according to Randy in another thread k in F= k Qq/r^2 remains the > > > > constant permittivity of space [quoted text clipped - 64 lines] > > - Randy So you're saying instead of applying the total charges in space to the equation and using "k for space", they are varying "k" instead so as to include the charges in the medium (as opposed to the charges outside the medium)? But if we were 100% blind and we had the k of space medium to work with, I think that we would never know that "k_space", as the original topic of this post suggested, isn't also(as the "k" of the other mediums) controlled (meaning it's own value determined) by perhaps other weaker(distant) uniform charges of the Aether? ------------------------------------- (continued from above reasoning) .....Unless perhaps the characteristic that velocity varies with "k_medium" and "k_space" BUT EM waves are only affected by "k_space" (therefore the hypothetical charges of an Aether could not be validated since EM waves should be affected by a K_medium as well)? The last statement isn't perfect since it only applies to high frequency EM waves, Wikepedia says that low frequency EM waves and static charges are affected by "k_medium", (I "THINK" they say because the charges of k_medium don't have sufficient TIME to react with the high frequency EM waves travelling through the medium)? snip > So you're saying instead of applying the total charges in space to the > equation and using "k for space", they are varying "k" instead so as > to include the charges in the medium (as opposed to the charges outside > the medium)? Whether charges exist ~outside the medium~ or not does not concern us. If they are absent, their charge is zero. If they are present, they exist in e+ e- pairs so their value is zero. Try to remember that on a grand scale, every positive charge is canceling out a negative charge. It is not the same a summing up the mass of all the atoms in the universe. So the universe has a total charge of zero, but its mass is non-zero. Sue... snip > snip > > [quoted text clipped - 7 lines] > zero. If they are present, they exist in e+ e- pairs so > their value is zero. Relativity is the word: In a way nothing is zero nor potential and all is kinetic (always in motion, never stops). Paired charges are neutral and not, but without question the field shrinks to it's smallest hypothetical value and the force will not be neutral but at it's maximum and if you apply energy to seperate the charges then the field is bigger volume wise but it's force decreases. (F_neutral = kQq/r^2 where r is the constant distance e and p remain when neutral and Epotential = (F - F_neutral)*r^1) -------------------------------------- Since nothing is neutral and only the field shrinks as Randy said the residual force of dipole charges is 1/r^3 same as a magnetic force..... If you have two dipole charges, does the residual remain at 1/r^3 or is it more or is it less? (meaning inside the field of two paired charges F at 1/r^2, outside these dipoles the field F at 1/r^3....therefore in between TWO individual dipoles (2 paired charges) F at 1/r^3 and outside these TWO individual dipoles F at ????? if F residual remains with Two dipoles above then perhaps these determine the permittivity of space, since the velocity of EM waves is affected by k_dielectric medium even if the dielectric medium's charges are neutral, as well LOW (not high) frequency EM waves and static charges are also affected force wise by k_dielectric (even if the dielectric medium's charges are neutral)....therefore likewise even if the total charges in space are neutral they may affect(determine) the permittivity of space and the velocity of light? > Try to remember that on a grand scale, every positive charge is > canceling out a negative charge. It is not the same a summing [quoted text clipped - 5 lines] > Sue... > snip > > snip > > > [quoted text clipped - 26 lines] > > Since nothing is neutral Why do you say nothing is neutral? If you have equal amounts of positive and negative charge, the net charge is zero. >and only the field shrinks as Randy said the > residual force of dipole charges is 1/r^3 Yes, if: (a) your + and - charges are separated so that you have a net polarization. Just because you have + and - charges doesn't mean they have to separate. (b) r is large compared to the separation between the charges. > same as a magnetic force..... > > If you have two dipole charges, does the residual remain at 1/r^3 or is > it more or is it less? It depends on how they are arranged and where you are situated. You can have two dipoles which will give a field that looks like one stronger dipole. You can have two dipoles cancelling each other out so what is left is the quadrupole moment (1/r^4 force). You could be located near to the dipoles in which case you have to use the exact expression, sum(kQq/r^2), not the approximation 1/r^3. - Randy snip I see a lot of insightful statements in both posts but there is a train wreck hidden in there somewhere. No good deed goes unpunished. Randy, If you're agreed these these: http://newton.umsl.edu/~philf/triplet.html http://hyperphysics.phy-astr.gsu.edu/Hbase/electric/dipole.html#c2 ...illustrate the same thing, then it might be productive to see which if any statements are off key and why. Something guskz wrote about ~a distant hydrogen atom~ ~shrinking fields~ ...is troubling me and I can't put my finger on it. Analysing it as a triplet will probably reveal why but I have some chores to analyse frist. :-( Sue... > snip > [quoted text clipped - 5 lines] > If you're agreed these these: > http://newton.umsl.edu/~philf/triplet.html Electric potential and field magnitude around a triplet of charges. You remember that potential and field are two different things, right? The potential far away from a dipole varies as 1/r^2, the field as 1/r^3. But these plots show those quantities CLOSE IN, not far away. > http://hyperphysics.phy-astr.gsu.edu/Hbase/electric/dipole.html#c2 Electric potential far away from a dipole. > ...illustrate the same thing, then it might be productive to > see which if any statements are off key and why. No, those don't show the same things. First link: - close in - 3 charges Second link: - far away - 2 charges. > Something guskz wrote about > > ~a distant hydrogen atom~ > ~shrinking fields~ > ...is troubling me and I can't put my finger on it. A hydrogen atom is not polarized. Water has a dipole moment (it polarizes in the presence of electric fields) because the H atoms are not located symmetrically relative to the O atom. > Analysing it as a triplet will probably reveal why I don't know why you'd analyze hydrogen as a triplet. "Triplet" means three charges. Hydrogen has one proton, one electron. That's two. But analyzing it as a dipole is also wrong, since a dipole requires polar SEPARATION of charges. - Randy > > snip > > [quoted text clipped - 13 lines] > But these plots show those quantities CLOSE IN, > not far away. OK I can consider that. > > http://hyperphysics.phy-astr.gsu.edu/Hbase/electric/dipole.html#c2 > [quoted text clipped - 12 lines] > - far away > - 2 charges. OK... I can state it another way. > > Something guskz wrote about > > [quoted text clipped - 6 lines] > because the H atoms are not located symmetrically > relative to the O atom. True. The triplet description won't work well. > > Analysing it as a triplet will probably reveal why > [quoted text clipped - 4 lines] > But analyzing it as a dipole is also wrong, since > a dipole requires polar SEPARATION of charges. It'll work. I'll post following the troubling statement. Sue... > - Randy > > snip > > [quoted text clipped - 10 lines] > are two different things, right? The potential far away > from a dipole varies as 1/r^2, the field as 1/r^3. Sounds logical since Ep = V = F r > But these plots show those quantities CLOSE IN, > not far away. [quoted text clipped - 26 lines] > because the H atoms are not located symmetrically > relative to the O atom. charges are always moving and in orbits, I think they should be called dipole frequencies instead of moments which perhaps would explain why according to Wikepedia a dielectric's dipolar forces only affect low frequency EM waves and static charges instead of high frequency EM waves?? If so the dipoles must have a SPECIFIC frequency and a force wave (the orbiting velocity, thus frequency must vary with the dielectric since "c" does? and therefore also the force between the dipoles since P =Qd where as "d" may vary with the orbiting velocity of the charges in the dielectric or not)? > > Analysing it as a triplet will probably reveal why > [quoted text clipped - 6 lines] > > - Randy > > > snip > > > > [quoted text clipped - 30 lines] > amounts of positive and negative charge, the net > charge is zero. Ok, the dipole moment then aint zero, P = Qd and it's Q aint zero. Otherwise H2O would never form? > >and only the field shrinks as Randy said the > > residual force of dipole charges is 1/r^3 [quoted text clipped - 6 lines] > (b) r is large compared to the separation between > the charges. Ok but if "r" was the same distance as the separation then the forces and charges would not be neutral. I see "d" (P=Qd) as the related to the normal velocity of the charges only and that all fields simply shrink with "d", The charges may be neutral but not the electric field otherwise if another charge whose "r" is almost the same as "d" would feel no effect. But in true nature: if q = neutral = 0 then Electric field would = 0 (since E= kQq/r^2) = none sense? I think the velocity of the charges is also related to the permittivity of space, and if so would this velocity of the charges change in a dielectric (since light's "c" does)? ....and therefore would "d" also change? > > same as a magnetic force..... > > [quoted text clipped - 9 lines] > to use the exact expression, sum(kQq/r^2), not the > approximation 1/r^3. I read that normally the quadpole is spherical thus neutral...maybe for hydrogen but for other atomic mediums the arangement due to the nucleus in any cross-section of the sphere (excluding neutrons) is: - - - - - - - - + + + + + + + + - - - - - - - - - pole (protons grouped together in nucleus)? What would be the force for those (1/r^4??) -------------------------------------------------- Not really a dipole moment but a dipole wave As well for hydrogen, negative charges(e) must always be orbiting around the heavier charge (p) ??.... therefore creating dipole moment pulses (a constant oribiting frequency of dipolar force as the electron orbits the proton, therefore the dipolar moment would be a wave force of a specific frequency and not a specific value....but Perhaps why this dipolar frequency in a dielectric cannot affect high frequency EM waves (since the dipolar frequency is slower)? (Wikepedia says dielectric mediums (force wise) only affect static charges and slow frequency EM waves) --------------------------------------------- Also I don't know if this frequency would be a nice sine wave since I believe in a previous post with you Randy long ago, that the force varies considerably with each angle between dipoles??? (Forgot which post). ....or was it that the force would be neutral if the 3rd charge was exactly perpendicularly in between the dipoles at distance "r"? > - Randy > > Why do you say nothing is neutral? If you have equal > > amounts of positive and negative charge, the net > > charge is zero. > > Ok, the dipole moment then aint zero, P = Qd and it's Q aint zero. > Otherwise H2O would never form? What? > > >and only the field shrinks as Randy said the > > > residual force of dipole charges is 1/r^3 [quoted text clipped - 9 lines] > Ok but if "r" was the same distance as the separation then the forces > and charges would not be neutral. What do you mean by forces being neutral? The expression for the force from a charge Q and a charge -Q on a charge q is F = kQq*(1/r1^2 - 1/r2^2). If r is the same distance as the separation, then the approximation used to convert (1/r1^2 - 1/r2^2) to 2*sin(theta)/r^3 (or whatever the correct expression is) breaks down. Instead you can use the exact value of (1/r1^2 - 1/r2^2). > I see "d" (P=Qd) as the related to the normal velocity of the charges You shouldn't, since it isn't a velocity. > only and that all fields simply shrink with "d", What? > The charges may be neutral but not the electric field Electric fields aren't charged. > otherwise if > another charge whose "r" is almost the same as "d" would feel no > effect. What? > But in true nature: if q = neutral = 0 then Electric field would = 0 > (since E= kQq/r^2) = none sense? With large objects at ordinary macroscopic differences, this is usually true. The force between you and the earth is dominated by the gravitational force. There's no net charge on you or the earth, so no electric force between the two of you. > I think the velocity of the charges is also related to the permittivity > of space, and if so would this velocity of the charges change in a > dielectric (since light's "c" does)? ....and therefore would "d" also > change? You can think stuff which is totally at odds with all of our experience and theory, but that won't make it true or even plausible. > > > If you have two dipole charges, does the residual remain at 1/r^3 or is > > > it more or is it less? [quoted text clipped - 9 lines] > > I read that normally the quadpole is spherical thus neutral... That's a little confused, but I can see what you read buried in there. 1. You can have a net charge, in which case you'll see monopole fields, E ~ 1/r^2. 2. You can have an arrangement of charges with net charge zero but a dipole moment, in which cases at large distances E ~ 1/r^3. 3. You can have an arrangement of charges with net charge zero and dipole moment zero but a nonzero quadropole moment, in which cases at large distances E ~ 1/r^3. And so forth. > maybe for > hydrogen but for other atomic mediums the arangement due to the nucleus > in any cross-section of the sphere (excluding neutrons) is: > - - - - - - - - + + + + + + + + - - - - - - - - - pole (protons > grouped together in nucleus)? The protons are in the nucleus. I don't know what else you're trying to say. The nucleus is very small compared to the atom, so the protons are "grouped together". If those "---" symbols mean you think there are electrons in the nucleus, that's wrong. > What would be the force for those (1/r^4??) I can't figure out what "those" refers to. > -------------------------------------------------- > > Not really a dipole moment but a dipole wave What? > As well for hydrogen, negative charges(e) must always be orbiting > around the heavier charge (p) ?? No, electrons are not in orbit in atoms. >... therefore creating dipole moment > pulses (a constant oribiting frequency of dipolar force as the electron > orbits the proton, therefore the dipolar moment would be a wave force > of a specific frequency and not a specific value....but No. Not in an atom. However, it is true you could create a situation like this with a macrosopic arrangement, and you'd get an oscillating dipole field. I don't know what you mean by "not a specific value". Of course it has a value. > Perhaps why this dipolar frequency in a dielectric cannot affect high > frequency EM waves (since the dipolar frequency is slower)? There's no such dipolar frequency due to orbits since they don't exist. There are resonant frequencies when dipoles are free to oscillate, such as in H2O. H2O resonates very nicely with frequencies in the GHz, which is why microwave ovens work. > (Wikepedia says dielectric mediums (force wise) only affect static > charges and slow frequency EM waves) That would indeed have something to do with how fast the dipoles wiggle, but it's not connected to electrons orbiting around nuclei. > Also I don't know if this frequency would be a nice sine wave since I > believe in a previous post with you Randy long ago, that the force > varies considerably with each angle between dipoles??? Yes, "wiggles" are typically "nice sine waves". > (Forgot which > post). ....or was it that the force would be neutral if the 3rd charge > was exactly perpendicularly in between the dipoles at distance "r"? Dipole force varies with sine of the angle (or cosine, depending on how you define the angle). That's why a rotating dipole would give a force which varies as a sine wave. But an atom is not a rotating dipole. - Randy > > > Why do you say nothing is neutral? If you have equal > > > amounts of positive and negative charge, the net [quoted text clipped - 4 lines] > > What? My mistake, if a dipole moment is never zero why does a quadrupole moment = zero when it's spherical? > > > >and only the field shrinks as Randy said the > > > > residual force of dipole charges is 1/r^3 [quoted text clipped - 19 lines] > correct expression is) breaks down. Instead you can use the > exact value of (1/r1^2 - 1/r2^2). When Q and -Q are neutral, due to the constant known velocity of charges in space thus "d" between them is a known constant (when they're paired into a dipole), so say "r" (of q) is = 1.5 d at first: 1. Does q orbit Q and -Q? 2. Does r remain at 1.5d when it returns to the same angle in it's orbit or does it keep moving farther and farther away for the dipoles? > > I see "d" (P=Qd) as the related to the normal velocity of the charges > > You shouldn't, since it isn't a velocity. Bad writing I ment: I see "d" as a constant distance which is related (regulated) by the normal velocity of the charges in space (since velocity is an anti-attraction phenomena...the higher the velocity of the attracting charges the bigger "d" between them). > > only and that all fields simply shrink with "d", > > What? The Electric field shrinks volume wise as the charges get closer together? > > The charges may be neutral but not the electric field > > Electric fields aren't charged. No but their entire value is related to the charges times the permittivity constant: (Qxq and r^2 which is determined by the velocity and polarities of Q &q) > > otherwise if > > another charge whose "r" is almost the same as "d" would feel no > > effect. > > What? What....does charges being neutral "mean" to you? It means they no longer have the same sex appeal (attraction repulsion forces) as they once did as when they where single. This sex appeal is the attraction/repulsion Force Field also known as the Electric field. > > But in true nature: if q = neutral = 0 then Electric field would = 0 > > (since E= kQq/r^2) = none sense? [quoted text clipped - 3 lines] > the gravitational force. There's no net charge on you or the earth, > so no electric force between the two of you. > > I think the velocity of the charges is also related to the permittivity > > of space, and if so would this velocity of the charges change in a [quoted text clipped - 3 lines] > You can think stuff which is totally at odds with all of our experience > and theory, but that won't make it true or even plausible. It was a question not a statement: 1. light and other waves slow down in dielectrics therefore does the velocity of charges also slow down? 2. I think the distance between dipoles is related to BOTH their attraction force and velocity, therefore if the velocity in #1 changes therefore does the distance between dipoles also change?? > > > > If you have two dipole charges, does the residual remain at 1/r^3 or is > > > > it more or is it less? [quoted text clipped - 12 lines] > That's a little confused, but I can see what you read buried > in there. It's one of Sue's links, Electric Quadrupoles at: http://hyperphysics.phy-astr.gsu.edu/Hbase/electric/elequad.html#c1 (ellipsoidal quadpoles are non-neutral, spherical are neutral) > 1. You can have a net charge, in which case you'll see monopole > fields, E ~ 1/r^2. [quoted text clipped - 7 lines] > > And so forth. Ok but how would quadpoles (as the link above says) be 100% neutral ...doesn't that mean no "r" at all (no 1/r^2...no 1/r^3...etc) > > maybe for > > hydrogen but for other atomic mediums the arangement due to the nucleus [quoted text clipped - 6 lines] > so the protons are "grouped together". If those "---" symbols mean > you think there are electrons in the nucleus, that's wrong. no no, forgeting the neutrons...you have a sphere with protons in the core and electrons around it, slice the sphere into a cross-section you have a circle with electrons surrounding protons in the center, ......then take *ANY* straight line across that circle and you have a line of e e e e e e e p p p p p p p p e e e e e e e e e (the "- -" were the arrangement of electrons). > > What would be the force for those (1/r^4??) F = k Qq/r^2 where r is the LINEAR distance between charges, we were talking about quadpoles (or more), so I took a cross section of an atom and the most ****POPULAR**** position(orientation) for charges in space (say only quadpoles) would be this orientiation: e p p e (so what is the force relation for such an orientation 1/r^2 or 1/r^4 or less or more?) (Therefore again doing my cross-section above, an eigth pole would be e e p p p p e e..... all though this arrangement is due to the presence of gluons forcing the protons together in the center of the atom) > I can't figure out what "those" refers to. > [quoted text clipped - 3 lines] > > What? The charges are not fixed in space, they are always in motion and not only in motion but orbiting each other, therefore the dipole moment is a dipole wave BECAUSE it is always varying in space due to it's orientation in TWO WAYS: #1: The paired charges are orbiting each other thus the dipole moment is also spinning. #2: pretend "q" is not moving in space and at a constant distance "r" from the dipole (Q & -Q), as the dipole is spinning (as in #1) the force (P= Qd thus the magnitude of Q instead) of the dipole moment changes also and this variation is in the form of a wave (example when q is exactly between Q & -Q...the dipole moment or force is neutral, when q is closest to Q then the dipole moment is at it's peak....all these variations with time cause a dipole wave of a specific frequency(frequency's value related to the spin velocity of the dipole). therefore q will not under go a dipole moment but a dipole WAVE with a specific FREQUENCY, a specific AMPLITUDE (related to q's "r" distance from dipole) ? > > As well for hydrogen, negative charges(e) must always be orbiting > > around the heavier charge (p) ?? [quoted text clipped - 16 lines] > There's no such dipolar frequency due to orbits since they don't > exist. ??? Newton's law says that velocity requires a straight path but since the paired charges are attracted to each other therefore they are orbiting each other due to their attraction force and due to their velocity (velocity = anti attraction and forms and causes them to orbit (spin around ) each other??? And for that reason the dipole moment is always spinning in space (meaning the batteries polarities are not fixed in space but rotate)?? > There are resonant frequencies when dipoles are free to oscillate, > such as in H2O. H2O resonates very nicely with frequencies in [quoted text clipped - 5 lines] > That would indeed have something to do with how fast the dipoles > wiggle, but it's not connected to electrons orbiting around nuclei. ?? electrons do not travel in straight paths ...they in an atom ORBIT around a nucleus or a proton??? > > Also I don't know if this frequency would be a nice sine wave since I > > believe in a previous post with you Randy long ago, that the force [quoted text clipped - 9 lines] > how you define the angle). That's why a rotating dipole would give > a force which varies as a sine wave. thus not a dipole moment but a dipole wave....correct?? > But an atom is not a rotating dipole. ???? > - Randy > > > > Why do you say nothing is neutral? If you have equal > > > > amounts of positive and negative charge, the net [quoted text clipped - 6 lines] > > My mistake, if a dipole moment is never zero Any moment can be either zero or nonzero. > why does a quadrupole moment = zero when it's spherical? If it's spherically SYMMETRIC there are no higher moments than monopole, because higher-order moments measure various ways in which a distribution is not symmetric. > > The expression for the force from a charge Q and a charge -Q > > on a charge q is F = kQq*(1/r1^2 - 1/r2^2). [quoted text clipped - 5 lines] > > When Q and -Q are neutral, You mean Q = 0? What else would it mean for a charge Q to be "neutral"? > due to the constant known velocity of > charges in space What? What "constant known velocity of charges in space"? What velocity would that be? Charges can be not moving, or moving at any velocity less than c. > thus "d" between them is a known constant (when > they're paired into a dipole), Are you just trying to say "Q and -Q are located a fixed distance d apart"? > so say "r" (of q) is = 1.5 d at first: > > 1. Does q orbit Q and -Q? What? It could be fixed, it could be in motion. If in motion, I don't think that the motion could be a closed orbit as that is a property of central 1/r^2 forces. The force in this case is not 1/r^2 but something more complicated, and it isn't central (it doesn't point toward a fixed center). > 2. Does r remain at 1.5d when it returns to the same angle in it's > orbit There's no orbit. See above. > or does it keep moving farther and farther away for the dipoles? There's not even any guarantee it could "return to the same angle". It might just fly by on some curved path. > > > I see "d" (P=Qd) as the related to the normal velocity of the charges > > > > You shouldn't, since it isn't a velocity. > > Bad writing I ment: I see "d" as a constant distance which is related > (regulated) by the normal velocity of the charges in space No such thing. > (since > velocity is an anti-attraction phenomena What? > ...the higher the velocity of > the attracting charges the bigger "d" between them). What? > > > only and that all fields simply shrink with "d", > > > > What? > > The Electric field shrinks volume wise as the charges get closer > together? What? What is the volume of an electric field? > > > The charges may be neutral but not the electric field > > [quoted text clipped - 11 lines] > > What....does charges being neutral "mean" to you? The sentence I can't decode is "another charge whose r is almost the same as d would feel no effect". Nothing about "neutral" there. I have no idea what situation you're trying to describe. > It means they no longer have the same sex appeal (attraction repulsion > forces) as they once did as when they where single. This sex appeal is > the attraction/repulsion Force Field also known as the Electric field. No clue what "they" refers to here. > > > But in true nature: if q = neutral = 0 then Electric field would = 0 > > > (since E= kQq/r^2) = none sense? [quoted text clipped - 13 lines] > > It was a question not a statement: There was a statement: "I think the velocity of the charges is also related to the permittivity of space, " If this is a general statement, it's not true. It seems to be connected to your belief that there is a constant natural velocity to charges, which is also not true. > 1. light and other waves slow down in dielectrics Only on average and in the continuous approximation. In reality light always travels at c. > therefore does the velocity of charges also slow down? Not "therefore", but matter is slowed down when passing through other matter, due to a host of effects which can be lumped together as "friction". > 2. I think the distance between dipoles is related to BOTH their > attraction force and velocity, The dipoles that appear in dielectrics, which are due to the polarizing of molecules, don't generally have a velocity. The "attraction force" in question depends on the chemical bonds in the molecule. > therefore if the velocity in #1 changes > therefore does the distance between dipoles also change?? I think that you have added the false beliefe "dipoles in a dielectric move with some constant velocity" to your false belief "charges everywhere move with characteristic constant velocity". There's no such velocity. > > > I read that normally the quadpole is spherical thus neutral... > > [quoted text clipped - 3 lines] > It's one of Sue's links, Electric Quadrupoles at: > http://hyperphysics.phy-astr.gsu.edu/Hbase/electric/elequad.html#c1 There is nothing there that says "a quadrupole is spherical". > (ellipsoidal quadpoles are non-neutral, spherical are neutral) No, it doesn't say that. That's just an example of a charged ellipsoidal distribution. It says that example can be represented as a CHARGED uniform sphere and a NEUTRAL quadropole consisting of four point charges. > > 1. You can have a net charge, in which case you'll see monopole > > fields, E ~ 1/r^2. [quoted text clipped - 5 lines] > > and dipole moment zero but a nonzero quadropole moment, > > in which cases at large distances E ~ 1/r^3. The "elementary quadropole" in the picture (four charges) has net charge zero, dipole moment zero, quadropole moment nonzero. The ellipsoid has net charge nonzero, dipole moment zero, quadropole moment nonzero IN THAT EXAMPLE. It's not a general statement about charge distributions. The spherical "component" IN THAT EXAMPLE has net charge nonzero, dipole moment zero, quadropole moment zero. > > And so forth. > > Ok but how would quadpoles (as the link above says) be 100% neutral > ...doesn't that mean no "r" at all (no 1/r^2...no 1/r^3...etc) If you have the same amount of + and - charge, your net charge is zero. But no, having NET charge zero is not the same as having no moments. It doesn't mean there are no individual charges, it means that the total charge is zero. Something else to keep in mind is that a 1/r^3 field falls off faster than a 1/r^2 field. So if you get very far from a distribution with no 1/r^2 field, it does look like there's no field at all. But as you get closer you start to perceive the nonuniformities, the individual charges. You'll see a 1/r^3 field if there is one. If there isn't, then you have to get closer still to see a 1/r^4 field, if there is one. > > > maybe for > > > hydrogen but for other atomic mediums the arangement due to the nucleus [quoted text clipped - 9 lines] > no no, forgeting the neutrons...you have a sphere with protons in the > core and electrons around it Yes. > , slice the sphere into a cross-section you > have a circle with electrons surrounding protons in the center, Electron shells are clouds that actually penetrate the nucleus, but yes you still have the protons confined to a tiny space in the middle. > ......then take *ANY* straight line across that circle and you have a > line of e e e e e e e p p p p p p p p e e e e e e e e e (the "- -" were > the arrangement of electrons). Not exactly, it's more like a positive point in the middle of a negative cloud. The nucleus is VERY small. > > > What would be the force for those (1/r^4??) > > F = k Qq/r^2 where r is the LINEAR distance between charges, we were > talking about quadpoles (or more), so I took a cross section of an atom > and the most ****POPULAR**** position(orientation) for charges in space > (say only quadpoles) would be this orientiation: I have no idea what you're saying. > e p p e (so what is the force relation for such an orientation > 1/r^2 or 1/r^4 or less or more?) I guess you're asking which moments an arbitrary atom has, all by itself. I'd say no net charge, no dipole moment, so only quadrupole and higher moments. But if the atom is ionized, then it has a net charge and so its behavior is dominated by 1/r^2 forces. If the atom is in a neutral molecule like H2O, then you get dipole moments and 1/r^3 forces dominate. > (Therefore again doing my cross-section above, an eigth pole would be e > e p p p p e e No, you don't get these poles by considering what happens along a straight line. Multipoles describe the distribution in 3-space. > > > Not really a dipole moment but a dipole wave > > > > What? > > The charges are not fixed in space, No, this is misguided, no matter how many times you say it. - Randy snip because it's all related to the motion of charges > > > > Not really a dipole moment but a dipole wave > > > [quoted text clipped - 5 lines] > > - Randy There's hydrogen or atoms: For hydrogen: Do you have any evidence or links that electrons and protons don't orbit each other (because in another post you said they don't nescessarily orbit but rather vibrate together)? For atoms: Do you have any evidence or links that electrons don't orbit the nucleus (does the electrons velocity increase or decrease the farther from the nucleus?) ? --------------------------------------------------------- If in both cases above there is orbit therefore there is a dipole wave with a specific frequency (related to the velocity of the charges)? > snip because it's all related to the motion of charges > [quoted text clipped - 17 lines] > the nucleus (does the electrons velocity increase or decrease the > farther from the nucleus?) ? Sorry I keep forgetting about the "other" quantum numbers, for S P D F orbitals, and even by looking at the pictures of them I still don't recognize the orbitals: .the "p" orbital is two spheres...since they're speaking of electron clouds therefore I presume both spheres are of the orbits of two electrons (and in the center is the protons and the nucleus)?...or is it the orbits of two groups of electrons in a shell? > --------------------------------------------------------- > > If in both cases above there is orbit therefore there is a dipole wave > with a specific frequency (related to the velocity of the charges)? > > snip because it's all related to the motion of charges > > [quoted text clipped - 26 lines] > electrons (and in the center is the protons and the nucleus)?...or is > it the orbits of two groups of electrons in a shell? all though they are ALL even the hydrogen still "orbitals" and therefore all generate dipole waves? ------------------------------------- I haven't found any so called "static motionless electrons" in nature that you mentioned prior (the penning trap is not a natural phenomena). > > --------------------------------------------------------- > > > > If in both cases above there is orbit therefore there is a dipole wave > > with a specific frequency (related to the velocity of the charges)? > > snip > > > [quoted text clipped - 9 lines] > > > Relativity is the word: > In a way nothing is zero nor potential and all is kinetic (always in > motion, never stops). [quoted text clipped - 5 lines] > and if you apply energy to seperate the charges then the field is > bigger volume wise but it's force decreases. Right... neutral and not neutral. :o) Consider a small region of space whose charge you can alter positive to negative. Consider a hydrogen atom in the not too farfield. It is neutral. When the space is positive the atom shifts its electron field toward the positive space. When the space is negative, the atom shifts its positive nucleus toward the space. The atoms angular momentum is participating in our attempt to alter the charge in the small region of space so it is not neutral... in a different sense of the word. If you remove the atom. The net charge won't be affected but the eps and mu for the space we have been wiggling will change. Yep... Neutral but not neutral. :o) Sue... > (F_neutral = kQq/r^2 where r is the constant distance e and p remain > when neutral and Epotential = (F - F_neutral)*r^1) [quoted text clipped - 30 lines] > > Sue... > > snip $$ Sue... wrote: > > guskz@hotmail.com wrote: > > snip > > > > So you're saying instead of applying the total charges in space to the > > equation and using "k for space", they are varying "k" instead so as [quoted text clipped - 5 lines] > zero. If they are present, they exist in e+ e- pairs so > their value is zero. $$ Clearly, the SUM of all the + & - charges isN'T ZERO. $$ This is where you dimwits need a lesson in "ENTHALPY". $$ Total Hamiltonian ENTHALPY includes all ..whatever sign. $$ YOU have to provide a mathematical way to CHANGE the sign. $$ [This has been provided to you over & over & over ..again]. $$ $$ Again ..SUM Enthalpy = (LaGrangian L) + Gibb's eG $$ = (m*v^2/2)*(m/M + 1) + Newton's G*M*m/(n-1)*r. $$ $$ Where NOTE (n - 1) exactly dictates what will be the proper sign. $$ Where NOTE (n - 1) exactly dictates what is the AMBiENT media #. $$ Where NOTE (n - 1) exactly dictates what is the CHARGE density. $$ [The GUESS iSS modified Couloumb~Newton equation already done]. $$ [Re: GUESS iSS modified Couloumb~Newton < REST Math=eM/c^2 > ]. $$ $$ How many more times? Duh, here it is again: $$ Hamiltonian ENTHALPY E - Kinetic energy eK = E - L - eV = REST eM $$ $$ | G*M1*m1 k_c*q1*q2| $$ Force*rA = |-- -- - -- + -- -- - --| = |(LaGrangian L) + eF| = eM $$ |(n - 1)*rA (n - 1)*rA| $$ $$ | m1*v1^2 | |3*m1*rA*(g + a)| $$ = |Gibbs eG + (-- - --)| = |-- -- --- -- --| = (eG - eV) = m*c^2 $$ | 2 | | 2 | $$ $$ | m1^2*v1^2 | | m1*v1^2 m1 | $$ = E - eK = |E - (-- --- --)| = |E - {(-- - --)*(-- + 1)} - eV| $$ | 2*M1 | | 2 M1 | $$ |E - ( eK )| |E - { LaGrangian L } - eV| $$ $$ = iNTRiNSiC REST energy of the ground state, for iSOLATED systems. $$ $$ E=eG, eM=eF and eK=eV ONLY if Ti in LaGrangian L COULD be zero (0). $$ The SAME rA for M1*m1 ORBiT distance ..is ESCAPE radius, for q1*q2. $$ SAME M1*m1 ORBiT rA = v1^2/g = ESCAPE vesc^2/2(n - 1)*a, for q1*q2. $$ [Note +(m1*v1^2/2) CANNOT be EQUAL to -eV at ONCE per calculation]. $$ Also, note ..ALL is the SAME, in GUESS PLANCK Temperature Tp units. $$ And finally, icon * is a MULTiPLiCATiON SiGN which SEPARATEs icons. $$ $$ This ought help a bit. Sincerely, Brian. > Try to remember that on a grand scale, every positive charge is > canceling out a negative charge. It is not the same a summing > up the mass of all the atoms in the universe. > > So the universe has a total charge of zero, but its mass > is non-zero. > > > > Sue... > snip Re: Since k varies but not G suggests an Eather. Re: [Electro] REST Math=eM/c^2. End of POST. > Therefore "k" indeed varies with the medium (dielectric medium versus > copper medium) and the main difference between mediums is the electron [quoted text clipped - 4 lines] > > F= k_medium (Qq/r^2) F = G_invariant (Mm/r^2) Wrong! You would need to take into account the mass densities surrounding the source body also, as in Gauss's law. Or, if it was a uniform medium, you could come up with something akin to the dielectric constant for such a medium and use it to modify G just like k. Either way should be valid. > > Therefore "k" indeed varies with the medium (dielectric medium versus > > copper medium) and the main difference between mediums is the electron [quoted text clipped - 10 lines] > constant for such a medium and use it to modify G just like k. Either > way should be valid. G is constant and never varies only the Force varies depending on the density of the medium. Permittivity is not constant and varies does vary with each medium, and the main difference between mediums is the density of the medium and mostly the density of the charges since the main difference between mediums and elements is the number of electrons & protons( number of charges) per volume of space = density of charge?? > > > Therefore "k" indeed varies with the medium (dielectric medium versus > > > copper medium) and the main difference between mediums is the electron [quoted text clipped - 13 lines] > G is constant and never varies only the Force varies depending on the > density of the medium. Even though you're responding to it, you obviously did not read my previous post. > Permittivity is not constant and varies does vary with each medium, and > the main difference between mediums is the density of the medium and > mostly the density of the charges since the main difference between > mediums and elements is the number of electrons & protons( number of > charges) per volume of space = density of charge?? It could vary, but then you really lose the concept of permittivity in the first place. If the medium has variable charge density, the only way to attack the problem is with Gauss's law. > > Therefore "k" indeed varies with the medium (dielectric medium versus > > copper medium) and the main difference between mediums is the electron [quoted text clipped - 10 lines] > constant for such a medium and use it to modify G just like k. Either > way should be valid. OK then how do you explain the following information retrieved from the web: permittivity = electric field strength / dielectric displacement ELECTRIC (electrostatic) FIELD STRENGTH terminology used instead of Force for static charges: E = k Qq/r^2 http://www.plus2physics.com/electrostatics/study_material.asp?chapter=2 QUOTE: Electric Field An electric field is a region where electric charges experience Coulombian forces. It exists in the space around a charged object. Lines of force Lines of force are a convenient way of visualizing an electric field The total number of lines of force (flux) from a charge is directly proportional to the charge (Likewise capacitance proportional to surface area of the plates, more surface = more lines of forces = more charges) MOST IMPORTANT: ******************************The total number of lines of force is inversely proportional to the permittivity of the medium in which the charge is located*********************************************************** k = 1/permittivity, as permittivity increases the FORCE OF CHARGES DECREASES (As well as capacitance decreases). > > > F= kQq/r^2 F = GMm/r^2 > > > [quoted text clipped - 13 lines] > But k will certainly have different values in different media, since it > depends on the total permittivity of the medium. Not by the definition of k I know. Can you provide a cite for this view? The electrostatic force felt by a particle q in a medium is given by the sum of kQq/r^2 over all other charges Q in that medium, using the standard, unchanging value of k. - Randy > > > > F= kQq/r^2 F = GMm/r^2 > > > > [quoted text clipped - 16 lines] > Not by the definition of k I know. Can you provide a cite > for this view? Jackson, "Classical Electrodynamics". > The electrostatic force felt by a particle q in a medium is > given by the sum of kQq/r^2 over all other charges Q > in that medium, using the standard, unchanging value > of k. All we know is that c^2/2 = k_e/k_m. c will change in different media therefore k_e and k_m will change in different media. For SI units, c = 1/sqrt(eps*mu) where eps and mu are permittivity and permeability for the media. FrediFizzx http://www.vacuum-physics.com > > > > > F= kQq/r^2 F = GMm/r^2 > > > > > [quoted text clipped - 25 lines] > > in that medium, using the standard, unchanging value > > of k. Jackson, 2nd ed, p. 29, (Chapter 1, "Introduction to Electrostatics"). "The CONSTANT k is determined by the unit of charge chosen... thus with cgs units k=1. We will use esu. ... E(x) = sum(i=1 to n) q_i * (x-x_i)/|x-x_i|^3" Now can you give me a reference within Jackson for YOUR view that the expression for electrostatic force is medium dependent? > All we know is that c^2/2 = k_e/k_m. c will change in different media > therefore k_e and k_m will change in different media. The speed of light in the medium will change in different media, and the values of k_e and k_m will change in different media. However, the value of the constant k in the expression for electrostatic force is the one based on vacuum values. - Randy > > > > > > F= kQq/r^2 F = GMm/r^2 > > > > > > [quoted text clipped - 47 lines] > However, the value of the constant k in the expression > for electrostatic force is the one based on vacuum values. The k is the same as k_e in your expression. Your expression is for the "vacuum". Not generalized for other media. It actually become much more clear in SI units. See Jackson 3rd edition. FrediFizzx http://www.vacuum-physics.com > > > > > > > F= kQq/r^2 F = GMm/r^2 > > > > > > > [quoted text clipped - 58 lines] > FrediFizzx > http://www.vacuum-physics.com It is generalized for other media. If you find the electric field inside of a dielectric, you'd better use the dielectric constant, otherwise known as the relative permittivity. > > > > > > > > F= kQq/r^2 F = GMm/r^2 > > > > > > > > [quoted text clipped - 62 lines] > inside of a dielectric, you'd better use the dielectric constant, > otherwise known as the relative permittivity. Correct. FrediFizzx Quantum Vacuum Charge papers; http://www.vacuum-physics.com/QVC/quantum_vacuum_charge.pdf or postscript http://www.vacuum-physics.com/QVC/quantum_vacuum_charge.ps http://www.arxiv.org/abs/physics/0601110 http://www.vacuum-physics.com > The k is the same as k_e in your expression. Your expression is for the > "vacuum". Not generalized for other media. Medium equations are merely rewritings of the vacuum equations (which are always valid). Maxwell's vacuum equations can equally well be applied in a medium, but it is just more convenient to consider D and H than E and B. Similarly, it is always valid, ALWAYS, that the total electrostatic force felt by a charge q is the sum of k*Q*q/r^2 over all other charges Q, including those which make up the medium, where k is the VACUUM value. Whatever equation you are looking at was derived from this. It isn't a separate equation, just one which is more convenient for charges embedded in dielectrics. > It actually become much > more clear in SI units. See Jackson 3rd edition. That's still not a reference. How about a chapter name or number, a section name or number that I can correlate to my 2nd ed so we can discuss the same equations in Jackson? Meanwhile, perhaps you want to review the derivation of the equation you're looking at. - Randy > > The k is the same as k_e in your expression. Your expression is for the > > "vacuum". Not generalized for other media. [quoted text clipped - 10 lines] > was derived from this. It isn't a separate equation, just one > which is more convenient for charges embedded in dielectrics. If your test charge q is in the linear, isotropic and uniform dielectric medium with the other charges, then it is silly to use the "vacuum" value of k. Use the dielectric eps. ;-) div E = rho/eps in SI units where eps is the electric permittivity of the medium. > > It actually become much > > more clear in SI units. See Jackson 3rd edition. [quoted text clipped - 4 lines] > that I can correlate to my 2nd ed so we can discuss the same > equations in Jackson? You won't find it in the 2nd ed at all since it is not in SI units. It is on page 154 in the 3rd ed. Section is 4.3 "Elementary Treatment of Electrostatics with Ponderable Media". But I am curious as to how he treats it in cgs. And you can also look at Griffith's "Intro. to Electrodynamics" 3rd ed. starting on page 178 thru 185. > Meanwhile, perhaps you want to review the derivation of the > equation you're looking at. Like I was saying, it is much more clear to deal with this in SI units. The force between two charges in the above describe medium in SI units will be, F = q1*q2/(4pi*eps*r^2) (that is eps and not eps0) FrediFizzx Quantum Vacuum Charge papers; http://www.vacuum-physics.com/QVC/quantum_vacuum_charge.pdf or postscript http://www.vacuum-physics.com/QVC/quantum_vacuum_charge.ps http://www.arxiv.org/abs/physics/0601110 http://www.vacuum-physics.com > > > > > > F= kQq/r^2 F = GMm/r^2 > > > > > > [quoted text clipped - 37 lines] > YOUR view that the expression for electrostatic force is > medium dependent? I don't know about jackson or micheal jackson books but I do know about logic: If the electrostatic force between charges changes by the dieletric in between them in a capacitor then it must surely change if they are with the dielectric instead of only in between. 1. is ELECTORSTATIC DIPOLE FORCE = 1 / R (dielectric thickness) in a capacitor but if the dielectric is not only IN BETWEEN (2D) but all around two charges (3D) then: 2. the ELECTORSTATIC DIPOLE FORCE = 1 / R^2 (= meaning porportional) **********1/R versus 1/R^2****** Case 1: Capacitance = 1/k * AREA /r (r = thickness of dielectric and gap between charges) Case 2: F = k Qq/r^2 I'm not sure how to calculate the electrostatic force between two opposite charges with a dielectric in between them (is it proportional to 1/r^2 or 1/r)? > > All we know is that c^2/2 = k_e/k_m. c will change in different media > > therefore k_e and k_m will change in different media. [quoted text clipped - 7 lines] > > - Randy > > > > F= kQq/r^2 F = GMm/r^2 > > > > [quoted text clipped - 23 lines] > > - Randy ? k = 1/permittivity 1. and each medium has it's own permittivity 2....as well the only difference between mediums is mostly the density of the charges in it (meaning # of charges per volume since elements are simply formed by the number of electrons and protons) 3. And as you said prior with me, the presence of other charges affects the force between two charges and their DISTANCE from these two charges (which is related to the density of the medium (density of the charges per volume ...since since elements are simply formed by the number of electrons and protons) 4. The charge force between a dielectric plates varies by the permittivity of the dielectric, therefore I presume this force of charges doesn't only behave with the dielectric medium in between it but as well if the charges are inside this medium......for what is inside...inside means the medium is all around instead of only in between the two charges such a dielectric). 5. And the permittivity is not affected by the thickness of the material for all charges are usually dispersed evenly in a medium so the permittivity is related to the density of the charges instead (# of charges per volume of material since each element only varies by the NUMBER of electron and proton charges in it (and neutrons which are neutral)). > > The electrostatic force felt by a particle q in a medium is > > given by the sum of kQq/r^2 over all other charges Q [quoted text clipped - 6 lines] > > k = 1/permittivity This is not true. k is defined in terms of the vacuum permittivity. You do not use a different value of k in different media. Remaining material deleted as it proceeds from this false assumption. - Randy > > > The electrostatic force felt by a particle q in a medium is > > > given by the sum of kQq/r^2 over all other charges Q [quoted text clipped - 16 lines] > > - Randy Well, if I asked you to calculate the electric field inside a dielectric, would you ignore the dielectric constant? > > > > The electrostatic force felt by a particle q in a medium is > > > > given by the sum of kQq/r^2 over all other charges Q [quoted text clipped - 19 lines] > Well, if I asked you to calculate the electric field inside a > dielectric, would you ignore the dielectric constant? ANOTHER POINT FOR IGOR, ANOTHER POINT FOR IGOR!! > > > > > The electrostatic force felt by a particle q in a medium is > > > > > given by the sum of kQq/r^2 over all other charges Q [quoted text clipped - 21 lines] > > ANOTHER POINT FOR IGOR, ANOTHER POINT FOR IGOR!! electostatic field, why else would a capacitor hold more charges? (hey you charges over there come and join the rest of the charges over here) > > > > The electrostatic force felt by a particle q in a medium is > > > > given by the sum of kQq/r^2 over all other charges Q [quoted text clipped - 14 lines] > Well, if I asked you to calculate the electric field inside a > dielectric, would you ignore the dielectric constant? It could be done that way, and in fact I did have to do it that way for a final exam in E & M. The whole business of dielectric constants and D fields is just another more convenient representation of the vacuum Maxwell equations. The dielectric constant is CALCULATED from the vacuum equations. The vacuum description is always correct. Always. The "dielectric constant" is merely a shorthand for those vacuum expressions. - Randy > > > > > The electrostatic force felt by a particle q in a medium is > > > > > given by the sum of kQq/r^2 over all other charges Q [quoted text clipped - 24 lines] > correct. Always. The "dielectric constant" is merely a shorthand > for those vacuum expressions. Indeed... the homgenous dielectric has no effective conductors or magnetic material to produce nearfield effects so the treatment is the same as free space. << Thus, Maxwell's equations for the propagation of electromagnetic waves through a dielectric medium are the same as Maxwell's equations for the propagation of waves through a vacuum (see Sect. 4.7), except that , where n = sqrt eps >> http://farside.ph.utexas.edu/teaching/em/lectures/node98.html Sue... > - Randy > > > > > > The electrostatic force felt by a particle q in a medium is > > > > > > given by the sum of kQq/r^2 over all other charges Q [quoted text clipped - 37 lines] > > Sue... OK then how do you explain the following information retrieved from the web: permittivity = electric field strength / dielectric displacement ELECTRIC (electrostatic) FIELD STRENGTH terminology used instead of Force for static charges: E = k Qq/r^2 http://www.plus2physics.com/electrostatics/study_material.asp?chapter=2 QUOTE: Electric Field An electric field is a region where electric charges experience Coulombian forces. It exists in the space around a charged object. Lines of force Lines of force are a convenient way of visualizing an electric field The total number of lines of force (flux) from a charge is directly proportional to the charge (Likewise capacitance proportional to surface area of the plates, more surface = more lines of forces = more charges) MOST IMPORTANT: ******************************The total number of lines of force is inversely proportional to the permittivity of the medium in which the charge is located*********************************************************** k = 1/permittivity, as permittivity increases the FORCE OF CHARGES DECREASES (As well as capacitance decreases). > > - Randy |
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