I Proof of Q=CV for arbitrarily shaped capacitors

[hutchphd]

Q=CV is a definition. It is not an assertion that C is constant. That is a separate assertion.

That is a definition for a single pair of object(s). Some times the really big sphere is the second object. Maxwell showed the simple relation of coefficients of Potential for and arbitrary geometrical assemblage of N conductors from which capacitances can calculated for the assemblage

 

[yucheng]

Let me just jot down a few extra things to consider

Differential capacitance (nonlinearities between capacitance and voltage, this already shows that C=QV only holds in some special cases)
Self- and mutual- capacitance
Capacitance matrix

 

[yucheng]

If you consider bringing in some positive charge and placing it near the conductor, you will have increased the amount of work required to bring it close to the conductor yet the charge on the conductor has not changed. Therefore charge on the conductor does not correlate to the voltage on the conductor, at least not in the general case.

I believe doing so only adds an extra mutual capacitance term between the new charge and the conductor under consideration, but this would not affect the self capacitance term and mutual capacitance terms with other conductors, only until the extra charge is placed on ("absorbed", integrated onto) the conductor.

Maybe more clearly, one can think of a point charge as equivalent to a spherical conductor with the same charge.

 

[Srini314]

This follows from Green’s third identity applied to Laplace’s equation.

Let’s say $\Sigma$ is the surface of an isolated perfect conductor which represents an equipotential surface where the potential takes the constant value $\phi_0$.

Suppose $G(x,y)$ is the Green’s function for Laplace’s equation that vanishes on $\Sigma$. This always exists by the properties of the Laplacian operator for arbitrary shapes $\Sigma$.

Then, one can show

$\phi(x) = \int \phi(u) \frac{\partial G(x,u)}{\partial n} dS_u$.

Here $u$ lies on $\Sigma$.

But $\phi(u) =\phi_0$.

Thus,

$\phi(x) = \phi_0 \int \frac{\partial G(x,u)}{\partial n} dS_u$.

This implies the normal derivative of $\phi$ on $\Sigma$, denoted $\phi_n(u)$ is proportional to $\phi_0$.

Finally,

$Q= -\int \phi_n(u) dS_u \propto \phi_0$

And the proportionality constant depends purely on the geometry of $\Sigma$.

In fact, one can give an explicit formula for the Capacitance as

$C = - \int \int \partial_n \partial_{n’} G(u,u’) dS_u dS_{u’}$

QED