VC3.27

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Here $r^2=x^2+y^2+z^2,\frac{\partial r}{\partial x}=\frac{x}{r}\,$ --(1)

Now, $\nabla^2[\frac{x}{r^3}]=\sum [\frac{\partial^2}{\partial x^2}{\frac{x}{r^3}}]\,$ --(2)

But $\frac{\partial^2}{\partial x^2}[\frac{x}{r^3}]=\frac{\partial}{\partial x}[\frac{\partial}{\partial x}{\frac{x}{r^3}}]=\frac{\partial}{\partial x}[\frac{1}{r^3}-\frac{3}{r^4}\frac{\partial r}{\partial x}]\,$

= $\frac{\partial}{\partial x}[\frac{1}{r^3}-\frac{3}{r^4}\cdot\frac{x}{r}]\,$ ,by using (1)

= $\frac{\partial}{\partial x}[\frac{1}{r^3}]-3\frac{\partial}{\partial x}[\frac{x^2}{r^5}]\,$

= $-\frac{3}{r^4}\frac{\partial r}{\partial x}-3[\frac{2x}{r^5}-\frac{5x^2}{r^6}\frac{\partial r}{\partial x}]\,$

= $-\frac{3x}{r^5}-3[\frac{2x}{r^5}-\frac{5x^3}{r^7}]=-\frac{9x}{r^5}+\frac{15x^3}{r^7}\,$ --(3)

Also, $\frac{\partial^2}{\partial y^2}[\frac{x}{r^3}]=x\frac{\partial}{\partial y}[\frac{\partial}{\partial y}{\frac{1}{r^3}}]\,$

= $x\frac{\partial}{\partial y}[-\frac{3}{r^4}\frac{\partial r}{\partial y}]=-3x\frac{\partial}{\partial y}[\frac{y}{r^5}]\,$ by (1)

= $-3x[\frac{1}{r^5}-\frac{5}{r^6}\frac{\partial}{\partial y}y]=-3x[\frac{1}{r^5}-\frac{5y}{r^6}\cdot\frac{y}{r}]\,$

= $-3x[\frac{1}{r^5}-\frac{5y^2}{r^7}]=-\frac{3x}{r^5}+\frac{15xy^2}{r^7}\,$ --(4)

Similarly, $\frac{\partial^2}{\partial z^2}[\frac{x}{r^3}]=-\frac{3x}{r^5}+\frac{15xz^2}{r^7}\,$ --(5)

Using (3),(4),(5), Equation(2) reduces to

$\nabla^2[\frac{x}{r^3}]=-\frac{9x}{r^5}+\frac{15x^3}{r^7}-\frac{3x}{r^5}+\frac{15xy^2}{r^7}-\frac{3x}{r^5}+\frac{15xz^2}{r^7}\,$

= $-\frac{15}{r^5}+\frac{15x}{r^7}(x^2+y^2+z^2)=-\frac{15x}{r^5}+\frac{15x}{r^5}=0\,$ , which is to be proved.

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VC3.27

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