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In Mr. Purcell's Electricity and Magnetism, page 103, it is stated,

An isolated conductor carrying a charge $Q$ has a certain potential $\phi _{0}$, with zero potential at infinity. $Q$ is proportional to $\phi _{0}$. The constant of proportionality depends only on the size and shape of the conductor. We call this factor the capacitance of that conductor and denote it by C. $$Q=C \phi _{0}$$

I understand that for a given charge $Q_{0}$ and its corresponding potential $\phi_{0}$ we could define a $C_{0}$ as a function of the shape and size of the conductor such that $Q _{0} =C_{0}\phi_{0}$.

When we change the charge to $Q_{1}$, the potential will become $\phi_{1}$. How can we prove that it is the same constant $C_{0}$ that will link $Q_{1}$ and $\phi _{1}$ ? In other words is charge being linearly proportional to potential an experimental result or can we prove it?

If one argues that it is the same constant because it depends only on the shape and size of the conductor, then they must also prove that this constant does satisfy $$Q=C _{0} \phi$$ for every given charge and its corresponding potential.

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  • $\begingroup$ From Coulomb's law, we know that the electric field is linearly proportional to the net charge. Then, as $\mathbf{E}=-\nabla\Phi$, the electric potential must also be linear with the net charge. The capacitance $C_0$ is, in this scenario, the same for different charges by definition - it only depends on the conductor's geometry. $\endgroup$
    – Woe
    Commented Aug 31, 2019 at 0:12
  • $\begingroup$ @BrunoAnghinoni " the electric field is linearly proportional to the net charge" which formula backs this statement? and if $\nabla \phi$ is proportional to the net charge, why would $\phi$ be proportional too? $\endgroup$
    – Hilbert
    Commented Aug 31, 2019 at 17:01
  • $\begingroup$ Coulomb's law backs the statement, like I said (en.wikipedia.org/wiki/Coulomb%27s_law). Besides, the net charge is not affected by the (spatial) divergence operator - then, if $\nabla\Phi$ is linear in $Q$, so will be $\Phi$, after any necessary integrations. $\endgroup$
    – Woe
    Commented Sep 1, 2019 at 16:02
  • $\begingroup$ But on the surface of a conductor we have $\nabla \phi =0$, so we can't really extract any information with regard to $\phi$ except for the fact that it's constant, i.e., $\phi =\phi_{0}$ on $S$. $\endgroup$
    – Hilbert
    Commented Sep 1, 2019 at 18:29
  • $\begingroup$ True. But we are talking about capacitors, and the capacitor is not just the surface of one conductor - rather, it is typically composed of two conductors and the media between them. The electrical potential you are looking for is the difference between the two conductors. Analisying just one of them is not very meaningful. $\endgroup$
    – Woe
    Commented Sep 1, 2019 at 21:42

3 Answers 3

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TLDR: All the equations are linear relations, so their results have that too.

Informally:

Consider some solution to Maxwell’s equations. This relates the fields (hence potentials) and charges in some particular geometry of interest.

Now multiply all the charges by some common constant. Since Maxwell’s equations have a linear relation between charge and electric field, the field at every point is multiplied by that same constant.

Since the potential is just a spatial line integral (a linear operator) of the electric field, it to is multiplied by the same factor at every point.

Hence there’s a linear relation between all-charges and potential.

Formally:

For an initial configuration given by a 0 subscript, $\phi_0(\vec{x})$ is given by $\vec{E_0}(\vec{x})$ over all space, which in turn is given by the arrangement of charge $q_0(\vec{x})$ (and perhaps the current density).

Now if charge everywhere where to increase by the same factor $k = q_1/q_0$, then the linearity of Maxwell's equations means that $\vec{E_1}(\vec{x}) = k \vec{E_0}(\vec{x})$ is a solution. Since it's a solution, it's the only solution. Then the potential comes from $\vec{E_1}$ in the same way, such that $\phi_1(\vec{x}) = k \phi_0(\vec{x})$.

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  • $\begingroup$ I respectfully disagree, I don't think it has to do with the equations themselves. We specify $Q_{0}$ or, equivalently, $\phi _{0}$ before solving Maxwell's equation, they are boundary conditions. The question can be reformulated as, given a boundary condition $\phi_{0}$ (equivalently $Q_{0}$) on the surface of a conductor, and another boundary condition $\phi_{1}$ (equivalently $Q_{1}$), why does one have $Q_{0}/ \phi_{0} = Q_{1} / \phi_{1} =C$? $\endgroup$
    – Hilbert
    Commented Aug 30, 2019 at 19:07
  • $\begingroup$ @hilbert I think you're saying the same thing I'm saying, so I don't see where the disagreement lies. Perhaps I wasn't clear. I've added a more formal paragraph to try to address that. $\endgroup$
    – Bob Jacobsen
    Commented Aug 30, 2019 at 19:17
  • $\begingroup$ In the question, $\phi _{0}$ (or $Q_{0}$) is the constant potential (charge) on the surface of the conductor, it is not a function of position. $\endgroup$
    – Hilbert
    Commented Aug 30, 2019 at 19:20
  • $\begingroup$ If something is true for an arbitrary function of position then it’s true for a constant function of position I.e not depending on position. $\endgroup$
    – Bob Jacobsen
    Commented Aug 30, 2019 at 19:40
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Here's a non-finished, non-rigorous attempt:

Let us say we have a conductor having a surface $S$ at a potential $\phi_{0}$ and having a charge $Q_{0}$.

Supposing all the charges are located on the surface, the potential can be defined at any point in space as:$$\phi (\textbf{x}) =\int _{S} \frac{ \sigma(\textbf{x}')}{\vert\lvert {\textbf{x}-\textbf{x}'\lvert\lvert }} da'$$

$\phi (\textbf{x})$ is constant inside and on the surface of the conductor. Choosing an arbitrary $\textbf {x}_{0}$ within or on the surface of the conductor, we find:

$$\phi (\textbf{x}_{0})=\phi_{0} =\int _{S} \frac{ \sigma(\textbf{x}')}{\vert\lvert {\textbf{x}_{0}-\textbf{x}'\lvert\lvert }} da'$$

where $da'$ is a surface element of $S$.

Since $\vert\lvert \textbf{x}_{0}-\textbf{x}'\lvert\lvert $ is a continuous function of $\textbf{x}'$, we can use the intermediate value theorem for integrals to rewrite the expression above as:$$\phi_{0} =\int _{S} \frac{ \sigma(\textbf{x}')}{\vert\lvert {\textbf{x}_{0}-\textbf{x}'\lvert\lvert }} da'=C \int _{S} \sigma (\textbf{x}')da'=C \times Q_{0}$$

Where $C$ is a constant.

The flaw: We must prove that $C$ doesn't depend on the choice of $\textbf{x}_{0}$.

It'd be fantastic if someone could formalize this a little bit more.

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  • $\begingroup$ could you help me clarify on how you can apply the intermediate value theorem to remove the |x0-x’| term? thanks $\endgroup$
    – V T Naveen Mugundh
    Commented Jun 27 at 10:49
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we know charge and potential have linear relation , as charge is directly proportional to potential. so if you draw charge versus potential graph you'll get straight line ( notice: the value you put on graph are experimental ) . now if you compare the equation y=mx and Q=CV means C(capacitance) represent slope which comes constant . Means C doesn't depend on Q and V

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