- #1

- 58

- 6

## Main Question or Discussion Point

The potential of a dipole distribution at a point ##P## is:

##\psi=-k \int_{V'}

\dfrac{\vec{\nabla'}.\vec{M'}}{r}dV'

+k \oint_{S'}\dfrac{\vec{M'}.\hat{n}}{r}dS'##

If ##P\in V'##, the integrand is discontinuous (infinite) at the point ##r=0##. So we need to use improper integrals by removing a small cavity ##\delta## from ##V'##:

##\psi=k \left[ -\lim \limits_{\delta \to 0}

\int_{V'-\delta} \dfrac{\vec{\nabla'}.\vec{M'}}{r} dV'+

\lim \limits_{\delta \to 0}

\oint_{S'+\Delta} \dfrac{\vec{M'}.\hat{n}}{r}dS'

\right] \tag 1##

##\nabla\psi=k \left[ -\lim \limits_{\delta \to 0}

\int_{V'-\delta} (\vec{\nabla'}.\vec{M'})\ \nabla \left( \dfrac{1}{r} \right) dV'+

\lim \limits_{\delta \to 0}

\oint_{S'} (\vec{M'}.\hat{n})\ \nabla \left( \dfrac{1}{r} \right)dS'

\right] \tag 2##

\begin{align}

\nabla^2 \psi&=k \left[ -\lim \limits_{\delta \to 0}

\int_{V'-\delta} (\vec{\nabla'}.\vec{M'})\ \nabla^2 \left( \dfrac{1}{r} \right) dV'+

\lim \limits_{\delta \to 0}

\oint_{S'} (\vec{M'}.\hat{n})\ \nabla^2 \left( \dfrac{1}{r} \right)dS'

\right]\\

&\bbox[yellow]{-4\pi\ k\ (\vec{\nabla}.\vec{M'})}\\

&=-4\pi\ k\ (\vec{\nabla}.\vec{M'})\\ \tag 3

\end{align}

##\psi=-k \int_{V'}

\dfrac{\vec{\nabla'}.\vec{M'}}{r}dV'

+k \oint_{S'}\dfrac{\vec{M'}.\hat{n}}{r}dS'##

If ##P\in V'##, the integrand is discontinuous (infinite) at the point ##r=0##. So we need to use improper integrals by removing a small cavity ##\delta## from ##V'##:

##\psi=k \left[ -\lim \limits_{\delta \to 0}

\int_{V'-\delta} \dfrac{\vec{\nabla'}.\vec{M'}}{r} dV'+

\lim \limits_{\delta \to 0}

\oint_{S'+\Delta} \dfrac{\vec{M'}.\hat{n}}{r}dS'

\right] \tag 1##

**It must be noted that we ignored the contribution to the volume integral from the point ##r=0## (if there is any).**##\nabla\psi=k \left[ -\lim \limits_{\delta \to 0}

\int_{V'-\delta} (\vec{\nabla'}.\vec{M'})\ \nabla \left( \dfrac{1}{r} \right) dV'+

\lim \limits_{\delta \to 0}

\oint_{S'} (\vec{M'}.\hat{n})\ \nabla \left( \dfrac{1}{r} \right)dS'

\right] \tag 2##

**Here also, it must be noted that we ignored the contribution to the volume integral from the point ##r=0## (if there is any).**\begin{align}

\nabla^2 \psi&=k \left[ -\lim \limits_{\delta \to 0}

\int_{V'-\delta} (\vec{\nabla'}.\vec{M'})\ \nabla^2 \left( \dfrac{1}{r} \right) dV'+

\lim \limits_{\delta \to 0}

\oint_{S'} (\vec{M'}.\hat{n})\ \nabla^2 \left( \dfrac{1}{r} \right)dS'

\right]\\

&\bbox[yellow]{-4\pi\ k\ (\vec{\nabla}.\vec{M'})}\\

&=-4\pi\ k\ (\vec{\nabla}.\vec{M'})\\ \tag 3

\end{align}

**Here, it must be noted that we *did not* ignore the contribution to the volume integral from the point ##r=0##.***So why is it that we are ignoring the contribution to the volume integral from the point ##r=0## in equation ##(1)## and ##(2)##?*