Let $\mathbf{I}$ be a continuous vector valued function in a volume $V'$ having boundary $S'$

Let $\mathbf{r}$ be the distance vector from points in $V'$ to a point $P$

Then the potential at point $P$ inside $V'$ due to a dipole distribution is:

$$\Phi = \int_{V'} \dfrac{\mathbf{I}.\mathbf{\hat{r}}}{r^2} dV' \tag1$$

Using spherical coordinate system, it can be seen that there is no integrand discontinuity.

$$\Phi = \int_{V'} \dfrac{\mathbf{I}.\mathbf{\hat{r}}}{r^2}\ r^2 \sin \theta'\ d\theta'\ d\phi\ dr'$$

Similarly, the field at point $P$ inside $V'$ due to a dipole distribution is:

$$\mathbf{E}=-\nabla \Phi=2 \int_V \dfrac{\mathbf{I}.\mathbf{\hat{r}}}{r^3}\ (\mathbf{\hat{r}})\ dV'$$

Using spherical coordinate system,

$$\mathbf{E}=-\nabla \Phi=2 \int_V \dfrac{\mathbf{I}.\mathbf{\hat{r}}}{r^3}\ (\mathbf{\hat{r}})\ r^2 \sin \theta'\ d\theta'\ d\phi\ dr'$$

the integrand discontinuity remains. So we have:

Statement I: $\mathbf{E}$ blows up at points inside $V'$

If $\delta$ is a small volume around point $P$, then applying vector identity $\nabla.(a \mathbf{A})=\mathbf{A}.(\nabla a)+a (\nabla.\mathbf{A})$ and the divergence theorem to equation (1):

\begin{align} \Phi &=-\int_{V'} \dfrac{\nabla.\mathbf{I}}{r} dV' + \lim \limits_{\delta \to 0} \int_{V'-\delta} \nabla . \left( \dfrac{\mathbf{I}}{r} \right) dV'\\ &=-\int_{V'} \dfrac{\nabla.\mathbf{I}}{r} dV' + \oint_{S'} \dfrac{\mathbf{I}.\mathbf{\hat{n}}}{r} dS' + \lim \limits_{\delta \to 0} \oint_{\partial \delta} \dfrac{\mathbf{I}.\mathbf{\hat{n}}}{r} dS'\\ &=-\int_{V'} \dfrac{\nabla.\mathbf{I}}{r} dV' + \oint_{S'} \dfrac{\mathbf{I}.\mathbf{\hat{n}}}{r} dS' \end{align}


\begin{align} \mathbf{E}&=-\nabla \Phi\\ &=-\int_{V'} \dfrac{\nabla.\mathbf{I}}{r^2} \mathbf{\hat{r}}\ dV' + \oint_{S'} \dfrac{\mathbf{I}.\mathbf{\hat{n}}}{r^2} \mathbf{\hat{r}}\ dS' \end{align}

Using spherical coordinate system, we can show that there is no integrand discontinuity in the first term.

Since $P$ is a point inside $V'$, the second term also has no discontinuity. So we have:

Statement II: $\mathbf{E}$ does not blow up at points inside $V'$

Statetements I and II are derived from equation (1). Yet there is a contradiction. Why?


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