Electric field integral: Convergence where ρ is nonzero

  • #1
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Hi.

I know how to use Gauss' Law to find the electric field in- and outside a homogeneously charged sphere. But say I wanted to compute this directly via integration, how would I evaluate the integral
$$\vec{E}(\vec{r})=\frac{1}{4\pi\varepsilon_0}\int\rho(\vec{r}')\frac{\vec{r}-\vec{r}'}{|\vec{r}-\vec{r}'|^3} dx' dy' dz'$$
if ##\vec{r}## is inside the sphere? This doesn't seem to converge for ##\vec{r}'\rightarrow\vec{r}## and ##\rho(\vec{r}')\neq\ 0##.
 

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  • #2
andrewkirk
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The non-convergence is because the electric force increases without limit as ##\vec r'\to\vec r##. This increase has no effect as long as ##\vec r## is strictly interior to the sphere - ie not on the surface. Why? - because for every incremental rectangle pushing a test particle at ##\vec r## in one direction there is an antipodal (relative to ##\vec r## as centre) rectangle pushing in the opposite direction, and thus cancelling out the force.

Let B be the subset of the sphere that is the largest spherical area centred on ##\vec r## that is entirely contained within the large sphere. Then, by the above reasoning, the field generated at ##\vec r## by the particles in B is zero. So we only need integrate over the large sphere excluding the area B - and that means that for all ##\vec r'## we have ##\|\vec r'-\vec r\|\geq \mathrm{radius}(B)>0##.

I think the calculation of electric forces inside a sphere can use the Shell Theorem, which is usually presented for gravity but should work just as well for electrostatic forces. The wiki article presents a geometric proof by Newton that is different from the one using Gauss's law (which it presents as well).
 
  • #3
andrewkirk
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Actually, now I think of it, it converges anyway, because as ##\vec r\to \vec r'## the product ##dx'dy'dz'##, which is the incremental volume element, is of order ##\|\vec r-\vec r'\|^3##, and that cancels out the denominator that was causing the concern. The apparent non-convergence is an illusion generated by using the shortcut notation ##\vec r'## in the integrand rather than the actual function of ##dx',dy',dz'## for which it stands.
 
  • #4
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I was actually talking about a filled sphere, not a spherical shell. Using Gauss' Law, one can show that the electric field inside increases linearly with increasing distance from the center and exhibits the usual ##\frac{1}{r^2}##-decay on the outside.

But you're right about the convergence, it becomes even more apparent in spherical coordinates.
 
  • #5
vanhees71
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Yes you can :-). I'd not bother with the combersome integral for the field directly but rather calculate the potential first. In Heaviside-Lorentz units we have
$$\phi(\vec{x})=\frac{1}{4 \pi} \int_{\mathbb{R}}^3 \mathrm{d} \vec{x}' \frac{\rho(\vec{x}')}{|\vec{x}-\vec{x}'|}.$$
Since it's clear that ##\phi## will depend on ##r=|\vec{x}|## only by symmetry, we can as well choose ##\vec{x}=r \vec{e}_z## and introduce the usual spherical coordinates, then for ##\rho(\vec{x})=\text{const}## for ##r<R## and ##0## for ##r \geq R## you have
$$\phi(r)=\frac{\rho}{4 \pi} \int_0^{R} \mathrm{d} r' \int_{0}^{\pi} \mathrm{d} \vartheta \int_0^{2 \pi} \mathrm{d} \varphi \frac{r^{\prime 2} \sin \vartheta}{\sqrt{r^2+r^{\prime 2} -2 r r' \cos \vartheta}}.$$
The integral over ##\varphi## just gives a factor ##2 \pi##, and then we substitute ##\cos \vartheta =u##, leading to
$$\phi(r)=\frac{\rho}{2} \int_0^{R} \mathrm{d} r' \int_{-1}^{1} \mathrm{d} u \frac{r^{\prime 2}}{\sqrt{r^2+r^{\prime 2} -2 r r' u}}.$$
Now we do the ##u## integral, leading to
$$\phi(r)=\frac{\rho}{2r} \int_0^{R} \mathrm{d} r' r'(r+r'-|r-r'|).$$
If now ##r>R## you can simply omitt the absolute sign and get
$$\phi(r)=\frac{\rho}{2r} \int_0^R \mathrm{d} r' 2 r^{\prime 2}=\frac{\rho R^3}{3 r}=\frac{Q}{4 \pi r} \quad \text{for} \quad r>R.$$
In the last step, I've used $$Q=4 \pi \rho/3$$. So this reproduces indeed the Coulomb law outside of the sphere, as it must be.

For ##r<R## you have to split the integral into two parts
$$\int_0^{R} \mathrm{d} r' r'(r+r'-|r-r'|)=\int_0^{r} \mathrm{d} r' 2 r^{\prime 2}+ \int_{r}^R \mathrm{d} r' 2 r r'=\frac{2}{3} r^3 + r (R^2-r^2)=r R^2-\frac{r^3}{3},$$
i.e.,
$$\phi(r)=\frac{\rho}{2} \left (R^2-\frac{r^2}{3} \right) \quad \text{for} \quad 0 \leq r \leq R.$$
 
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  • #6
andrewkirk
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I was actually talking about a filled sphere, not a spherical shell.
I understand that. The shell theorem starts with shells and then deduces results about filled spheres, by integrating over concentric spherical shells, to get the result that the field at any point p inside a solid sphere and distance r from the centre is the same as the field that would obtain if the entire charge (for electricity) or mass (for gravity) of the concentric sub-sphere of radius r was concentrated at the centre and there were no other charge or mass.
 
  • #7
vanhees71
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Yes, and in my calculation I just did that: First integrate over "shells", i.e., doing the angle integrals in spherical coordinates, i.e., first integrating over the spherical shells.
 

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