91Ó°ÊÓ

The force field are mentioned below.

$\begin{array}{rcl}F& =& -\frac{C}{r^2}{e}_r\\ & =& \frac{C}{r^2}\frac{r}{\left|r\right|}\\ & =& -\frac{C}{r^3}r\\ F& =& -mg\text{k}\\ & & \end{array}$

A force is said to be conservative if $\mathbf{∇}\mathbf{×}\mathit{F}\mathbf{=}\mathbf{0}$.

The scalar potential is independent of the path. The scalar potential is the sum of potential in all the 3 dimensions calculated separately.

The formula for the scalar potential is$\mathit{W}\mathbf{=}\mathbf{∫}\mathit{F}\mathbf{.}\mathit{d}\mathit{r}$ .

The force is said to be conservative if$∇×F=0$.

$\begin{array}{ccc}i& j& k\\ \frac{∂}{∂x}& \frac{∂}{∂y}& \frac{∂}{∂z}\\ 0& 0& -mg\end{array}$

$$\begin{gathered} i\left(0-0\right)-j\left(0-0\right)+k\left(0-0\right) \\ 0 \end{gathered}$$

Solve for other equation.

$\begin{array}{ccc}i& j& k\\ \frac{∂}{∂x}& \frac{∂}{∂y}& \frac{∂}{∂z}\\ \frac{-Cx}{{\left(x^2+y^2+z^2\right)}^{\frac{3}{2}}}& \frac{-Cy}{{\left(x^2+y^2+z^2\right)}^{\frac{3}{2}}}& \frac{-Cz}{{\left(x^2+y^2+z^2\right)}^{\frac{3}{2}}}\end{array}$

$\begin{array}{rcl}\left(∇×F\right)i& =& \left(\frac{3Cyz{\left(x^2+y^2+z^2\right)}^{\frac{-3}{2}}}{\left(x^2+y^2+z^2\right)}-\frac{3Cyz{\left(x^2+y^2+z^2\right)}^{\frac{-3}{2}}}{\left(x^2+y^2+z^2\right)}\right)\\ & =& 0\end{array}$

$\begin{array}{rcl}\left(∇×F\right)j& =& \left(\frac{3Cxz{\left(x^2+y^2+z^2\right)}^{\frac{-3}{2}}}{\left(x^2+y^2+z^2\right)}-\frac{3Cxz{\left(x^2+y^2+z^2\right)}^{\frac{-3}{2}}}{\left(x^2+y^2+z^2\right)}\right)\\ & =& 0\end{array}$

$\begin{array}{rcl}\left(∇×F\right)k& =& \left(\frac{3Cyz{\left(x^2+y^2+z^2\right)}^{\frac{-3}{2}}}{\left(x^2+y^2+z^2\right)}-\frac{3Cyz{\left(x^2+y^2+z^2\right)}^{\frac{-3}{2}}}{\left(x^2+y^2+z^2\right)}\right)\\ & =& 0\end{array}$

Hence, the equations are conservative.