Why is charge $q$ chosen as the variable of integration when converting from summation to integral form in electric field calculations?

Question

If we have several point charges $q_1, q_2, q_3, \cdots$, and their distances from a test charge $Q$ are respectively ${\imath}_1, {\imath}_2, {\imath}_3$, $\cdots$, then the electric field intensity at $Q$ is given by $E(r) \equiv \frac{1}{4 \pi \varepsilon_0} \sum_{i=1}^n \frac{q_i}{{\imath}_i^2} \hat{{\imath}}_i$

This equation can be written in integral form as: $E(r) = \frac{1}{4 \pi \varepsilon_0} \int \frac{1}{\imath^2} \hat{\imath} dq$.

I don't quite understand why, when converting from the summation form $E(r) \equiv \frac{1}{4 \pi \varepsilon_0} \sum_{i=1}^n \frac{q_i}{{\imath}_i^2} \hat{{\imath}}_i$ to the integral form, $q$ is chosen as the variable of integration. Both $q$ and ${\imath}$ have the subscript $i$, so why not choose ${\imath}$?

Answer 1

The idea is that you are breaking up a charge distribution into many small chunks, small enough so that each chunk can be treated as a point charge. We will then take the limit of more and more, smaller and smaller chunks, in which case the charge $q_i$ on each chunk goes to zero. The distances $\imath_i$ from source charge $q_i$ to the field point does not go to zero during this process. According to the meaning of an integral as a Riemann sum, this means that you have to take $q_i$ as your $dq_i$, and you can't use $\imath_i$.

Answer 2

I think the formula should be:

$$ \vec E(\vec r) = \frac 1 {4\pi\epsilon_0}\int_V \frac {\rho(\vec r')(\vec r' - \vec r)}{||\vec r - \vec r'||^3} d^3r' $$

where $\rho(\vec r)$ is the charge density distribution, which is contained in $V$.

Traditionally, $q$ is used for point charges, while $\lambda, \sigma, \rho$ are used for line, surface, volume densities, respectively.

Regarding the distance variable in your posted equation..I have no idea what that is. It is common to use $\vec r$ for affine points (positions in an arbitrary coordinate system), and $\vec r'$ for the dummy variable in the integral.

The resulting formula should be independent of origin, and depend only on the vector difference ($\vec r' - \vec r$) of affine points ($\vec r', \vec r$).