Topic: Electronics Engineering
If $\varphi=2x^{2}y-xz^{3}.$ then the Laplacian of $\varphi$ is
- A. $4yz-6xz$
- B. $4z-6xy$
- C. $4y-6xz$
- D. $2xy-6yz$
Answer: C. $4y-6xz$
Explanation: Laplacian of $\phi = \nabla^{2}\phi = \frac{\partial^{2}\phi}{{\partial x}^{2}}+\frac{\partial^{2}\phi}{\partial y^{2}}+\frac{\partial^{2}\phi}{\partial z^{2}}$.
$\frac{\partial^{2}\phi}{\partial x^{2}}=4y$, $\frac{\partial^{2}\phi}{\partial y^{2}}=0$, $\frac{\partial^{2}\phi}{\partial z^{2}}=-6xz$.
$\frac{\partial^{2}\phi}{\partial x^{2}}+\frac{\partial^{2}\phi}{\partial y^{2}}+\frac{\partial^{2}\phi}{\partial z^{2}} = 4y-6xz$.