- #1
Mike400
- 59
- 6
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##
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)##?
##\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)##?