# Electric field on the boundary of a continuous charge distribution

In Purcell and Morin's Electricity and Magnetism, 3rd Edition, the claim is made that the magnitude of the electric field on the boundary of a continuous charge distribution is finite (assuming the charge distribution is finite everywhere). I have a question similar to one asked here, but I do not think my question was fully answered.

Equation ($$1.22$$): $$\vec{E}(x,y,z)=\dfrac{1}{4 \pi \epsilon_0} \int \dfrac{ρ\ (x^\prime, y^\prime, z^\prime)\ \hat{r}\ dx^\prime, dy^\prime, dz^\prime}{r^2}.\tag{1.22}$$

Looking at Equation (1.22), I can see how, when using spherical coordinates, the $$r^2$$ in the denominator of Equation (1.22) is canceled and, consequently, the integral does not become infinite when $$r=0$$. However, I do not see how performing the same integral using Cartesian coordinates would not blow up. The limits of integration for this example would contain $$(x,y,z)=(0,0,0)$$ making the denominator 0. There is no $$r^2$$, when using Cartesian coordinates, to cancel out the $$r^2$$ in the denominator. To my understanding, an integral cannot be finite in spherical coordinates and infinite in Cartesian. So, how is the Cartesian integral finite?

• Does your reference mention Cauchy principal values? Commented Jan 29, 2019 at 1:05
• It does not, unfortunately
– dts
Commented Jan 29, 2019 at 1:35
• The denominator should go to zero in whatever coordinate system you use, so I think you made a mistake in what you did for spherical coordinates. The singular integrand will always be there, but using the Cauchy principal value, the infinities from opposing sides of the point where you are calculating will cancel. My answer to the similar question was not very detailed & received no votes. A more complete answer would probably be too long for here & would require better familiarity with MathJax than I have. Commented Jan 29, 2019 at 1:44

If you use the Cartesian coordinate, you should recall the definition of "improper integral". Let me give you a simpler example here, consider

$$\int_0^1\frac{1}{\sqrt{x}}dx$$,

the integrand approaches infinity as $$x\to 0$$, however, the "improper integral " defines it as a limit, i.e.,

$$\int_0^1\frac{1}{\sqrt{x}}dx=\lim_{\epsilon\to0}\int_{\epsilon}^1\frac{1}{\sqrt{x}}dx=\lim_{\epsilon\to0}2\sqrt{x}\big|_{\epsilon}^1=2$$,

thus the integral will still give you a finite value.

Note that if the singularity is even stronger, you may consider the Cauchy principle value. But this integrand has the singularity type $$1/r^2$$, in 3D, this integral is just weakly singular, so probably you do not even need to use the principle value.

• Your example has a much weaker singularity that what the question asks about. To do the integral in the question, the Cauchy principal value will certainly be needed. Commented Jan 29, 2019 at 17:25
• Maybe. His integral is in 3D, higher dimension will weaken the singularity. Commented Jan 29, 2019 at 20:02