Capacitance of a single charged plate?
Lets say we have a single plate that has a charge of $+Q$ on it. A plate with charge $-Q$ is infinite distance away. Will the plate with $+Q$ have a capacitance associated with it? Why or why not? I was thinking that because the distance between the plates is infinity, the capacitance is zero, but according to my TA, that is not the case. There is always some sort of "absolute" capacitance for the plate with $+Q$ (a capacitance will exist between the plate and air?). Can someone help me out in understanding this?
electrostatics charge potential capacitance
John BurgermanJohn Burgerman
Consider the case of a conducting sphere with no counter electrode.
The capacitance of the conducting sphere can be found with
$$C=Q/V$$
and we know that at a distance $r$ from the centre of the sphere the potential $V$ is given by
$$V={1 \over 4\pi\epsilon_0}{Q\over r}$$
provided that $r$ is greater than the radius of the sphere. Now if we consider the potential at the surface and subsititute into the equation above we get for a conducting sphere
$$C= 4\pi\epsilon_0 r$$
with no counter electrode and now $r$ is the radius of the sphere.
Now the plate that you are talking about is not a sphere, but still will have some similar, but more complicated expression for the capacitance.
Thus any conductor with a charge on it generates some potential even without a counter electrode and we can calculate the the capacitance with $Q/V$.
$\begingroup$ $V=\frac{Q}{4\pi €_○ r}$ should give the potential at a point at distance r from the surface of sphere in stead of its centre due to the formula derivation. But I have seen everywhere the $r$ is taken as the distance wrt center and not surface of charged sphere. why? $\endgroup$ – user104909 Dec 8 '16 at 12:40
$\begingroup$ Oh it is due to Gauss' law, I see. Izzenit? $\endgroup$ – user104909 Dec 8 '16 at 13:03
$\begingroup$ @SufyanNaeem - yes - gauss' law works really well in a spherically symmetrical situation like this where we consider an isolated charged sphere $\endgroup$ – tom Dec 8 '16 at 20:44
Tom's answer is completely correct, but I wanted to add some detail to it.
First, why is it that your logic fails? The answer is that the familiar formula for the capacitance of two parallel plates relies on the approximation that the electric field between the two plates is completely uniform. In such a case, we have $E = V/d$ and $E = Q/\epsilon_0 A$; setting the two expressions equal to each other and solving for $C = Q/V$ yields the familiar formula for the capacitance of a pair of parallel plates.
However, if we take the two plates and start to move them apart, then at some point our approximations about the electric field are going to fail; the electric field will no longer be uniform, and it will definitely not be related to the charge on the plates in the above-mentioned way. Usually this approximation will really start to fail when $d$ gets to be a significant fraction of the size of the plates; once this happens, then the parallel-plate formula is a bad approximation. To put it another way, the familiar parallel-plate capacitance formula was derived on the assumption that $d \ll \sqrt{A}$, and therefore you can't take its limit as $d \to \infty$ and still expect it to hold.
So imagine moving the two discs a huge distance apart. At some point, they're going to be sufficiently far apart that they're not going to "feel" each other any more, and at that point each plate acts like an independent charged conductor sitting alone in space. But there's still going to be a potential difference between each conductor and infinity, and at that point we can define a capacitance of this lone conductor and another "electrode at infinity". Since we usually define $V = 0$ at infinity, we just need to calculate the potential of the conductor when it carries a charge $Q$, and take the ratio of $Q/V$ again. This is the process described by @tom in his answer.
Now, for an arbitrarily shaped conductor, you're going to have trouble getting an exact result for $C$. @tom has done the sphere, which is the simplest result. However, it turns out to be possible to do this for an infinitesimally thin conducting disc as well! To do this, we view the disc as the limit of an ellipsoid as one of the principal axes goes to zero. The charge density on the surface of an ellipsoid can be found in Section 5.02 of Smythe's Static and Dynamic Electricity; the results are also cited in a problem at the end of Chapter 2 of Griffiths (4th edition). When we take the limit as one of the principal axes goes to zero, the result is that the surface charge density is $$ \sigma(s) = \frac{Q}{2 \pi R} \frac{1}{\sqrt{R^2 - s^2}} $$ where $s$ is the distance from the center, $Q$ is the total charge on the disc, and $R$ is its radius. We can then find the potential on the disc pretty easily; just split the disc up into concentric rings, calculate the potential at the center of each one, and integrate. The result, after some math, is $$ V = \frac{Q}{8 \epsilon_0 R} $$ and we therefore have $C = Q/V = 8 \epsilon_0 R$.
Michael SeifertMichael Seifert
Both good answers. Note that the Earth has a self-capacitance by the same arguments. The Earth carries a net charge and really does have a capacitance. Remember that capacitance is defined in terms of the work that you must do to take a charge from one plate to the other. Two charged objects are involved. Even if the plates are nearly an infinite distance apart you still will have to do some work to, say, move a positive charge from the negative plate to the positive plate. The electric field due to the destination plate's positive charges will produce a repelling force that you will have to work against to move the test charge. This after all what we mean by potential difference. That is, work done to move a unit charge between two points. Very close to either charged plate the electric field only depends upon the charge density on that part of the plate and is not directly dependent upon the shape or location of the second plate. The proximity of the second plate may affect the value of this charge density. In idealized parallel plate capacitors the field happens to be uniform throughout between the plates. But this local field is always there in all geometries and results in a finite capacitance. C=Q/V. You can always calculate self-capacitance of anything by simply moving a second plate, or rock, or sphere to infinity and do the work calculation.
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