# Electric Field of a Continuous Charge Distribution problem

1. Jan 18, 2010

### LucasGB

The most general way of calculating the value of the vector electric field at a certain point P is given by the formula E = k times Integral of (dq/r² times unit vector). That means you break the charge distribution into infinitesimal elements dq and vectorially add the contributions of each at that point P.

1. But right now I'm studying calculus and trying to figure out what kind of integral is this? Is it a line integral, an area integral or a volume integral? It doesn't seem to be any of these, since I'm just adding values at a single point.

2. And is it a definite or indefinite integral?

2. Jan 18, 2010

### diazona

$$\vec{E} = k\int \frac{\mathrm{d}q}{r^2}\hat{r}$$
1. It could actually be any of the above, depending on what kind of charge configuration you have. If it's a line charge, you do a line integral. If it's a surface charge, you do a surface integral. And if it's a volume charge you do a volume integral. Typically you'll have an expression for $\mathrm{d}q$ that looks like one of

$$\mathrm{d}q = \lambda\mathrm{d}s$$ (line)
$$\mathrm{d}q = \sigma\mathrm{d}A$$ (surface)
$$\mathrm{d}q = \rho\mathrm{d}V$$ (volume)

Keep in mind that there are two points involved, the point where the charge element is and the point where you're trying to compute the electric field. The latter of these, point P, is just a fixed point, but the former point (the location of the charge element) is varied along the line/surface/volume of the charge, as you're used to doing in an integral.

2. Depends on whether you want a numeric or symbolic answer Why, does it matter?

3. Jan 18, 2010

### LucasGB

1. It makes sense, but this just looks unusual to me. It seems strange that you take an integral over a line, a surface or a volume in order to calculate a value in a very distant point. It does make sense, though, and I'll get my head around it.

2. But the integral sign is written without bounds. Doesn't this mean that in this case it is an indefinite integral? (I understand that if I want a numeric answer, all I have to do is assign bounds, turning it into a definite integral and do the math.) It doesn't really matter, it's just that I'm studying calculus more seriously now and trying to make an effort to understand everything.

4. Jan 18, 2010

### elect_eng

I agree with the above answer for question 1, but i disagree on question 2. The answer is a definite integral. The question as to whether you want symbolic or numerical answer is an independent one and not relevant to the question. Either way the integral is a definite integral. Definite integrals are integrals with specified limits, which is required in a physical problem. Those limits may be represented symbolically to give a general form for the answer, but it's still a definite integral. Indefinite integrals are just antiderivatives and have no limits.

5. Jan 18, 2010

### LucasGB

How can that be, if, as you said "definite integrals are integrals with specified limits"? That integral has no limits, and must be, therefore, an indefinite integral. When you calculate the integral as presented in that formula, you get a function, and not a number. And then, using the second part of the fundamental theorem of calculus, you use that function you just obtained and get a numerical answer.

6. Jan 18, 2010

### diazona

elect_eng makes a good point, actually - in physics all integrals are definite, in the sense that you can't apply an integral to a real physical situation without having limits or boundary conditions of some sort.

So I misspoke. That is, in fact, a definite integral, but it's in disguise because it's part of a very general formula. We don't write the limits because in general, we don't know what they are. It depends on the situation. Part of applying the formula to a particular problem or situation is choosing the limits (or boundary conditions) for the integral, just like another part of applying it is plugging in the appropriate sort of charge density.

7. Jan 18, 2010

### elect_eng

Perhaps I'm misuderstanding your question, but it seems to me that whenever you set up a field problem (or any physical problem for that matter), there will be defined limits for the integration. Even if these limits are infinity, numbers, zero, or even a symbolic representation for generality, there should be defined limits. Even if you don't specify the problem, you will say things like integrate over the "boundary", or over a closed surface or closed volume etc. These are all descriptions of definite integrals in my view. That's all I'm trying to say.

If you are asking whether the integral (as you wrote it) is an indefinite one, then yes it is because you wrote no limits.

EDIT: Yes, I see diazona's response now and agree with that.

8. Jan 18, 2010

### espen180

I think that in physics, there is a little more slack on notation than in mathematics.

You must spesify yourself the bounds you want to integrate between, though they are often inplied. For example, in the case of a finite rod of distributed charge, it should be obvious that you need to integrate over the whole rod.

For infinite lines, planes, cylinders etc. I reccomend you drop Coloumb's law and use Gauss's law instead;

$$\oint \vec{E}\cdot \vec{dA}=\frac{Q_{ins}}{\epsilon_0}$$

For an arbitrarily chosen gaussian surface.

9. Jan 18, 2010

### LucasGB

OK, I think I get it now. In the real world, whenever one uses a physical formula like that there will always be specified limits, and therefore, that's a definite integral. But, mathematically speaking, the formula, as it is written, presents no bounds, and is, therefore, an indefinite integral. That's what I make out from what you guys are saying. :)

10. Jan 18, 2010

### jambaugh

You can think of it as a definite integral over all space and the restriction to a surface or line or point being a matter of invoking Dirac delta functions.

Note you can always internalize the bounds of integration by invoking a characteristic function which is a function with value 1 inside a region and 0 outside. One then is formally always integrating over all of space (and thus don't bother to write limits) while the characteristic function forces contributions outside the given region to be zero.

$$\int_{\Omega} f(x)d^3x = \int_{\mathbb{R}^3} K_{\Omega}(x)f(x)d^3x\equiv \int K_{\Omega}(x)f(x)d^3x$$
K begin the characteristic function for the region Omega.
K(x) = 1 for x in Omega, otherwise K(x)=0.

If we do this universally we don't bother indicating the integration over all space but rather leave it as implied.
One can then absorb this characteristic function into the definition of our function to be integrated or our definition of the differential (the measure of the space):
$$\int_{\Omega}f(x)d^3 x = \int f(x) dV$$
with$$dV = K_{\Omega}(x)d^3x$$

Typically it is absorbed into the density (in this case charge density). That is a natural way to do this as one understands that the physical charge density in question typically is only non-zero in the region of interest.

Its all in the dm or dq of the integrand. One may in fact view these as Riemann-Stieltjes or Lebesgue integrals with the density distribution defining the measure used. It includes restrictions to a region, or even invocation of Dirac delta functions to restrict to surfaces, curves, discrete set of points or any combination thereof.

11. Jan 18, 2010

### diazona

I guess you could say that. But I still don't see why you think it matters. It's been many years (since high school) since I cared about the difference between a definite vs. an indefinite integral. As far as I'm concerned, they're all just integrals. The boundary conditions on any given integral might be explicit or symbolic or implied or unspecified or whatever, but there's no fundamental difference with respect to how you evaluate the integral or what it physically means. (The mathematicians might disagree, of course, but then again they're mathematicians)

12. Jan 18, 2010

### LucasGB

I agree. And that was funny.

13. Jan 18, 2010

### elect_eng

That one made my day. I can't stop laughing.

14. Jan 18, 2010

### jambaugh

No, they are specifically definite integrals and yes it does matter.

Indefinite integrals are only meaningful in one variable (anti-derivatives). When you speak of volume or surface integrals you are always defining definite integrals.

15. Jan 18, 2010

### LucasGB

That's an interesting idea I hadn't though about yet. I didn't know indefinite integrals were only meaningful in one variable. Why is that?

PS.: I understand why volume or surface integrals are always definite. And, from what I understand, even line integrals CAN be definite.

16. Jan 18, 2010

### jambaugh

"Meaningful" may not be the right term. You can stretch the definition possibly but not practically meaningful is the more appropriate phrase.

The "Why" has to do with the nature of the derivatives. In one dimension you have a single derivative operator and thus one meaningful differential equation defining the indefinite integral.

In essence the indefinite integral notation is analogous to the square root notation:

$$\int f(x)dx$$ is the set of solutions to $$y'=f$$
in the same way as
$$\pm\sqrt{a}$$ is the set of solutions to $$x^2 = a$$

When you get into multi-variable calculus you have gradients, curls, and divergences and a host of possible "simplest" differential equations to then solve. It just isn't that simple to define an anti-curl, or an anti-divergence, or an anti-gradient especially as not all scalar or vector functions can be written as curls of or gradients of or divergences of an appropriate function.

Instead we stick to definite multivariate integrals and deal with solving specific equations explicitly.

If I could change history I'd even get rid of the one variable indefinite integral. It is too confusing and as we see inelegant and inconsistent in its generalization. And it is not essential to the practical application of calculus nor to the general mathematical theory.

17. Jan 18, 2010

### espen180

Another point to make is that the bounds of one variable may depend on another, such that the integral isn't solvable for arbitrary bounds.

18. Jan 18, 2010

### diazona

But you can apply a 1D antiderivative several times in succession with possibly different variables, leading to a multidimensional analogue of an indefinite integral. And anyway, from a physics perspective, an antiderivative like
$$\int \frac{1}{x^2 + 1}\mathrm{d}x$$
with appropriate boundary conditions is functionally identical to the definite integral
$$\int_{-\infty}^{\infty} \frac{1}{x^2 + 1}\mathrm{d}x$$
The same mathematical techniques go into solving either of them, their physical interpretations are the same, etc. So the difference doesn't matter. A physicist looks at them the same way.

To put it another way, a definite integral has its boundary conditions specified, an indefinite integral doesn't (and thus represents a whole set of solutions with various possible boundary conditions), but as far as I'm concerned, that's the extent of the difference. I'm still wondering whether LucasGB thought there was more to it than that.

19. Jan 18, 2010

### LucasGB

No, I understand what you guys are saying. It's just that I'm trying to be as rigorous as I can learning these things, since I'm self-taught, and when I first wrote this post I had just read the section in my calculus book regarding definite and indefinite integrals and wasn't very sure about their differences and resemblances, so this seemed a somewhat important question. I think I can see clearly now the rain is gone.

20. Jan 18, 2010

### jambaugh

That's why I hedged a bit. It isn't that we can't its that it isn't very useful. Recall that the order of your integration will matter since the constant you get from integrating w.r.t. x will then be cy+d when you integrate w.r.t. y, and then cyz+dz+e when you integrate w.r.t. z (if you integrate in that order).

Of course (with appropriate boundary conditions) since both are functionally constants and not function i.e. not indefinite integrals. If you meant the first to be an indefinite integral then you're not correct in their functional equivalence.

But that's treating both as definite integrals (I'm interpreting your "with appropriate boundary conditions" as such). That's not the point (or rather that is the point, to notationally drop the limits to imply integration over the whole space as opposed to indicating an indefinite integral.).

Nope, that's not it. A definite integral maps function to number. An indefinite integral maps a function to a class of functions parametrized by the constant of integration. They are very distinct types of actions though you use antiderivatives to find definite integrals.