ধরি, অবস্থান ভেক্টর $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$। সুতরাং, এর মান, $$ r = \sqrt{x^2 + y^2 + z^2} $$ এবং, $$ r^2 = x^2 + y^2 + z^2 $$ আমাদের প্রমাণ করতে হবে যে, $\vec{\nabla} \times \left( \frac{\vec{r}}{r^2} \right) = \vec{0}$। বামপক্ষ, $$ \begin{aligned} \vec{\nabla} \times \left( \frac{\vec{r}}{r^2} \right) &= \left( \frac{\partial}{\partial x}\hat{i} + \frac{\partial}{\partial y}\hat{j} + \frac{\partial}{\partial z}\hat{k} \right) \times \left( \frac{x\hat{i} + y\hat{j} + z\hat{k}}{x^2 + y^2 + z^2} \right) \end{aligned} $$ নির্ণায়ক আকারে লিখলে পাই, $$ \begin{aligned} \vec{\nabla} \times \left( \frac{\vec{r}}{r^2} \right) &= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ \frac{x}{r^2} & \frac{y}{r^2} & \frac{z}{r^2} \end{vmatrix} \end{aligned} $$ এখন, $\hat{i}$ এর সহগ, $$ \begin{aligned} \frac{\partial}{\partial y} \left( \frac{z}{r^2} \right) - \frac{\partial}{\partial z} \left( \frac{y}{r^2} \right) &= z \frac{\partial}{\partial y} (r^{-2}) - y \frac{\partial}{\partial z} (r^{-2}) \\ &= z (-2r^{-3}) \frac{\partial r}{\partial y} - y (-2r^{-3}) \frac{\partial r}{\partial z} \end{aligned} $$ যেহেতু $r^2 = x^2 + y^2 + z^2$, তাই $2r \frac{\partial r}{\partial y} = 2y \Rightarrow \frac{\partial r}{\partial y} = \frac{y}{r}$ এবং $\frac{\partial r}{\partial z} = \frac{z}{r}$। মান বসিয়ে পাই, $$ \begin{aligned} \hat{i} &= z (-2r^{-3}) \left( \frac{y}{r} \right) - y (-2r^{-3}) \left( \frac{z}{r} \right) \\ &= -\frac{2yz}{r^4} + \frac{2yz}{r^4} = 0 \end{aligned} $$ একইভাবে, $\hat{j}$ এবং $\hat{k}$ এর সহগগুলোও শূন্য হবে। সুতরাং, $$ \vec{\nabla} \times \left( \frac{\vec{r}}{r^2} \right) = 0\hat{i} + 0\hat{j} + 0\hat{k} = \vec{0} $$ প্রমাণিত।