Filling the gaps in the “Electromagnetics” textbook: Surface Integral of Electric vector field (part 2: Surface Integration)

Previously I expressed x-component of E on surface A with Taylor expansion. (η and ζ are the displacement in the y- and z-directions.)

Ex on the surface A with Taylor expansion

The First Step: y-direction Integration

We firstly integrate the expansion with respect to η (displacement of y-direction) from 0 to Δy.

(1) Integration of the 0th-order:

Since E_x is treated as a constant during the y-integration, the result is simply E_x Δy.

(2) Integration of the 1st-order:

(3) Integration of the 2nd-order:

Finally, Integration in the y-direction of surface A is

The next step: z-direction Integration

Then, we integrate this y-integration over z (0 to Δz). (1) Integration of the 0th-order:

(2) Integration of the 1st-order:

(3) Integration of the 2nd-order:

Finally, we get an integral of E_x(E・n) on surface A as:

Integral of E・n through surface A.

E_x in the formula is same as E_x(x + Δx , y, z). (The Left-bottom point of surface A.) Now we can calculate the Integral of E_x(x, y+η, z+ζ) through surface B in the same way. The difference from surface A is that the reference point is (x, y, z) and entire integral has a negative sign since unit normal vector has opposite direction of that on surface A (-n).

Integral of E・n through surface B.

We have to add both integrals to derivate the “div”. because “div” represents the total amount of the field flowing out from the point.

The definition of “div” is obtained by dividng the total integral by ΔV(ΔxΔyΔz) and taking the limit of ΔV -> 0. As next operation, I would like to divide ΔV(ΔxΔyΔz).

This is the formula before taking limit Δx,Δy,Δz -> 0 with 1st and 2nd-order. But the textbook described only 0th-order.

How will 1st and 2nd-order disappear?

According to the textbook, 1st-order and 2nd-order have to disappear. But I wondered how it can disappear since the term like Δy/Δx does not towards not 0. Here, let us see the figure displayed in previous article that shows the Premise of integration again.

We are considering the integration of the electric field passing through the surface of a perfect cube. since the cube has edges of equal length (Δx=Δy=Δz ), no matter how fast Δy and Δz rush toward zero, their ratio Δy/Δx is locked at “1” the entire time.

Δy/Δx =1 although taking limit. Since Δy=Δx.

And the following derivative terms also vanish as Δx to 0.

For these reasons, all the first-order terms become zero.

2nd-order terms also can disappear. there are terms like:

this term can be rearranged:

It can be zero through taking this term limit to zero:

Furthermore, the following derivative terms also approach zero when we take the limit.

As a result, all the first- and second-order terms became zero.

The following terms (after third order) of Taylor expansion will also all disappear in same way since they contain factors like Δ³/Δ.

Finally, the formula becomes to be same as textbook. (This is the aggregate of integral of E・n through A/B surface that is parallel to the x-axis.

Ultimately, we can calculate the surface integral for the y- and z-components in the same way and get the definition of “div”.

Textbooks often make it look like we are simply taking the dot product of a uniform electric field E and the normal vector for each surface. However, I had a nagging doubt: shouldn’t the vector field E vary across the surface A? That single question led me through this long process to finally reach a deep, satisfying understanding.

I originally wrote this down as a note for my future self to remember this logic, but I truly hope it saves someone out there who is scratching their head over the exact same question.