Does scaling the charge also scale the charge density?

Question

In Griffiths' Introduction to Electrodynamics, 4th ed., he motivates the notion of capacitance by arguing that the electric field $\bar{E}$ is proportional to the charge, and hence so is the potential $V$; the constant of proportionality being the capacitance $C$.

Now, the electric field for a charge distribution is defined: $$\bar{E} = \frac{1}{4\pi\epsilon_{0}}\int\frac{\rho}{r^{2}}\hat{r}d\tau.$$ Griffiths argues that doubling the total charge $Q$ doubles the charge density $\rho$ as well. As in, doubling the charge does not asymmetrically distribute this new net charge; it distributes itself in such a way that the density $\rho$ doubles at every point, independent of the geometry or property (conductor, insulator etc) of the charge distribution.

Why is this always true?

Answer 1

I suppose you are referencing the following section that I found in the 3rd edition in Chapter 2, section 2.5.4, p. 103-104 where Griffiths is discussing capacitors.

$\mathbf{E}$ is proportional to $Q$. For $\mathbf{E}$ is given by Coulomb's law [the equation you write above]. So if you double $\rho$, you double $\mathbf{E}$. (Wait a minute! How do we know that doubling $Q$ (and also $-Q$) simply doubles $\rho$? Maybe the charge moves around into a completely different configuration, quadrupling $\rho$ in some places and halving it in others, just so the total charge on each conductor is doubled. The fact is that this concern is unwarranted - doubling $Q$ does double $\rho$ everywhere; it doesn't shift the charge around. The proof of this will come in Chapter 3; for now you'll just have to believe me.)

Note that Griffiths is here discussing what happens when you double the total charge $Q$ on a conductor - not any medium - and let the charge distribute itself freely. The proof he refers to in Chapter 3 is the Second uniqueness theorem:

In a volume $\mathcal{V}$ surrounded by conductors and containing a specified charge density $\rho$, the electric field is uniquely determined if the total charge on each conductor is given. (The region as a whole can be bounded by another conductor, or else unbounded.)

I encourage you to read more about the proof of the second uniqueness theorem in the book. But one consequence of it is, that if you have found an electric field that satisfy that the total charge on all conductors is correct, then this is the only correct solution to the electric field.

I am not sure I can answer your question any better here, but to give some intuition, consider the following two examples (A) and (B). You start with a single conductor that has total charge $+Q$, for which you have found the electric field $\mathbf{E}_0$ around it:

Example (A) Now add a negative charge $-Q$, so that the total charge is zero. Due to the second uniqueness theorem above, there is a single unique electric field which is zero everywhere, and hence you know that the $+Q$ and $-Q$ charges has distributed themselves in the same way such that $\rho=0$ everywhere.

Example (B) Now add a positive charge $+Q$, so that the total charge is $+2Q$. One way of obtaining an electric field for this conductor is to simply double the electric field $\mathbf{E}_0$ found from the conductor with total charge $+1Q$. But by the second uniqueness theorem, you know that this is in fact the only possible electric field, and therefore, the second $+1Q$ charge has distributed itself exactly as the initial $+1Q$.