I don't understand the logic/concept of $\mathrm dQ=\lambda\,\mathrm dx$. How did we arrive at this expression?

Question

So I've been learning Electrostatics. So while solving for the Electric Field Due to an infinite positively charged rod, I encountered the following expression on the internet wherein the following expression was used to reach the conclusion of the above-mentioned problem. I'm having a hard time understanding how did we get that expression and is there any other approach to the same question without the use of the expression; $\mathrm dQ=\lambda\,\mathrm dx$. All help is appreciated.

Answer 1

$Q$ is electric charge, $X$ is position along the rod and $\lambda$ is linear charge density, charge per unit length. $\lambda = \frac{dQ}{dX}$ by definition because differentiating charge with respect to distance gives the rate of change of electric charge with respect to length. Using the chain rule, $dQ=\frac{dQ}{dX} dX$, which is the same as $dQ=\lambda dX$

Answer 2

$dQ/dx$ is the density of charge per length. In this case it is a constant, $\lambda$, and so is how you arrive to your expression. We usually use density in terms of volume, like $dQ/dV=\rho$, but in this case we take the space to be one-dimensional, because the rod is.

Answer 3

Imagine taking a small, but finite, "chunk" of the rod of length $\Delta x$. Since $\lambda$ is defined to be the charge per unit length of the rod, the amount of charge $\Delta Q$ on this chunk should be $\Delta Q = \lambda \, \Delta x$.

So why do we need to use the version $dQ = \lambda \, dx$ instead? Well, when we want to find the total electric field from all of the "chunks", we need to sum up the contributions from all of them together. But for a truly continuous distribution, we won't get the right answer unless the chunks are infinitesimally small: $\Delta x \to 0$. In this limit, the sum I mentioned above becomes an integral over $x$. The relationship between the "infinitesimal charge elements" $dQ$ and the "infinitesimal distance elements" $dx$ remains the same as the relationship between $\Delta Q$ and $\Delta x$.

Finally, you ask whether there are other ways to solve this problem without using this relationship. The answer is yes for this particular problem; look up "Gauss's Law for infinite line". (In fact, I would expect that any intro course that would lead you to ask your original question would also cover Gauss's Law.) However, the approach using Gauss's Law only works for a few particularly symmetric charge distributions. For example, the Gauss's Law approach doesn't work for a line of charge of finite length; it only works for an infinitely long line.

Answer 4

Imagine a small piece of the rod, of length $dx$. This small piece will contain a small about of charge $dQ$. The linear charge density $\lambda = \frac{dQ}{dx}$ so that the net charge $$ Q=\int dQ = \int \lambda dx\, . $$ If $\lambda$ is contant (i.e. does not depend on the location of your small piece of rod), then the net charge is just $$ Q= \lambda \ell $$ where $\ell$ is the length of your rod, and the linear density $\lambda$ is in Coulomb/meter.

An analogy can be made with the (mass) density of an object: a small volume $dV$ of this object will contain a small amount of mass $dM$, and the density $\rho$ is the ratio $\rho=dM/dV$ so that the net mass is just $$ M=\int dM = \int dV \rho\, . $$ If the density is contant, then the mass is just $M=\rho V$, i.e. the density multiplied by the volume, where the density is in kilograms/meter$^3$.

Once you have the linear charge density at every point of your rod, you can work out the small field $d\vec E$ created by the small amount $dQ$ of charge located at that point, considering your small piece of rod of size $dx$ to be a point carrying the small charge $dQ=\lambda dx$. If $\lambda$ is constant, the “neighboring” piece of the rod will carry the same charge $dQ$ but it will be little closer or father than the previous piece of rod, so the small field $d\vec E$ due to the neighbour will have a slightly different magnitude and direction than the previous piece. The net field $\vec E$ is the sum of all these small pieces: $$ \vec E=\int d\vec E\, . $$