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Quote:

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[itex]\mathbf{E(r)}=\int\;d\mathbf{r'}\;\frac{\mathbf{r-r'}}{|\mathbf{r-r'}|^3}\rho(\mathbf{r'})[/itex]

[itex]\mathbf{E(r)}=\int\;d\mathbf{r'}\;\frac{\mathbf{r-r'}}{|\mathbf{r-r'}|^3}\rho(\mathbf{r'})[/itex]

where [itex]d\mathbf{r'}[/itex] represents the three-dimensional volume element. Note that in spite of the singularity at [itex]\mathbf{r=r'}[/itex], the integral is finite for a finite charge distribution, even when the point [\itex]\mathbf{r}[\itex] is in the region containing charge. This is because the volume element [itex]d\mathbf{r'}[/itex] in the neighbourhood of a point [itex]\mathbf{r'}[/itex] goes like [itex]|\mathbf{r-r'}|^2[/itex] for small [itex]\mathbf{r-r'}[/itex], thereby cancelling the singularity.

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I have highlighted the part that I do not fully understand. Does that sentence means in the spherical coordinate, one can write the volume element as [itex]d\mathbf{r'}=\tilde{r}^2\sin\theta d\tilde{r}d\theta d\phi[/itex] where [itex]\tilde{r}=|\mathbf{r-r'}|[/itex], and the [itex]\tilde{r}^2[/itex] terms cancel? If so, what if I am not using the spherical coordinate but others such as Cartesian coordinate? There is no [itex]\tilde{r}^2[/itex] term to do the cancellation.

Thanks in advance for giving me a helping hand.

Reference: P.3, Classical Field Theory by Francis E. Low