VC3.35

From Exampleproblems

$\nabla^2(\frac{x}{r^2})=\sum \frac{\partial^2}{\partial x^2}[\frac{x}{r^2}]\,$ --(1)

Now, $\frac{\partial}{\partial x}(\frac{x}{r^2})=\frac{1}{r^2}-\frac{2x}{r^3}\frac{\partial r}{\partial x}=\frac{1}{r^2}-\frac{2x}{r^3}[\frac{x}{r}]=\frac{1}{r^2}-\frac{2x^2}{r^4}\,$ --(2) [Since $r^2=x^2+y^2+z^2\,$ and $\frac{\partial r}{\partial x}=\frac{x}{r}\,$ ]

Therefore, $\frac{\partial^2}{\partial r^2}=\frac{\partial}{\partial x}[\frac{\partial}{\partial x}(\frac{x}{r^2})]=\frac{\partial}{\partial x}[\frac{1}{r^2}-\frac{2x^2}{r^4}]\,$ , by using (2)

= $-\frac{2}{r^3}\frac{\partial r}{\partial x}-[\frac{4x}{r^4}-\frac{8x^2}{r^5}\frac{\partial r}{\partial x}]=-\frac{2}{r^3}(\frac{x}{r})-\frac{4x}{r^4}+\frac{8x^2}{r^5}\frac{x}{r}\,$

Therefore, $\frac{\partial^2}{\partial x^2}[\frac{x}{r^2}]=\frac{8x^3}{r^6}-\frac{6x}{r^4}\,$ --(3)

Now, $\frac{\partial}{\partial y}(\frac{x}{r^2})=-\frac{2x}{r^3}(\frac{y}{r})\,$

$\frac{\partial^2}{\partial y^2}=\frac{\partial}{\partial y}[-\frac{2xy}{r^4}]\,$

= $-2x[\frac{1}{r^4}-\frac{4y}{r^5}\frac{\partial r}{\partial y}]=-2x[\frac{1}{r^4}-\frac{4y^2}{r^6}]=\frac{8xy^2}{r^6}-\frac{2x}{r^4}\,$ --(4)

Similarly, $\frac{\partial^2}{\partial z^2}[\frac{x}{r^2}]=\frac{8xz^2}{r^6}-\frac{2x}{r^4}\,$ --(5)

Adding (3),(4),(5), we have

$\sum \frac{\partial^2}{\partial x^2}=\frac{8x}{r^6}[x^2+y^2+z^2]-\frac{10x}{r^4}=-\frac{2x}{r^4}\,$

or $\nabla^2[\frac{x}{r^2}]=-\frac{2x}{r^4}\,$ , by using (1)

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