# What is the intuition behind why $\vec{E} = \int \mathrm{d}\vec{E}$ works for summing up infinitesimal contributions of electric field?

Say you want to find the electric field, $$\vec{E}$$, at some point in space, $$P$$, which is induced by some uniformly charged rod, $$Q$$, of known length, $$L$$. What you would do is break this charged rod up into infinitesimally small portions of charge $$\mathrm{d}q$$, and sum up the infinite number of the infinitesimally small electric field contributions at point P induced by $$\mathrm{d}q$$.

One could then say that the electric field induced by $$Q$$ at point $$P$$ is $$$$\vec{E} = \int_a^b \mathrm{d}\vec{E}$$$$

Knowing that $$\mathrm{d}\vec{E} = \dfrac{k\mathrm{d}q}{r^{2}}\hat{r}$$ where $$r$$ is the distance from the infinitesimal charge, $$\mathrm{d}q$$, to the point $$P$$, we could then solve this question using known values.

Where my confusion lies is in the intuition behind this integral. I can make sense of the problem stated above by instead doing something like this where is sum all of the infinitely many infinitesimal electric field contributions

$$$$\vec{E} = \lim_{n \to \infty} \sum_{i = 1}^n \Delta \vec{E}_{i}$$$$

Using $$\Delta \vec{E} = \dfrac{k\Delta q}{r^2}\hat{r}$$, $$\Delta q = \dfrac{Q}{L} \Delta x$$ and $$\Delta x = \dfrac{b - a}{n}$$ where $$b$$, and $$a$$ are the beginning and ending points of the uniformly charged rod on a Cartesian x,y plane, we can say

$$$$\vec{E} = \lim_{n \to \infty}\sum_{i = 1}^{n} \frac{kQ\hat{r_{i}}}{Lr_{i}^{2}} \Delta x$$$$

where $$r$$ is a function of $$x$$ with $$x$$ representing the position of $$\Delta q$$ along the rod.

I can then recognize how that this is a Riemann sum, and convert it to an integral to solve the question.

With this method, I am able to easily conceptualize and understand what is happening when I arrive at the integral; however with $$\vec{E} = \displaystyle\int_{a}^{b} \mathrm{d}\vec{E}$$ I don't maintain the same level of intuition. What I see is that for some reason we are finding the area under a dimensionless curve f(x) (of the form $$F(x) = \displaystyle\int f(x) \mathrm{d}x$$), and for some reason this area equals the total electric field.

I fully understand whats going on when I think of it as a discrete sum of small electric fields, but I am unable to maintain the same level of intuition when thinking about it as a discrete integral where we are summing infinitesimal vector areas? What is this invisible $$f(x) = 1$$? What does the area under a constant function have to do with summing up infinitesimal vectors?

• One thing, though it's something trivial but I do wish to point out, it's representatively incorrect to have an integral of ${\rm d}\vec{E}$ and put upper and lower limits as lengths, locations or coordinates of the end points upon the integral sign. Either you reform the integral as $\int_a^b \vec{f(l)}{\rm d}l$, where ${\rm d}l$ is the infinitesimally small length unit of rod or you put limits in the form of values of $\vec{E}$. I don't blame you for this. I've seen many people using similar notations, but to say mathematically, it's an incorrect representation method. Jan 15 at 9:39
• @SteelCubes you are absolutely correct. I should have defined different limits of integration for the integral. I knew that the limits were different but I subconsciously kept to the standard $a$ and $b$ out of, I would say, nothing more than habit! Jan 15 at 9:50
• You can take arbitrary values for analytical purpose such a $\vec {E_{\small a}}$ and $\vec {E_{\small b}}$. Because at the end of the day, they either end up as $\Delta \vec E$ or break as a function of more easily measurable quantities. These things are nothing to worry about in general aspects, and I believe that's where we take them for so granted and make such mistakes ;). Jan 15 at 10:17
• Related question here. (It's about an integral containing $dm$, but it's the exact same idea as $d\vec{E}$.) Jan 16 at 8:49

$$\vec E$$ is identical to $$E_x \hat i +E_y \hat j+ E_z \hat k$$. Right?

So, $${\rm d}\vec{E}={\rm d} E_x \hat i +{\rm d}E_y \hat j+ {\rm d} E_z \hat k$$.

If I say now that your vector integral is actually sum of three different scalar integrals, would you agree?

$$\int{\rm d}\vec E = \int{\rm d} E_x \hat i +\int{\rm d}E_y \hat j+ \int{\rm d} E_z \hat k$$

Now take a look at the graph of $$\int_{\small A}^{\small B} 1\times {\rm d} E_x$$, where I've assumed $$A$$ and $$B$$ be the magnitudes of electric fields in the $$+^{ve} \space x$$ direction due to infinitesimally small elements of rod taken near points $$a$$ and $$b$$ respectively.

Don't you think the area represents $$1\times\Delta E_x=\Delta E_x$$?

Same would be the analysis of $$E_y$$ and $$E_z$$.

Hence, $$\int{\rm d}\vec E = \int{\rm d} E_x \hat i +\int{\rm d}E_y \hat j+ \int{\rm d} E_z \hat k\\ = \Delta E_x \hat i +\Delta E_y \hat j+ \Delta E_z \hat k = \\ \Delta \vec E$$

And if you take for granted, that there is no field in absence of the rod (which logically do makes sense), we would arrive at a place where $$\Delta \vec E = \vec E$$

## Edit

What I think from your comments is that you have understood why this integral $$\int_a^b {\rm d}\vec E$$ represents $$\vec E$$, but are unable to follow up with how this represents what it does. So here it goes,

Once you have successfully written a vector $$\vec A$$ in terms of its components in mutually perpendicular directions (say x, y and z), you arrive at

$$\Delta A_x = \int_{A_{a_x}}^{A_{b_x}} {\rm d}A_x\\ \Delta A_y = \int_{A_{a_y}}^{A_{b_y}} {\rm d}A_y\\ \Delta A_z = \int_{A_{a_z}}^{E_{A_z}} {\rm d}A_z$$

This is because of independency of vector components along mutually perpendicular directions. Also, you must keep in mind that component of a vector along any direction is a scalar quantity. They are magnitudes. And the unit vectors are just constants whose presence in any way do not interfere with the calculation of magnitudes in that particular direction. But to avoid any confusion, let's leave them alone.

Don't you think the above written three integrals are nothing more than just an abridged from of the Fundamental theorem of Calculus?

So, $$\Delta \vec A =\Delta A_x \hat i + \Delta A_y \hat j + \Delta A_z \hat k \\ = \int_{A_{a_x}}^{A_{b_x}} {\rm d}A_x \hat i + \int_{A_{a_y}}^{A_{b_y}} {\rm d}A_y \hat j+ \int_{A_{a_z}}^{E_{A_z}} {\rm d}A_z \hat k\\ \Rightarrow \Delta\vec A=\int_a^b {\rm d}\vec A$$

## About the constant function 1

Mathematics gives you power so that you can write any scalar quantity, say $$B$$ as $$B= \beta \space (some \space units)$$ Where $$\beta$$ is numerical magnitude and is accompanied by the units specifying what quantity it is and in what scale it is measured. Similarly, any vector $$\vec A$$ can be represented as $$\vec A= \vec\alpha \space (some \space units)$$ Where $$\vec\alpha$$ is a dimensionless vector which says how large is $$\vec A$$ and is pointing in which specific direction and is accompanied by some units specifying what quantity it is and in which scale is it measured. So, our scalar integral (say of x component) converts to

$$A_x = \int {\rm d}A_x\\ = \int 1 \times {\rm d}\alpha_x \space (some \space units)$$

So we are left with the integral of a dimensionless quantity $${\rm d}\alpha_x$$ to muddle with. Wouldn't it represent the sum of areas of small rectangles of width 1 and length $${\rm d}\alpha$$ (infinitesimally small)? Wouldn't this sum be represented as $$\lim_{n \to \infty}\sum_{i=1}^n 1\times {\rm d}\alpha_x\\ =1\times\alpha_x$$ So from this, what do we infer about "1".

We can clearly see that, "1" has no physical significance but it's presence could mathematically be justified and explained. It holds an importance in graphical representation of Riemann integral but is devoid of any importance in the physical aspects of integration. Even in graphical representation, it is there to formulate a dimensionless unitary area whose sum would yield the required Riemannian sum.

In simple words, the number "1" exists there as the universal factor and would keep on existing whether we like it or not.

Note: In the above graph, both the axes have quantities that are dimensionless.

• Oh! Okay, so the dimensionless f(x) = 1 comes from the unit vector (since unit vectors are dimensionless and have a magnitude of 1) inside the integral? Jan 15 at 10:56
• @Kalcifer- Actually, no! The 1 was always there. I've broken down the vector and now I'm integrating a scalar function and it's integration is only possible if it is a perfect differential of some other function. Say, if we have $\frac{{\rm d}F(x)}{{\rm d} x}=f(x)$ or ${\rm d}(F(x))= f(x){\rm d}x$ then only we can have $F(x) = \int f(x){\rm d}x = \int {\rm d}(F(x))$. The "1" on the graph is there just to visualise what this area speaks about. Jan 15 at 11:09
• @Kalcifer - Calculus of vectors is a different and a vast topic. At this level, we usually break down things and simplify them in scalar form by taking the components of the vector in the direction they will finally "add" up to some value and not become zero. This is because resultant vector is going to have a unique direction. We find that direction and solve the problem. If you intend to solve the problem in "vector" form, this is something I can't help with right now. I don't have any idea of vector calculus. But interestingly, we don't require that knowledge for high school physics. Jan 15 at 11:22
• Also one thing for clarifying your 1st comment @Kalcifer - we have that "1" there because integrals are defined that way. Have you ever thought why are integrals called anti-derivatives? Also, I've made a slight correction in my answer. The graph is of $E_x$ and not $\vec E$ Jan 15 at 11:25
• I've given your answer some more thought and I am still confused as to why this area represents total electric field. You replied to my first comment saying that the $f(x) = 1$ in actuality does not stem from the unit vector. So then what does this $f(x) = 1$ represent? If its not summing up portions of vector in the direction of the unit vector, then I arrive back at my original question "what does this integral represent?". What representation of $f(x)$ is there that is a constant 1 and is dimensionless? The idea of $f(x)$ being a unit vector felt like it worked so well! Jan 15 at 23:48