Question

F=(2xy +z³)i + x³j + 3xz²k find a scalar potential and work done in moving an object in the field from (1,-2,1) to (3,1,4)​

Answer 1

Step-by-step explanation:

Given:

$\textbf{F} = (2xy + z^3)\hat{\textbf{i}} + x^3\hat{\textbf{j}} + 3xz^2\hat{\textbf{k}}$

This field will have a scalar potential $\varphi$ if it satisfies the condition $\nabla \times \textbf{F}=0$. While the first x- and y- components of $\nabla \times \textbf{F}$ are satisfied, the z-component doesn't.

$(\nabla \times \textbf{F})_z = \left(\dfrac{\partial F_y}{\partial x} - \dfrac{\partial F_x}{\partial y} \right)$

$\:\:\:\:\:\:\:\:\: = 3x^2 - 2x \ne 0$

Therefore the field is nonconservative so it has no scalar potential. We can still calculate the work done by defining the position vector $\vec{\textbf{r}}$ as

$\vec{\textbf{r}} = x \hat{\textbf{i}} + y \hat{\textbf{j}} + z \hat{\textbf{k}}$

and its differential is

$\textbf{d} \vec{\textbf{r}} = dx \hat{\textbf{i}} + dy \hat{\textbf{j}} + dz \hat{\textbf{k}}$

The work done then is given by

$\displaystyle \oint_c \vec{\textbf{F}} • \textbf{d} \vec{\textbf{r}} = \int ((2xy + z^3)\hat{\textbf{i}} + x^3\hat{\textbf{j}} + 3xz^2\hat{\textbf{k}}) • (dx \hat{\textbf{i}} + dy \hat{\textbf{j}} + dz \hat{\textbf{k}})$

$\displaystyle = (x^2y + xz^3) + x^3y + xz^3|_{(1, -2, 1)}^{(3, 1, 4)}$

$= 422$