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

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.

• Do you know calculus? – Mark Eichenlaub Mar 24 at 7:19

$$Q$$ 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 $$dQ=\lambda dX$$

• This is perfect. Absolutely on point. Thanks, mate! – Vishwas Sharma Mar 24 at 12:57
• @Vishwas Sharma good to hear, no problem – bemjanim Mar 24 at 13:51

$$dQ/dx$$ 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.

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.

• Yes, now I get it. Thanks, buddy :). – Vishwas Sharma Mar 24 at 13:01

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: $$\vec E=\int d\vec E\, .$$