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Electric Field of a Continuous Charge Distributionby LucasGB
Tags: field calculus 
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#1
Jan1810, 12:33 AM

P: 181

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/rē 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
Jan1810, 02:38 AM

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P: 2,155

[tex]\vec{E} = k\int \frac{\mathrm{d}q}{r^2}\hat{r}[/tex]
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 [itex]\mathrm{d}q[/itex] that looks like one of [tex]\mathrm{d}q = \lambda\mathrm{d}s[/tex] (line) [tex]\mathrm{d}q = \sigma\mathrm{d}A[/tex] (surface) [tex]\mathrm{d}q = \rho\mathrm{d}V[/tex] (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
Jan1810, 12:44 PM

P: 181

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
Jan1810, 12:55 PM

P: 370

Electric Field of a Continuous Charge Distribution



#5
Jan1810, 01:05 PM

P: 181




#6
Jan1810, 01:16 PM

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P: 2,155

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
Jan1810, 01:18 PM

P: 370

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
Jan1810, 01:20 PM

P: 836

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; [tex]\oint \vec{E}\cdot \vec{dA}=\frac{Q_{ins}}{\epsilon_0}[/tex] For an arbitrarily chosen gaussian surface. 


#9
Jan1810, 01:37 PM

P: 181

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
Jan1810, 01:57 PM

Sci Advisor
PF Gold
P: 1,776

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. [tex]\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[/tex] 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): [tex] \int_{\Omega}f(x)d^3 x = \int f(x) dV[/tex] with[tex] dV = K_{\Omega}(x)d^3x[/tex] 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 nonzero in the region of interest. Its all in the dm or dq of the integrand. One may in fact view these as RiemannStieltjes 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
Jan1810, 02:15 PM

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#12
Jan1810, 02:34 PM

P: 181




#13
Jan1810, 02:41 PM

P: 370




#14
Jan1810, 02:42 PM

Sci Advisor
PF Gold
P: 1,776

No, they are specifically definite integrals and yes it does matter.
Indefinite integrals are only meaningful in one variable (antiderivatives). When you speak of volume or surface integrals you are always defining definite integrals. 


#15
Jan1810, 02:56 PM

P: 181

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


#16
Jan1810, 03:26 PM

Sci Advisor
PF Gold
P: 1,776

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: [tex]\int f(x)dx[/tex] is the set of solutions to [tex] y'=f[/tex] in the same way as [tex] \pm\sqrt{a}[/tex] is the set of solutions to [tex]x^2 = a[/tex] When you get into multivariable 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 anticurl, or an antidivergence, or an antigradient 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
Jan1810, 03:52 PM

P: 836

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
Jan1810, 06:23 PM

HW Helper
P: 2,155

[tex]\int \frac{1}{x^2 + 1}\mathrm{d}x[/tex] with appropriate boundary conditions is functionally identical to the definite integral [tex]\int_{\infty}^{\infty} \frac{1}{x^2 + 1}\mathrm{d}x[/tex] 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. 


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