Is it valid to write $\int{dq}$?

Question

While I was going through the expression for the electric field in continuous charge distributions (Section 2.1.4) in Griffiths' textbook, it was given as $${E(r)} = \frac{1}{4\pi\epsilon_o}\int{\frac{1}{r^2}\hat{r}dq}$$

So, would it be valid to write this expression because the quantization of charge implies that there exists a lower limit to the possible value that charge can take (i. e. the charge of an electron (or proton, if you consider the magnitude), which is about $1.6$ $\cdot{}$ $10^{-19}$ C). However, $dq$ denotes an infinitesimal piece of charge that we're using to integrate.

Also, is this why $dq$ is often computed by other means like using $\lambda$ (charge per unit length, in case of a one dimensional continuous charge distribution)?

After some thought, it appears to me that maybe it is in indeed incorrect to use the expression above and rather use $\Delta{q}$ instead and proceed to calculate the electric field by using Riemann sums instead. Quite tedious, but that's the only way I see.

Can you help me understand this and let me know if there is/are some flaws in my understanding?

Answer 1

The formula assumes a continuous charge distribution [1]. Such a thing may not actually exist.

However, before you throw away the formula, pause to consider that the alternative you're proposing -- a static distribution of quantized, point charges -- also does not actually exist. Real electrons are quantum objects. For example, as part of most ordinary materials, electrons live in the various orbitals and/or electronic bands and do not have well-defined positions, let alone static positions.

All physics is done with mathematical models of reality. The continuous distribution model is close enough to reality for most purposes, and it is easy to work with. In some cases you may find the continuous distribution model to be inadequate, and switch to a model with discrete unit charges. But it will still not be completely accurate.

[1] I suppose the charge density of a set of discrete point charges can be described using the Dirac delta function. If you do that, the integral formulation is still correct. In some sense, though, this is just a notational trick, because to compute the resulting integral, the first step is to convert it to a sum.

Answer 2

It is correct to suspect that something more fundamental is occurring when the integral is written with the quantity $\mathrm{d}q$. But consider the electric field created by a charge. How can one represent the $\vec D$ fields created by a particular charge in a defined shell volume $4\pi r^2\mathrm{d}r$? That comes down to Maxwell's Equation $\vec \nabla \cdot \vec D=\rho$ so that $$\int_\text{Inside Sphere Volume} \rho \left(\partial \text{Volume}\right)=\int_\text{Inside Sphere Volume} \rho \left(4\pi r^2\mathrm{d}r \right) \\ =q=\int_\text{Volume}\vec D \cdot \hat{n} \left(\partial \text{ Surface}\right).$$

So where exactly is this charge located to create a radially-symmetric $\vec D$ field pointing outwards, and why is a charge density $\rho$ used in Maxwell's equation instead of only the charge $e_c$ for an electron or $e_p$ for a proton? From this question, one can try to understand $\vec \nabla \cdot \vec D=\rho$ more critically. Sometimes there is such a great number of individual charges that contribute to the field, that so-to-speak, the pennies count up to dollars, and the change does not appreciably affect the result because the answer is close enough using the charge density approximation. At other times, the $\vec D$ field itself outside a certain radius may not change instantaneously with motion of an electron or proton within the radius. So then for accounting-on-the-radial-shell $\vec r$, the quantity $\rho$ can be used to make a calculation as an equivalent charge-density on the shell, and not as an individual electron or proton quantity (since the proton or electron motion within the shell may not have enough time to alter the expected charge density $\vec D$ on the outside of the shell).

Applying quantum mechanics applies a probability density to finding the electron in a particular volume, and therefore also a most probable $\vec D$ field until the electron wave-function is actually measured at a point. So quantum mechanics gives up on the point-view of the electron altogether when making probability calculations for important values such as the ionization energy or most probable location of finding the electron and the probability that it would be found there. In that sense, then the electron wave function is spread out across the volume of the universe, and only has an expected value of being measured within any restricted volume ($\partial\text{Volume}$) so it makes sense the quantity $\partial q=\int_\text{Volume} \rho \left(\partial \text{Volume}\right)$, where $\rho$ is taken from Maxwell's Electromagnetic Equation $\vec \nabla \cdot \vec D =\rho$.

Appendix: Maxwell Equations Reference as Quoted

From Maxwell's Electromagnetic Equation $\vec \nabla \cdot \vec D =\rho$

Answer 3

Yes, you can write $q=\int\mathrm dq$, even if the charge is discrete.

To include discrete charges in continuous calculus, we use Dirac's "delta density distribution," defined by $$ \delta(x)=0\text{ where }x\neq 0 \\ \int_{-\infty}^{+\infty}\mathrm dx\ \delta(x) = 1 $$ You can string these properties together to prove that $\int \mathrm dx\ \delta(x)f(x)=f(0).$

You can generally define the $\delta$ as the limit of some well-behaved family of spiky functions that are convenient to integrate, such as $\frac{\sin kx}{kx}$ or $e^{-x^2/a^2}$, with appropriate normalization. After you do the integral, you take the limit of your result as the appropriate parameter makes $\delta$ narrower, and you get the discretized result. If you don't care whether $\delta$ is "smooth," you can use a box function, $$ \delta_a(x)= \begin{cases} 1/a&\text{where }|x|<a/2 \\0&\text{elsewhere} \end{cases} $$ However, the particular model you use doesn't matter — the point is that in the limit of the delta function, the integrals get simple again.