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The electrical charge density, is the ratio of an infinitesimal electric charge $dQ$ to an infinitesimal piece of length on a line, $dl$. In other words $\rho = dQ/dl$

I understand this, but I don't understand how this formula $\rho = dQ/dl $ doesn't indicate that the density is the change in charge per unit length. How doesn't this notation imply that the density is the derivative of the charge?

This is obviously a derivative, but it's not. It is confusing me why $dQ/dl$ is not a derivative of Q with respect to l.

https://en.wikipedia.org/wiki/Charge_density

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    The density is the derivative of the charge, yes. This means that when you integrate that density over a segment of the line, you get the amount of charge in that line. In other words the FTC tells you that these two notions are the same (derivative of charge with respect to length, and charge per unit length). – 2017-02-16
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    @Ian Ok, assume Q is constant. Then the density is 0? – 2017-02-16
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    @Goldname In your link to wikipedia, you can see that the charge density is defined as $\frac{dQ}{dl}$ for **continuous** charge distributions. If you scroll down further, you will see it is defined differently for Homogenous charge densities (i.e the density is independent of the position). I believe this is where the confusion comes from, someone may correct me if I am wrong on this. – 2017-02-16
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    I think "Q is constant" doesn't mean what you think it means: "Q is constant" means that the amount of charge between $a$ and $x$ does not depend on $x$. That means there can't be any, because there is no charge between $a$ and $a$ by the definition of a continuous distribution. I think your intuition for "Q is constant" is actually "dQ/dl is constant" (which essentially means that the amount of charge "at" each point is constant). – 2017-02-16
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    @Ian y=4 is a continuous distribution, so I don't know what you mean. Hushus46 might be right, although I don't understand why you'd break up such a phenomenon into two cases mathematically. – 2017-02-16
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    What I mean is that when you talk about a uniform continuous distribution, that means the *density* is constant, not the accumulated amount of charge. If the accumulated amount of charge between $a$ and $x$ doesn't depend on $x$ then there can't be any charge there. There is no need to mathematically distinguish between non-uniform distributions and uniform distributions. Rephrasing, the point is that $Q(x)$ is not the amount of charge at $x$ (which is zero, if the charge distribution is continuous) but rather the amount of charge between $x$ and some reference point. – 2017-02-16
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    @Ian I see what you mean, but how is Q(x) not the amount of charge at x? Furthermore I don't see how a continuous distribution is 0 at a specific point. Imagine a distribution modeled by y = x^2. At x = 1, the charge is 1. (If you integrate this finite region, it is not infinite.) – 2017-02-16
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    "A continuous distribution modeled by $y=x^2$": again the *density* is $y=x^2$, and there is no charge at the single point $x=1$. In probability theory language, if $X$ is continuously distributed, then its CDF $F(x)=P(X \leq x)$ is a continuous function, which means that for any $x$, $P(X=x)=\lim_{y \to x^+} F(y)-F(x)=0$. – 2017-02-16
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    I think what is confusing you is the following misconception: "$Q(x)$ is a function of $x$ with the units of charge, so it must be the amount of charge at the point $x$". But it isn't, again it is the amount of charge between $x$ and a reference point $a$. – 2017-02-16
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    A related, important conceptual point: if the density *were* instead an amount of charge at each point (for example, maybe there is $x^2$ charge at each $x \in [0,1]$), then there would have to be an infinite amount of charge present, because any uncountable sum of positive numbers is infinite. This is an important difference between integration and summation, and is one of the reasons why we keep the "dx" around in integration notation (to remind us that we are essentially summing only after multiplying by the tiny number $\Delta x$). – 2017-02-16
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    @Ian I see, thanks. It looks like I misunderstood what Q represents, and the application of the integral. – 2017-02-17
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    @Ian Actually, this question has led me to be confused about integrals. Assume the density were the amount of charge at each point. Then, what is Q(x) * dx? Is it not the charge for each point, dx? – 2017-02-17
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    Strictly speaking, for a density $Q(x)$, $Q(x) dx$ is not meaningful as a quantity unto itself. It only really has meaning in an integral. The physical intuition for it, however, is that it is the amount of charge in the infinitesimally short interval $[x,x+dx]$ (which is an infinitesimal amount). – 2017-02-18

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