Answer
$div F=\dfrac{3-p}{(x^2+y^2+z^2)^{p/2}}$ ; div F=0 when p=3
Work Step by Step
Apply definition. $div F=\dfrac{\partial p}{\partial x}+\dfrac{\partial q}{\partial y}+\dfrac{\partial r}{\partial z}$
$div F=\dfrac{\partial p}{\partial x}+\dfrac{\partial q}{\partial y}+\dfrac{\partial r}{\partial z}=\dfrac{2x^2(\dfrac{-p}{2})}{(x^2+y^2+z^2)^{p/2+1}}+\dfrac{1}{(x^2+y^2+z^2)^{p/2}}+\dfrac{2y^2(\dfrac{-p}{2})}{(x^2+y^2+z^2)^{p/2+1}}+\dfrac{1}{(x^2+y^2+z^2)^{p/2}}+\dfrac{2z^2(\dfrac{-p}{2})}{(x^2+y^2+z^2)^{p/2+1}}+\dfrac{1}{(x^2+y^2+z^2)^{p/2}}$
or, $div F=\dfrac{-p(x^2+y^2+z^2)}{(x^2+y^2+z^2)^{p/2+1}}+\dfrac{3}{(x^2+y^2+z^2)^{p/2}}$
Hence, we get $div F=\dfrac{3-p}{(x^2+y^2+z^2)^{p/2}}$ ; div F=0 when p=3
