Verify that the force field is conservative. Then find a scalar potential φ such that$\mathit{F}\mathbf{=}\mathbf{-}\mathbf{\nabla }\mathit{\phi }\mathbf{,}\mathit{F}\mathbf{=}\mathbf{-}\mathit{k}\mathit{r}\mathbf{,}\mathit{r}\mathbf{=}\mathit{i}\mathit{x}\mathbf{+}\mathit{j}\mathit{y}\mathbf{+}\mathit{k}\mathit{z}$,

k= constant.

The force field is$F=-kr,r=ix+jy+kz$and$F=-\nabla \phi$ .

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

The formula for thescalarpotential is$\mathit{W}\mathbf{=}\mathbf{\int }\mathit{F}\mathbf{.}\mathbf{}\mathit{d}\mathit{r}$ .

The force is said to be conservative if $\nabla \times F=0$.

Putthe values given below in the above equation.

$$\begin{gathered} F=-k\left(r\right) \\ F=-kxi-kyj-kzk \end{gathered}$$

The equation becomes as follows.

$\begin{array}{rcl}\nabla \times F& =& \left|\begin{array}{ccc}i& j& k\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ -kx& -ky& -kz\end{array}\right|\\ & & \\ \nabla \times F& =& \left(\frac{\partial }{\partial y}(-kz)-\frac{\partial }{\partial z}(-ky)\right)i-\left(\frac{\partial }{\partial x}(-kz)-\frac{\partial }{\partial z}(-kx)\right)j+\left(\frac{\partial }{\partial x}(-ky)-\frac{\partial }{\partial y}(-kx)\right)k\\ \nabla \times F& =& 0\end{array}$

Hence the field is conservative.

The formula for the scalar potential is $W=\int F.dr$.

$$\begin{gathered} W=\int (-kxi-kyj-kzk).(dxi+dyj+dzk) \\ W=\int -kxdx-kydy-kzdz \end{gathered}$$

$W_1$ is from(0,0,0) to (x,0,0).

y = 0

dy = 0

z = 0

dz = 0

Substitute the above value in the equation mentioned below.

$$\begin{gathered} W=\int -kxdx-kydy-kzdz \\ {W}_1=\underset{0}{\overset{x}{\int }}-kxdx \\ {W}_1={\left[\frac{-k{x}^2}{2}\right]}_0^x \\ {W}_1=\frac{-k{x}^2}{2} \end{gathered}$$

$W_2$ is from (x,0,0) to (x,0,z) and is constant.

dx = 0

y = 0

dy = 0

Substitute the above value in the equation mentioned below.

$$\begin{gathered} W=\int -kxdx-kydy-kzdz \\ {W}_2=\underset{0}{\overset{z}{\int }}-kzdz \\ {W}_2={\left[\frac{-k{z}^2}{2}\right]}_0^z \\ {W}_2=\frac{-k{z}^2}{2} \end{gathered}$$

$W_3$ is from (x,0,z) to (x,y,z) and ,x,z are constant.

dx = 0

dz = 0

Substitute the above value in the equation mentioned below.

$$\begin{gathered} W=\int -kxdx-kydy-kzdz \\ {W}_3=\underset{0}{\overset{y}{\int }}-kydy \\ {W}_3={\left[\frac{-k{y}^2}{2}\right]}_0^y \\ {W}_3=\frac{-k{y}^2}{2} \end{gathered}$$

The formula states the equation mentioned below.

$$\begin{gathered} W=W_1+W_2+W_3 \\ W=\frac{-k{x}^2}{2}+\frac{-k{y}^2}{2}+\frac{-k{z}^2}{2} \\ W=-\left(\frac{k{x}^2}{2}+\frac{k{y}^2}{2}+\frac{k{z}^2}{2}\right) \end{gathered}$$

The formula states the equation mentioned below.

$F=\nabla W$

It is given that $F=-\nabla \phi$.

By both the values of $-\nabla \phi =\nabla W$, .

$\phi =-W$

Put the value of W in above equation.

$$\begin{gathered} \phi =-(-\frac{k{x}^2}{2}-\frac{k{y}^2}{2}-\frac{k{z}^2}{2}) \\ \phi =\frac{k{x}^2}{2}+\frac{k{y}^2}{2}+\frac{k{z}^2}{2}) \end{gathered}$$

Hence the scalar potential is $\frac{k{x}^2}{2}+\frac{k{y}^2}{2}+\frac{k{z}^2}{2}$.