Finding $\nabla^2V $ if $V=\frac{2\cos\theta+3\sin^3\theta \cos\phi}{r^2}$

Question

How do you find $\nabla^2V $ if $V=\frac{2\cos\theta+3\sin^3\theta \cos\phi}{r^2}$

The correct answer is supposedly $\frac{6\sin\theta \cos\phi(4-5\sin^2\theta)}{r^4}$, but I can't seem to get the answer. Can anyone help?

Answer 1

The Laplacian in spherical coordinates is given by: $$ \nabla^2{V}=\frac{1}{r^2}\frac{\partial}{\partial{r}} \left(r^2 \frac{\partial V}{\partial r} \right) + \frac{1}{r^2\sin{\theta}}\frac{\partial}{\partial{\theta}} \left(\sin{\theta} \frac{\partial V}{\partial \theta} \right) + \frac{1}{r^2\sin^2{\theta}}\frac{\partial^2{V}}{\partial{\phi^2}} = 0$$

Proceeding carefully, the first term gives: $$ \begin {align} \frac{\partial V}{\partial{r}} &=\frac{-4\cos\theta-6\sin^3\theta\cos\phi}{r^3} \\ {r^2}\frac{\partial V}{\partial{r}} &= \frac{-4\cos\theta-6\sin^3\theta\cos\phi}{r} \\ \frac{\partial}{\partial{r}} \left({r^2}\frac{\partial V}{\partial{r}}\right) &= \frac{4\cos\theta + 6\sin^3\theta\cos\phi}{r} \\ \Rightarrow \frac{1}{r^2}\frac{\partial}{\partial{r}} \left(r^2 \frac{\partial V}{\partial r} \right) &= \frac{4\cos\theta + 6\sin^3\theta\cos\phi}{r^4} \end{align} \nonumber $$

The second term gives: $$ \begin {align} \frac{\partial V}{\partial{\theta}} &=\frac{-2\sin\theta + 9\sin^2\theta\cos\theta\cos\phi}{r^2} \\ \sin\theta\frac{\partial V}{\partial{\theta}} &=\frac{-2\sin^2\theta + 9\sin^3\theta\cos\theta\cos\phi}{r^2} \\ \frac{\partial}{\partial{\theta}}\left( \sin\theta\frac{\partial V}{\partial{\theta}} \right) &= \frac{-4\sin\theta\cos\theta + 9\cos\phi \left(3\sin^2\theta - 4\sin^4\theta\right)}{r^2} \\ \Rightarrow \frac{1}{r^2 \sin\theta}\frac{\partial}{\partial{\theta}}\left( \sin\theta\frac{\partial V}{\partial{\theta}} \right) &= \frac{-4\cos\theta + 9\cos\phi \left(3\sin\theta - 4\sin^3\theta\right)}{r^4} \end{align} \nonumber $$

The third term gives: $$ \begin {align} \frac{\partial V}{\partial{\phi}} &= -\frac{3\sin^3\theta}{r^2}\sin\phi \\ \frac{\partial^2 V}{\partial{\phi}^2} &= -\frac{3\sin^3\theta}{r^2}\cos\phi \\ \Rightarrow \frac{1}{r^2 \sin^2\theta} \left(\frac{\partial^2 V}{\partial{\phi}^2} \right) &= -\frac{3\sin\theta\cos\phi}{r^4} \\ \end{align} \nonumber $$

Adding up all three terms gives: $$ \begin {align} \nabla^2{V} &= \frac{4\cos\theta + 6\sin^3\theta\cos\phi - 4\cos\theta + 9\cos\phi \left(3\sin\theta - 4\sin^3\theta \right) -3\sin\theta\cos\phi }{r^4} \\ &= \frac{6\sin^3\theta\cos\phi + 27\sin\theta\cos\phi -36\sin^3\theta\cos\phi -3\sin\theta\cos\phi }{r^4} \\ &= \frac{24\sin\theta\cos\phi -30\sin^3\theta\cos\phi }{r^4} \\ &= \frac{6\sin\theta\cos\phi \left(4 -5\sin^2\theta \right)}{r^4} \\ \Rightarrow \nabla^2{V} &= \frac{6\sin\theta\cos\phi \left(4 -5\sin^2\theta\right)}{r^4} \end{align} \nonumber $$