আমরা জানি, অবস্থান ভেক্টর $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$ এবং এর মান $r = \sqrt{x^2 + y^2 + z^2}$। সুতরাং, $r^2 = x^2 + y^2 + z^2$। এখন, আংশিক অন্তরীকরণের নিয়ম অনুযায়ী, $$ \begin{aligned} \frac{\partial r}{\partial x} &= \frac{1}{2\sqrt{x^2+y^2+z^2}} \cdot 2x = \frac{x}{r} \\ \frac{\partial r}{\partial y} &= \frac{y}{r} \\ \frac{\partial r}{\partial z} &= \frac{z}{r} \end{aligned} $$ বামপক্ষ, $$ \begin{aligned} \vec{\nabla} \cdot \left( \frac{\vec{r}}{r^3} \right) &= \left( \hat{i} \frac{\partial}{\partial x} + \hat{j} \frac{\partial}{\partial y} + \hat{k} \frac{\partial}{\partial z} \right) \cdot \left( \frac{x\hat{i} + y\hat{j} + z\hat{k}}{r^3} \right) \\ &= \frac{\partial}{\partial x} \left( \frac{x}{r^3} \right) + \frac{\partial}{\partial y} \left( \frac{y}{r^3} \right) + \frac{\partial}{\partial z} \left( \frac{z}{r^3} \right) \end{aligned} $$ এখন প্রতিটি পদ আলাদাভাবে নির্ণয় করি: $$ \begin{aligned} \frac{\partial}{\partial x} \left( \frac{x}{r^3} \right) &= \frac{r^3 \cdot 1 - x \cdot \frac{\partial}{\partial x}(r^3)}{r^6} \\ &= \frac{r^3 - x \cdot 3r^2 \frac{\partial r}{\partial x}}{r^6} \\ &= \frac{r^3 - 3xr^2 \cdot \frac{x}{r}}{r^6} \\ &= \frac{r^3 - 3x^2r}{r^6} = \frac{r^2 - 3x^2}{r^5} \end{aligned} $$ অনুরূপভাবে, $$ \begin{aligned} \frac{\partial}{\partial y} \left( \frac{y}{r^3} \right) &= \frac{r^2 - 3y^2}{r^5} \\ \frac{\partial}{\partial z} \left( \frac{z}{r^3} \right) &= \frac{r^2 - 3z^2}{r^5} \end{aligned} $$ যোগফল নির্ণয় করি, $$ \begin{aligned} \vec{\nabla} \cdot \left( \frac{\vec{r}}{r^3} \right) &= \frac{r^2 - 3x^2}{r^5} + \frac{r^2 - 3y^2}{r^5} + \frac{r^2 - 3z^2}{r^5} \\ &= \frac{3r^2 - 3(x^2 + y^2 + z^2)}{r^5} \end{aligned} $$ যেহেতু $x^2 + y^2 + z^2 = r^2$, $$ \begin{aligned} \vec{\nabla} \cdot \left( \frac{\vec{r}}{r^3} \right) &= \frac{3r^2 - 3r^2}{r^5} \\ &= 0 \end{aligned} $$ অতএব, $$ \vec{\nabla} \cdot \left( \frac{\vec{r}}{r^3} \right) = 0 $$ প্রমাণিত।