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The book defines the electric field at a point $P$ a distance $r$ due to a point charge $q$ as:

$$ E = \frac{1}{4\pi \epsilon _0} \frac{q}{r^2}$$

it then tells us that the electric field at a point $P$ due to $n$ point charges $q_i$ is:

$$ E = \sum^n_{i=1} \frac{1}{4\pi \epsilon _0} \frac{q}{r_i^2} $$

where $r_i$ is the distance between $q_i$ and $P$. As in the electric field follows the superposition principle.

It then tells us that when we have instead of discrete point charges, have a continuous charge (either a line charge, a surface charge, or volume charge) then the electric fields given for the respective situations are given as:

  1. $$ E = \frac{1}{4\pi \epsilon _0} \int_{}^{} \frac{\lambda(r')}{r^2}dl' $$

  2. $$ E = \frac{1}{4\pi \epsilon _0} \int_{}^{} \frac{\sigma(r')}{r^2}da' $$

  3. $$ E = \frac{1}{4\pi \epsilon _0} \int_{}^{} \frac{\rho(r')}{r^2}d \tau' $$

Where for a line charge: $dq= \lambda dl' $,

for a surface charge: $dq = \sigma da'$

for a volume charge: $dq = \rho d\tau'$


Here's where my question comes in, I am having a great trouble understanding how to 'interpret' these integrals.

Firstly, I tried viewing these integrals mathematically. For example for the surface charge integral (eq 2.) I tried to express it as the Electric field at a point $P$ due to a charged rectangle in cartesian coordinates of length $a$ and width $b$ :

$$ E = \frac{1}{4\pi \epsilon _0} \int_{area}^{} \frac{\sigma(r')}{r^2}da' = \frac{1}{4\pi \epsilon _0} \int_{area}^{} \frac{\sigma(r')}{r^2}dx'dy' = \frac{1}{4\pi \epsilon _0}(\int_{0}^{a} (\int_{0}^{b} \frac{\sigma(r')}{(\sqrt{(x_1 - x)^2 + (y_1 - y)^2)}^2}dx')dy') $$

Where $r$ is the distance of each point in the rectangle to the point $P = (x_1,y_1)$ from the arbitrary points to integrate over $(x,y)$, (assuming the charged rectangle and $P$ are on the same plane here.

I made a rough paint to illustrate

  • I tried to interpret the last expression on the RHS as the volume above a rectangle of length $a$ and width $b$, with a height of function $\frac{1}{4\pi \epsilon _0} \frac{\sigma(r')}{r^2}$. Where the 'height' of each point above this rectangle, is the magnitude of the contribution for a point charge located at that point. Kind of like the height above each point is $dE$ at $P$, and adding them all together gives us $E$ at $P$.

I made another rough paint.

But we're talking about 'surface charges' not point charges. So this interpretation shouldn't hold for a line charge, because it uses a different variable for some reason ($\lambda$ instead of $\sigma$). But I don't see why you can't similarly just interpret equation 1. as the area above a line with height $\frac{1}{4\pi \epsilon _0} \int_{}^{} \frac{\lambda(r')}{r^2}dl'$, similar to how one interprets definite integrals in Integral Calculus. Which leads to my second question

  • Why are we using different terms for the integration variable? At first I thought it was so you get the correct units but I'm not so sure how to justify that intuition. How does the integral 'know' I'm talking about areas when I say $da'$ and lines or volumes when I say $dl'$ and $d\tau '$?

  • The easiest way I've found to think of all this is to think of the function under the integral 'as a whole'. As in to think of equation 2. As taking an infinitesmal area of the charged rectangle (surface element $da'$) and multiplying it by the charge density ($\sigma(r')$), which would give you the total charge on the surface element ($dq$). Then picking a point on this infinitesmal surface area and measuring the distance from this point to $P$ (this distance would be $r$). Then multiplying this entire expression by $\frac{1}{4 \pi \epsilon _0}$. Which gives the expression

$$ dE = \frac{1}{4\pi \epsilon _0} \frac{\sigma(r')}{r^2}da' = \frac{1}{4\pi \epsilon _0} \frac{dq}{r^2}$$

as the electric field due to a singular surface element a distance $r$ away from a point $P$. And then when you take the integral of this above expression, since integrals represent infinite sums, the above integrated would represent adding up the contribution of all these surface elements, to give the total electric field of the entire surface. And this makes a lot of sense to me, however it is not actually how integration works to my knowledge. What I've described is multiplying the area of a bunch of squares ($da$) by a factor dependent on this squares position and charge density ($\frac{1}{4 \pi \epsilon _0} \frac{\sigma(r')}{r^2})$ . Which gives an associated $dE$ with each square, and adding them (integrating) to get an $E$ associated with the whole rectangle. To my knowledge integrating the above expression over a rectangle for example, would be to integrate the contribution of a line on this rectangle, with its height determined as $\frac{1}{4\pi \epsilon _0} \frac{dq}{r^2}$ , then sweeping it along the rectangle. My calculus professor made a demo of what I mean here.


Sorry if this question is extremely long or difficult to follow. Part of my question is me not really understanding what I'm missing about how to interpret these integrals. I suspect I also have some fundamental flaw in my understanding of integrals that I'm missing. Any answers appreciated.

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  • $\begingroup$ Maybe instead of asking this here, perhaps strip off the physics part and ask for clarification of what a line/surface/volumn integral means as opposed to your usual scalar single/double/triple integral over on mathematics stackexchange as it seems to be more of a mathematics misconception at first. $\endgroup$
    – Triatticus
    Commented Dec 11, 2022 at 8:15
  • $\begingroup$ There is only Coulomb's law F=q1q2/r^2 in electro theory. It might be easier to understand if we have faith that everything is can be explained by this; how E relates to F, how the line integral relates to F, etc. Electricity + magnetism is a lot more confusing, because they induce one another, and depends on relative motion from each other's point of view, etc. Electricity by itself is more straightforward, almost like gravity. $\endgroup$
    – James
    Commented Dec 11, 2022 at 8:44
  • $\begingroup$ "What I've described is multiplying the area of a bunch of squares (π‘‘π‘Ž) by a factor dependent on this squares position and charge density (14πœ‹πœ–0𝜎(π‘Ÿβ€²)π‘Ÿ2) . Which gives an associated 𝑑𝐸 with each square, and adding them (integrating) to get an 𝐸 associated with the whole rectangle." -- That's basically correct for surface charges. For line charges, replace "area of squares" with "length of line segments", "surface charge density" with "linear charge density", and "$dE$ associated with square" with "$dE$ associated with line segment." $\endgroup$
    – Andrew
    Commented Dec 11, 2022 at 10:05
  • $\begingroup$ The actual volume charge density for the latter integrals is infinity such that its integral about the infinitely thin sliver of space is finite, aka the surface charge density, aka line charge density. $\endgroup$ Commented Dec 11, 2022 at 12:50

1 Answer 1

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Your electric field equations are missing the very important script-r ${\mathscr{\hat r}}$ (see Griffith's preface).

The electric field is a vector field.

Your line-charge equation should be $$\color{red}{\vec E} = \frac{1}{4\pi \epsilon _0} \int_{}^{} \frac{\lambda(r')}{{\mathscr{ r}}^2} \color{red}{\mathscr{\hat r}}\ dl'$$

So, your second diagram describing the electric field as a "height" is not appropriate.
It would appropriate for (say) the x-component of the electric field. But you’ll need a total of three such plots for the three components of the electric field.

The "electric field at a point in space"
is a vector-sum of the "electric-field-vector contributions there from each point-charge".

I think it's helpful to distinguish "symbols (like $dq$, $dl$, $da$, $d\tau$ describing elements)" from "integration coordinates (used in $dx$, $dy$, $dz$, $dr$, $d\phi$, $d\theta$) to label and describe the size of those elements".

This equation (from https://books.physics.oregonstate.edu/LinAlg/deltadensity.html , "The Geometry of Linear Algbera" by Manogue and Dray) involving the dirac delta function may help connect (say) the volume-charge density with the surface-charge density:

\begin{equation} \iiint \rho(x,y,z) \,dz\,dx\,dy = \iiint \sigma(x,y) \,\delta(z) \,dz\,dx\,dy = \iint \sigma(x,y) \,dx\,dy\tag{11.8.2} \end{equation}

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