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I searched for this question but I cannot get satisfactory answer. Say if we have a parallel plate or cylindrical capacitor and if I supply unequal charge to it how could I find its Capacitance. We know $Q=CV$. I have the potential difference (i can calculate that). So what should I take the value of Q to find its capacitance?

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  • $\begingroup$ How could you calculate the potential difference without using the capacitance relation? $\endgroup$
    – Rol
    Dec 22, 2015 at 13:28
  • $\begingroup$ By electric field $\endgroup$
    – Vaibhav
    Dec 22, 2015 at 13:50
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    $\begingroup$ Very good. So, what would the electric field look like for two plates with different amounts of charge on them? $\endgroup$
    – Jon Custer
    Dec 22, 2015 at 13:57
  • $\begingroup$ I realised i was wrong...How could we have two unequal charge plates. Wouldn't Electric field pierce into the conductor which shouldn't. Correct me if I'm wrong. I am horribly Confused.. $\endgroup$
    – Vaibhav
    Dec 22, 2015 at 14:00
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    $\begingroup$ This is the same question, in essence, as: physics.stackexchange.com/q/519925 $\endgroup$ Apr 22, 2021 at 14:35

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Remember that the word "capacitance" is just that, a word. The standard definition of capacitance given in introductory textbooks only applies when the two objects have charges $Q$ and $-Q$, in which case we define $$C = \frac{Q}{V}$$ where $V$ is the potential difference between the objects. There is no inherently "correct" way to generalize this definition to unequal charges; there are just more and less useful ways. Furthermore, you never need the idea of capacitance. You can just solve for everything directly using Coulomb's and Gauss's laws. Capacitance is merely a definition, not a law.

That said, there can certainly be bad definitions. When you have unequal charges $Q_1$ and $Q_2$, it seems many people in this thread and others would like to define capacitance so that it stays the same as before, which they claim happens if you define $C = |Q_1 - Q_2| / 2 V$. However, this claim is wrong. As a simple example, consider two distant spheres of different radii $r_i$. Their pairwise capacitance is certainly nonzero, but if you give the spheres the same charge, $Q_1 = Q_2 = Q$, then $V_1 \approx k Q_1 / r_1$ and $V_2 \approx k Q_2 / r_2$. As a result, their claimed definition of capacitance yields zero!

The problem is that you can't summarize the response of two conductors to general charges with a single number; instead, you need three, which are roughly the original pairwise capacitance and the self-capacitances of each one alone. Only by using all three of these quantities can you compute the voltages. Similarly, for $n$ conductors you need to specify $\binom{n}{2} + n$ numbers.

In electrical engineering courses, this information is summarized in an elegant way. To motivate it, suppose you had $n$ conductors, which had charges $Q_i$. The charge on any one conductor affects the potentials and charge distributions on all of the other conductors in a complicated way. However, the equations governing the system are still linear, which means the superposition principle works. In particular, that means the potentials $V_i$ of the conductors in the general case can be found by adding together the potentials you get by only charging the first conductor, and then only charging the second conductor, and so on, assuming the potential is set to zero at infinity.

Therefore, the $Q_i$ and $V_i$ are always related by a linear transformation, and we define $$Q_i = \sum_j C_{ij} V_j.$$ The matrix of $C_{ij}$ is called the Maxwell capacitance matrix. For example, in the special case where there are only two conductors with opposite charges, you can show that the "usual" simple definition of pairwise capacitance is related to these coefficients by $$C = \frac{C_{11} C_{22} - C_{12}^2}{C_{11} + C_{22} + 2 C_{12}}.$$ Furthermore, it can be shown that $C_{ij} = C_{ji}$, and that the usual self-capacitances are the $C_{ii}$. For more about this, see section 3.6 of Purcell and Morin, Electricity and Magnetism.

Once again, you never need this idea to solve problems, because it is derived from more basic things, such as Gauss's law. But it's the most useful way to parametrize things in certain contexts.

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  • $\begingroup$ Why did you say electrical engineering way? What other ways are there to introduce this topic? $\endgroup$
    – Babu
    Apr 29, 2021 at 10:18
  • $\begingroup$ @Buraian I've edited to expand the answer. Actually, the electrical engineering way is the only one I've seen that works for a general number of conductors. Of course, if you start doing things in more depth (such as caring about time-dependent situations with electromagnetic waves) you can instead just abandon the idea of capacitance entirely and work directly with Maxwell's equations. $\endgroup$
    – knzhou
    Apr 29, 2021 at 15:33
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If you have a capacitor where one plate has a charge $Q_1$ and the other has a charge $Q_2$, you can think of this as a capacitor with a differential charge $Q_1-Q_2$ that has an additional charge $\frac{Q_1+Q_2}{2}$ added to both plates.

Now the voltage difference between the two plates is given by the differential voltage, while the entire capacitor would be at a non-zero potential due to the net charge on the entire capacitor.

The relationship would then be

$$C = \frac{Q_1-Q_2}{2V}$$

Why the factor $2$? Well - the total difference in charge is $Q_1-Q_2$; but normally, when a capacitor is charged with $Q$, that means one plate is at $Q$ and the other plate at $-Q$. So the differential charge would be $2Q$...

Alternatively, you can compute the capacitance from the geometry - but you were not given enough information to do that (and that calculation is fraught with complications).

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  • $\begingroup$ Thanks for the answer. Actually I'm trying to solve a problem of three concentric cylindrical shells having radii R, 2R and 2√2 R. The innermost and outermost shell is connected and I have to find capacitance Across middle and inner shell. I assumed charge -Q on 2nd and Q on first shell then I equated potentials of first and third shell which give me Q/3 Charge on first shell and 2Q/3 on outermost. Now I tried as you have said by using differential charge and tried calculating capacitance but I am not able to get at answer. Any hints? $\endgroup$
    – Vaibhav
    Dec 22, 2015 at 14:19
  • $\begingroup$ That's a very different question from the one you posed. Why don't you ask the question you are actually trying to answer... It comes down to having to find the charge distribution on the plates and the fields between them (which you integrate to get voltage) $\endgroup$
    – Floris
    Dec 22, 2015 at 14:22
  • $\begingroup$ I have distributed the charge. I now have two cylindrical shell with charge -Q and Q/3. I too have calculated the potential difference between them. What next?Now I only have to find capacitance which I couldn't so that's why I have asked this. Any Help will be appreciated! $\endgroup$
    – Vaibhav
    Dec 22, 2015 at 15:13
  • $\begingroup$ If the first and third shells are connected, it makes sense to say that the sum of their charges is $Q$. And then the problem should be much simpler. $\endgroup$
    – Floris
    Dec 23, 2015 at 18:29
  • $\begingroup$ So can I directly apply then $Q=CV$ to find Capacitance. Anyway I'd done like that and got the answer. I just want to make sure is is correct? $\endgroup$
    – Vaibhav
    Dec 24, 2015 at 5:28
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What would be the capacitance of an unequal charged capacitor?

The same as the equally (and opposite) charged capacitor. Capacitance is a geometrical property.

...if we have a parallel plate or cylindrical capacitor and if I supply unequal charge to it ...

... then this will happen. For three cylindrical plates the outermost one plays the role of the "environment". But, in accordance with Floris, that's another question.

what should I take the value of Q to find its capacitance?

The increase in charge of the more-positive plate due to the increase in potential difference between that and another plate.

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    $\begingroup$ Could you please explain your last sentence with an example $\endgroup$
    – Vaibhav
    Dec 22, 2015 at 19:15
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This is an interesting question but is asked in a rather vague way as the context given is limited.

The standard definition of capacitance $C=\frac QV$ is useful because in many examples the capacitance is a constant and so $Q \propto V$. However there is the possibility of defining an incremental capacitance $C=\frac {\Delta Q}{\Delta V}$ when the charge stored is not proportional to the potential difference in the same way increment resistance is defined for non-ohmic circuit elements.

The analysis for a parallel plate arrangement is not quite the same as that for a concentric cylinder arrangement in that in the latter case the outer conductor shields the inner conductor.

Let's consider an ideal parallel plate capacitor made up of two parallel plates with charges $Q$ and $q$ on them.
The charges on the plates distribute themselves as shown in the diagram below.

enter image description here

The final arrangement of charges ensures that all electric field lines which start on one plate finish on the other plate - field lines start on a positive charge and finish on a negative charge. Also the equal charge on the outside of the plates is the arrangement which minimise the energy stored in the electric field outside the plates.

This capacitance of the arrangement relating to the inner charges is the usual $C=\frac {\epsilon_0 A}{d}$.
You can show this by the electric field equals charge over (area $(A)$ \times permittivity $\epsilon$) and the electric field is equal to the potential difference divided by the separation of the plates $(d)$.

Now let's connect this charged capacitor to a resistor.

What happens?

The capacitor discharges (charges of magnitude $\frac {Q-q}{2}$ neutralise one another) with a time constant of $CR$ and left behind are the $\frac {Q+q}{2}$ charges on the outside of the plates.

So what about adding a charge $\Delta q$ to the plate which had charge $q$ on it?
If you go through the analysis as shown above the capacitance stays the same but now there is a charge $\frac{Q+q+\Delta q}{2}$on each of the outer sides of the plates.

It would appear that outer sides of the plates do not contribute to the capacitance but that cannot be so because adding charge $\Delta q$ has raised the potential of the outer two plates relative to infinity.
This complication is not easy to deal with but the post Capacitance of a single charged plate? gives an insight on how one proceed.


The cylindrical case is different because the inner conductor has only one surface.
Also changing the charge on the outer conductor does not influence the charge stored on the inner conductor.
Again the capacitance is determined by the geometry of the arrangement but now if there is a charge of $Q$ on the inner cylinder outer surface there must be a charge of $-Q$ on the inner surface of the outer cylinder and the residual charge, $q+Q$, on the outer surface of the outer cylinder which to the first approximation does not really affect the capacitance of the arrangement.


Finally your three concentric cylindrical shells having radii R, 2R and 2√2 R with innermost and outermost shell connected can be thought of as two capacitors in parallel, inner conductor and inner surface of middle conductor, and inner surface of outer conductor and outer surface of middle conductor.

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This is a frequently asked question by college physics students.

Since the situation in the question deviates from the usual definition and use of capacitors, we will need to generalize our definition of capacitance somewhat. Here, a useful generalization is the differential capacitance. Differential capacitance is the derivative, $C = dQ/dV$, and is normally used when the charge and voltage have a nonlinear relationship. An example of such a nonlinear relationship might be to use a flexible dielectric material between the plates that compresses so that when an electric potential is applied the plates are brought closer together thereby changing the geometric parameters. The differential capacitance is used in this answer because it is well-defined for the question at hand.

Sanity check: Neutral capacitors have a geometrically defined capacitance even when there is no charge. Whatever we define here as the capacitance should reproduce this result when the two charges vanish.

Quick answer: The differential capacitance of non-neutral capacitors is the same as neutral ones.

The general reason for this is simple. The voltage in this capacitor still has a linear relationship with the amount of charge on the plates. Therefore the differential capacitance is constant, geometrically defined, and independent of the existing charge on the capacitor.

Proof for parallel plate capacitor

Without loss of generality, let $Q_1 \le 0 \le Q_2$ be the charges stored on the plates of a parallel plate capacitor. Using Gauss's law, we know the electric field strength immediately surrounding a single charged electric plate with a surface area $A$ is $$E={Q\over 2\epsilon_0 A}$$ so the electric field strength between two plates of the same surface area with different charges, $Q_1$ and $Q_2$ is $$E={Q_2 - Q_1 \over 2 \epsilon_0 A}$$ The electric potential increase going from the negatively charged plate to the positively charged plate separated a distance $l$ is $$V={Q_2 - Q_1 \over 2 \epsilon_0 A}l$$ We see the linear relationship between potential and charge here. To check, and compute the derivative $dQ/dV$, let's remove a small additional amount of charge, $\Delta Q>0$, from the negative plate and put it on the positive plate to raise the potential an amount, $\Delta V$. This change in potential is $$\Delta V={\left( Q_2 + \Delta Q \right) - \left( Q_1 - \Delta Q \right) \over 2 \epsilon_0 A}l−{Q_2 - Q_1 \over 2 \epsilon_0 A}l={\Delta Q \over \epsilon_0 A}l$$ Thus the differential capacitance $C = \Delta Q / \Delta V = \epsilon_0 A / l$ is the same geometric factor as the capacitance of a neutral capacitor.

Note about circuits

If a circuit makes use of a non-neutral capacitor, then the potential over this component used in Kirchhoff's Voltage (loop) law can be written as $V=\left( Q_2 - Q_1 \right) / 2C + Q/C$, where Q is any additional charge put onto the capacitor during the operation of the circuit. This can also be written as $V=V_0 + Q/C$, where $V_0=\left( Q_2 - Q_1 \right) / 2C$ is the initial potential in the capacitor due to the initial charges. You can expect some abnormal behavior from such a circuit due to the fact it is not electrically neutral. For example, a simple RC circuit will still discharge until the potential difference and electric field between the plates vanishes, but there will be a residual charge $\left( Q_2 + Q_1 \right)/2$, now equal (not opposite) on each plate. At the end of the discharge, the potential through each component of the RC loop will be zero, but the potential between the loop and ground will be non-zero and it may end up shocking someone who touches any part of the circuit.

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  • $\begingroup$ Hmm fair this makes sense, but could you provide a source for further reading on this idea? $\endgroup$
    – Babu
    Apr 29, 2021 at 10:16
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Capacitance does not depend on the charge you put on it. Just as the resistance of a resistor does not depend on the current flowing through it. Capacitance is a measure of "how much charge a system can hold before it reaches breakdown voltage". It solely depends on the geometry of the system.

To adress another issue, when the two plates of a capacitor is inequally charged the relation Q=CV does not hold if you are trying to find capacitance.

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    $\begingroup$ Sorry, that's just plain wrong. Capacitance is the proportionality factor between charge and voltage. It has absolutely nothing to do with the breakdown voltage. $\endgroup$
    – Hilmar
    Dec 22, 2015 at 13:50
  • $\begingroup$ So how could I find capacitance in that case? Is there any other method? $\endgroup$
    – Vaibhav
    Dec 22, 2015 at 13:53
  • $\begingroup$ And depends on the dielectric constant of the material (vacuum counts as a 'material' here) between the plates. Yes, geometry is important, but not everything (and geometry can certainly impact the observed breakdown voltage, but other factors come in to play as well). $\endgroup$
    – Jon Custer
    Dec 22, 2015 at 13:55
  • $\begingroup$ @Vaibhav The capacitance of a parallel plate capacitor is C=ϵ0A/d. The larger the area, the grater is the capacitance. The smaller the gap between the plates, the greater the capacitance. The capacitance has nothing to do with the charges you put on. The formula Q = CV indicates only that the potential diifference between the plate is directly proportional to the charges on the plates. $\endgroup$ Apr 28, 2021 at 7:57
  • $\begingroup$ I am surprised why this gets a negative like for the correct answer! $\endgroup$ Apr 28, 2021 at 8:00

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