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Natural Science Forum / Physics / Relativity / June 2006Why does Capacitance decrease with "r" but F decreases with "r^2"?
Wheter the capacitor's dielectric is space itself or another material, why does Capacitance decrease with "r" but F decreases with "r^2"? It would seem logical that if the attractive force between the charges SUBSTANTIALLY decreases by r^2 then it doesn't seem logical that the capacity to retain these charges(capacitance) in a capacitor decrease only by r (even if space is the dielectric)? F= kQq/r^2 (and capacitor's energy field: E = KQq/r^2) where as Capacitance = Area/ (k * r) > Wheter the capacitor's dielectric is space itself or another material, > why does Capacitance decrease with "r" but F decreases with "r^2"? [quoted text clipped - 6 lines] > F= kQq/r^2 (and capacitor's energy field: E = KQq/r^2) where as > Capacitance = Area/ (k * r) << These formulae are valid for any type of capacitor, since the arguments we used to derive them do not depend on any special property of parallel plate capacitors. Where is the energy in a parallel plate capacitor actually stored? Well, if we think about it, the only place it could be stored is in the electric field generated between the plates. This insight allows us to calculate the energy (or, rather, the energy density) of an electric field. Consider a vacuum-filled parallel plate capacitor ... >> http://farside.ph.utexas.edu/teaching/302l/lectures/node34.html Sue... > Where is the energy in a parallel plate capacitor actually > stored? Well, if we think about it, the only place it could be [quoted text clipped - 3 lines] > Consider a vacuum-filled parallel plate capacitor ... >> > http://farside.ph.utexas.edu/teaching/302l/lectures/node34.html The concept of a stored 'field energy' is actually flawed. The energy balance is fully taken into account by considering the potential and kinetic energy of the charges in the capacitor. If no current is flowing there is just potential energy. Otherwise the latter is gradually turned into kinetic energy of the current which in turn is then lost from the system through resistive losses. Thomas > > Where is the energy in a parallel plate capacitor actually > > stored? Well, if we think about it, the only place it could be [quoted text clipped - 10 lines] > gradually turned into kinetic energy of the current which in turn is > then lost from the system through resistive losses. When you speak of the energy of a charge I expect to see some numbers like 0.511 MeV. Conductor resistance is not a part of the equation. I fear you are confusing the electric field which moves along the surface at the speed of light, with the motion of the loose electrons within the conductor. They are found at accelerator laboratories when moving near the speed of light. Sue... > Thomas > When you speak of the energy of a charge I expect to see > some numbers like 0.511 MeV. What's the mass of the electron have to do with it? > > When you speak of the energy of a charge I expect to see > > some numbers like 0.511 MeV. << What's the mass of the electron have to do with it? >> It should have nothing to do with it. I was perhaps a bit to subtile in responding to a suggestion we work with the kinetic and potential energy of the charge. Looking over the reference again: http://farside.ph.utexas.edu/teaching/302l/lectures/node34.html ...the poster may have seen the word potential and misread the kappa symbol as kinetic energy. Anyway, I think I made the point we that we see few electrons outside of an accelerator whose mass or kinetic energy concerns us. Sue... Thomas Smid: >> Where is the energy in a parallel plate capacitor actually >> stored? Well, if we think about it, the only place it could be [quoted text clipped - 5 lines] > >The concept of a stored 'field energy' is actually flawed. Actually, it's your understanding that's flawed. >The energy >balance is fully taken into account by considering the potential and >kinetic energy of the charges in the capacitor. If no current is >flowing there is just potential energy. >Otherwise the latter is gradually turned into kinetic energy of the >current which in turn is then lost from the system through resistive >losses. Uh, the potential energy _is_ the stored field energy. What do you think causes the leakage current to flow? The electric field. > Uh, the potential energy _is_ the stored field energy. The total energy of a system of charged particles consists of their potential energy (as given by their Coulomb interaction) + their kinetic energy. If you would additionally assume a 'field energy' equal to the potential energy you would violate energy conservation exactly by this amount. Thomas > The total energy of a system of charged particles consists of their > potential energy (as given by their Coulomb interaction) + their > kinetic energy. This is plain and simply not true. If some of those charges are accelerated, EM radiation will carry energy and momentum away, and for energy and momentum conservation to hold they must be included. For an _isolated_ set of charged particles their total kinetic energy will diminish asymptotically to some value, because of this radiation; if the system is bound the ultimate value of the kinetic energy will be zero (relative to the final c-o-m frame). This is classical electrodynamics. Quantum effects make a profound difference.... > If you would additionally assume a 'field energy' equal > to the potential energy you would violate energy conservation exactly > by this amount. Hmmm. The total energy in the EM field is more than just the Coulomb potential energy for the charged particles -- see above. Note that the energy in the EM field _includes_ the Coulomb potential energy, so you must not double count. Tom Roberts Thomas Smid: >> Uh, the potential energy _is_ the stored field energy. > [quoted text clipped - 3 lines] >to the potential energy you would violate energy conservation exactly >by this amount. Uh, this is a simple electrostatics problem. Let me direct you to sears & zemansky or halliday & resnick. Look up ``Capacitor, energy stored in,'' in the index. > Thomas Smid: > > [quoted text clipped - 9 lines] > Uh, this is a simple electrostatics problem. Let me direct you > to sears & zemansky or halliday & resnick. Look up ``Capacitor, I went to sears and all they had was clothes, they laughed when I asked for capacitors. They never heard of zemansky, only simpson & sears. Is Bige Bald? > energy stored in,'' in the index. guskz@hotmail.com, who is beneath broken toasters on the evolutioanry ladder: >> Thomas Smid: >> > [quoted text clipped - 12 lines] >I went to sears and all they had was clothes, they laughed when I asked >for capacitors. Why are you surprised? Everyone laughs at you. >They never heard of zemansky, only simpson & sears. > >Is Bige Bald? > >> energy stored in,'' in the index. > guskz@hotmail.com, who is beneath broken toasters on the evolutioanry ladder: > [quoted text clipped - 17 lines] > > Why are you surprised? Everyone laughs at you. Ok then you go to sears and ask for capacitors or zemansky and see if they don't laugh at you. Why are you bald? > >They never heard of zemansky, only simpson & sears. > > [quoted text clipped - 3 lines] > >> energy stored in,'' in the index. > > > > Wheter the capacitor's dielectric is space itself or another material, > > why does Capacitance decrease with "r" but F decreases with "r^2"? [quoted text clipped - 19 lines] > > Sue... It doesn't explain the question? what it does at the end is w=cv^2/2 to give a distance "r" instead of "r^2". It doesn't explain why c (capacitance) and v( potential difference) are only affected by distance "r" instead of "r^2" like F and the Energy field? Actually it says the Electric field = k Q/A where as other locations E = kQq/r^2? > > > Wheter the capacitor's dielectric is space itself or another material, > > > why does Capacitance decrease with "r" but F decreases with "r^2"? [quoted text clipped - 29 lines] > Actually it says the Electric field = k Q/A where as other locations > E = kQq/r^2? Spherical capacitors and granite slabs halfway to the earth's centre are both pretty rare items but see if that visual and the later exchange with Thomas Smid, will iron out where your question is really just a matter of the geometry you are assuming. diatance 1 area 2 volume 3 Your expected exponent is off by one so your mental image condsiders too few or too many dimensions somewhere. Sue... BTW the slab is weighless at the earths centre. > > > > Wheter the capacitor's dielectric is space itself or another material, > > > > why does Capacitance decrease with "r" but F decreases with "r^2"? [quoted text clipped - 39 lines] > area 2 > volume 3 How come the force between charges is AREA since Q & q are linearly distant (= line) from each other where as the potential difference force (V) is a line of force instead of an AREA of force? Even if you had only one Q & q charge in total (instead of a plurality of charges) the potential difference force (V) across Space (as a dielectric) is a line of force (meaning your "distance 1" above)? > Your expected exponent is off by one so your > mental image condsiders too few or too many > dimensions somewhere. > > Sue... > BTW the slab is weighless at the earths centre. > > > > > Wheter the capacitor's dielectric is space itself or another material, > > > > > why does Capacitance decrease with "r" but F decreases with "r^2"? [quoted text clipped - 43 lines] > distant (= line) from each other where as the potential difference > force (V) is a line of force instead of an AREA of force? The force is newtons. LOL See the red interaction regions here: http://newton.umsl.edu/~philf/trich_ex.gif The stubby horizontal 'weiner' represents a volume. When you slice the weiner in half you get a cross sectional 'area' that quantifies the interaction of the two charges. > Even if you had only one Q & q charge in total (instead of a plurality > of charges) the potential difference force (V) across Space (as a > dielectric) is a line of force (meaning your "distance 1" above)? We speak of the force as acting along a *line of motion*. But the charge interaction occurs in a 'volume' of space. Slicing the volume quantifies the interaction as an 'area'. where we might count imaginary flux lines whose quantity would diminish by 1/r^2. ( 2 is exponenet for area Eh? ) You remember some visuals where the flux lines are ~countable~. O + (()) O - Move the charges apart in free space: \ / O + () O - / \ Two of the flux lines found other partners out in the 'sea of charges' that represents the ~vacuum~. Image from: http://newton.umsl.edu/~philf/triplet.html Easier to visualize the volumes of space that are operative, than is illustrated here: http://hyperphysics.phy-astr.gsu.edu/Hbase/electric/dipole.html#c2 As the advertisment says, it is easier to read in color. Sue... > > Your expected exponent is off by one so your > > mental image condsiders too few or too many > > dimensions somewhere. > > > > Sue... > > BTW the slab is weighless at the earths centre. > > > > > > Wheter the capacitor's dielectric is space itself or another material, > > > > > > why does Capacitance decrease with "r" but F decreases with "r^2"? [quoted text clipped - 89 lines] > > Sue... Ok I got mixed-up with forces and energy (sometimes called a force itslef: electromotive force EMF or volts or force per unit charge). The electrical potential = V = Fd (gives 1/r^1) where as the electrical field = F/q = (gives 1/r^2)= V/r So an Electrical field is a force field and not an energy field. Still strange that Energy and Power Capacity (capacitance) only decrease by a thousanth (1/r^1) when the total force that generates them decreases by a millionth (1/r^2) > > > Your expected exponent is off by one so your > > > mental image condsiders too few or too many > > > dimensions somewhere. > > > > > > Sue... > > > BTW the slab is weighless at the earths centre. > > > > > > > Wheter the capacitor's dielectric is space itself or another material, > > > > > > > why does Capacitance decrease with "r" but F decreases with "r^2"? [quoted text clipped - 97 lines] > > So an Electrical field is a force field and not an energy field. Well... It took some energy to build the fundamenal particles and some energy to polarize the field. But we don't have to to consider those after they are formed. So... yes. It is a static field where every thing remains in equilibrium 'till some experimenter wiggles something. > Still strange that Energy and Power Capacity (capacitance) only > decrease by a thousanth (1/r^1) when the total force that generates > them decreases by a millionth (1/r^2) Ugggh. That is rather a tough word/symbol salad to swallow. Unless you are building some capacitors for your video monitor I'll avoid that side excursion. :o) Thinking about 'Power Capacity' sounds like Maxwell's mistake of assuming a 'displacement current'. Let's not go there. Correctly modeled 'free space' will radiate light, instead of sloshing liquid. BTW, I don't have any association with University of Texas. I favor the web site because I know it is maintained for consistancy and gets a lot of testing from soon-to-be plasma and fusion researchers. http://farside.ph.utexas.edu/teaching/302l/lectures/node34.html The author choose to derive the space in terms of energy because he knows from insight and experience that approach will produce the most generalized expressions. That is what we value so that we are not playing see-saw with eps and mu. I realise a model that tracks individual charge fields might seem more descriptive, but from Maxwell's experience, that approach will leave us in search of an electrcal fluid that can't exist instead of the EM coupling structure that we are measuring in the lab. Sue... > > > > Your expected exponent is off by one so your > > > > mental image condsiders too few or too many > > > > dimensions somewhere. > > > > > > > > Sue... > > > > BTW the slab is weighless at the earths centre. > > Wheter the capacitor's dielectric is space itself or another material, > > why does Capacitance decrease with "r" but F decreases with "r^2"? [quoted text clipped - 8 lines] > > << These formulae are valid for any type of capacitor, Wrong! Learn some basic electronics. > > > Wheter the capacitor's dielectric is space itself or another material, > > > why does Capacitance decrease with "r" but F decreases with "r^2"? [quoted text clipped - 10 lines] > > Wrong! Learn some basic electronics. > > > Wheter the capacitor's dielectric is space itself or another material, > > > why does Capacitance decrease with "r" but F decreases with "r^2"? [quoted text clipped - 10 lines] > > Wrong! Learn some basic electronics. If you think Universiy of Texas is teaching bad capacitors then unsnip the URL and cite the equation number. Sue... > > > > Wheter the capacitor's dielectric is space itself or another material, > > > > why does Capacitance decrease with "r" but F decreases with "r^2"? [quoted text clipped - 15 lines] > > Sue... But the formula for parallel plate capacitance is not true for all capacitors. And I don't understand what the University of Texas has to do with it. But then again, you love to spew out non sequitors. > > > > > Wheter the capacitor's dielectric is space itself or another material, > > > > > why does Capacitance decrease with "r" but F decreases with "r^2"? [quoted text clipped - 19 lines] > capacitors. And I don't understand what the University of Texas has to > do with it. But then again, you love to spew out non sequitors. Perhaps English in not your first languge. 'These' is plural... more than one. I have found most schools are very appreciatvie if you report errors on their web pages. That site is a part of several active degree programs and is constantly being improved. http://www.as.utexas.edu/ Which equation is missing and where would you suggest they put it? http://farside.ph.utexas.edu/teaching/302l/lectures/node34.html Sue... > > > > > > Wheter the capacitor's dielectric is space itself or another material, > > > > > > why does Capacitance decrease with "r" but F decreases with "r^2"? [quoted text clipped - 34 lines] > > Sue... The post was a direct response to a comment about parallel plate capacitors. Can you tell me how such formulae are derived? I seriously doubt it. You need to stop hiding behind web links and learn some actual physics. > > > > > > > Wheter the capacitor's dielectric is space itself or another material, > > > > > > > why does Capacitance decrease with "r" but F decreases with "r^2"? [quoted text clipped - 39 lines] > seriously doubt it. You need to stop hiding behind web links and learn > some actual physics. I suspect you would be perfectly happy if I offered references that could be weasel-worded into a twins conundrum. Thank heavens they are getting harder to find on web sites run by acedemia. Sue... > > > Wheter the capacitor's dielectric is space itself or another material, > > > why does Capacitance decrease with "r" but F decreases with "r^2"? [quoted text clipped - 10 lines] > > Wrong! Learn some basic electronics. Apparently you didn't look at the web page Sue was citing. Looks OK to me. "These formulae" are ones on the web page. FrediFizzx http://www.vacuum-physics.com > Wheter the capacitor's dielectric is space itself or another material, > why does Capacitance decrease with "r" but F decreases with "r^2"? I think you are getting something wrong here: the electric field E of a plate is actually constant for distances small compared to its size. Now the capacitance C is defined as Q/V where Q is the total charge and V the potential. On the other hand, V is (for a constant electric field) given by E*d (where d is the distance), and thus C=Q/(E*d) . Since E is constant, this means that C decreases like 1/d as you mentioned. Thomas > > Wheter the capacitor's dielectric is space itself or another material, > > why does Capacitance decrease with "r" but F decreases with "r^2"? [quoted text clipped - 9 lines] > > Thomas Yes... the 1/r^2 is the geometric dilution (spherical), we use out in free space for either gravity or Coulomb force. The 1/r^1 geometric dilution is used below the surface for gravity and can be seen similary appropriate for a confined volume as between capacitor plates. Sue... > > Wheter the capacitor's dielectric is space itself or another material, > > why does Capacitance decrease with "r" but F decreases with "r^2"? [quoted text clipped - 9 lines] > > Thomas Ok, I think I read theat V (=Ed) is also the voltage and "potential force" (= total force)?, if so it's strange that it varies by "r" where as the force and Energy between charges decreases by "r^2" (F=E = kQq/r^2)? > > > Wheter the capacitor's dielectric is space itself or another material, > > > why does Capacitance decrease with "r" but F decreases with "r^2"? [quoted text clipped - 14 lines] > as the force and Energy between charges decreases by "r^2" (F=E = > kQq/r^2)? No, your really confused here. The force between two parallel plates is constant for small separations. This is because the electric field is a constant for that configuration. You should be asking why that field for the parallel plate capacitor is constant and the field due to a point charge fall of as the square of the distance. And the answer would be Gauss's law. > Wheter the capacitor's dielectric is space itself or another material, > why does Capacitance decrease with "r" but F decreases with "r^2"? [quoted text clipped - 6 lines] > F= kQq/r^2 (and capacitor's energy field: E = KQq/r^2) where as > Capacitance = Area/ (k * r) That's just the formula for a parallel plate capacitor. And in each of your formulas, the r means virtually different things. In your E equation, r is the radial coordinate, in your C formula, it's the distance between the plates. They're completely different concepts. In general, the capacitance is defined as the charge q divided by the potential difference V. For a parallel plate V = E d = k q d/A, so the Capacitance is C = q/V = A/kd. For different types of capacitors, we would have different formulas for C, but it is always determined by the geometry. > > Wheter the capacitor's dielectric is space itself or another material, > > why does Capacitance decrease with "r" but F decreases with "r^2"? [quoted text clipped - 14 lines] > potential difference V. For a parallel plate V = E d = k q d/A, so > the Capacitance is C = q/V = A/kd. For different types of Strange I think the A for E is r^2 so A is the radial distance from q where as for C the A is the Area of the plates which is perpendicular to "r" (since C=Q/V means only the ellimination of Q). I thought I read tha V is also know as a total potential force (volts), strange that this force is related to "r" where as the force between charges and the electric field's Energy is "r^2" ? ( F = E = kQq/r^2) > capacitors, we would have different formulas for C, but it is always > determined by the geometry. > > > Wheter the capacitor's dielectric is space itself or another material, > > > why does Capacitance decrease with "r" but F decreases with "r^2"? [quoted text clipped - 24 lines] > charges and the electric field's Energy is "r^2" ? ( F = E = > kQq/r^2) I thought I read that apples were bananas and peaches were pears. When you ask a question. it helps to know a little of what you are asking about. I don't know where to begin to address your ignorance in this response, so I won't even start. |
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