Q16P - Introduction to Electrodynamics

Question

Sketch the vector function

$\mathbf{v}\mathbf{=}\frac{\hat{\mathbf{r}}}{{\mathbf{r}}^{\mathbf{2}}}$

and compute its divergence. The answer may surprise you ... can you explain it?

Step-by-Step Solution

Answer

The sketch of the divergence is shown as follows:

                              

                             

The divergence of the vector function $V$ is 0, that is ${}^{\nabla .\mathrm{v}=0}$ . The divergence of the vector is outwards which means it is positive and the surface is in the form of concentric spheres.

Step 1: Describe the given information.

The given function is ${}^{\mathrm{v}=\frac{\hat{\mathrm{r}}}{{\mathrm{r}}^2}}$

Step 2: Define the divergence.

Consider the vector point function  

$\mathit{F}\mathbf{(}\mathit{X}\mathbf{,}\mathit{Y}\mathbf{,}\mathit{Z}\mathbf{)}\mathbf{=}{\mathit{F}}_{\mathbf{1}}\mathbf{(}\mathit{X}\mathbf{,}\mathit{Y}\mathbf{,}\mathit{Z}\mathbf{)}\mathbf{+}{\mathit{F}}_{\mathbf{2}}\mathbf{(}\mathit{X}\mathbf{,}\mathit{Y}\mathbf{,}\mathit{Z}\mathbf{)}\mathbf{,}\mathit{w}\mathit{h}\mathit{e}\mathit{r}\mathit{e}\mathbf{ }{\mathit{F}}_{\mathbf{1}\mathbf{,}}{\mathit{F}}_{\mathbf{2}\mathbf{,}\mathbf{ }}{\mathit{F}}_{\mathbf{3}\mathbf{ }}\mathit{a}\mathit{r}\mathit{e}\mathbf{ }\mathit{c}\mathit{o}\mathit{m}\mathit{p}\mathit{o}\mathit{n}\mathit{e}\mathit{n}\mathit{t}\mathit{s}\mathbf{ }\mathit{o}\mathit{f}\mathbf{ }\mathit{F}\mathbf{(}\mathit{X}\mathbf{,}\mathit{Y}\mathbf{,}\mathit{Z}\mathbf{)}$

The divergence of function ${}^{\mathbf{F}\mathbf{ }\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{,}\mathbf{z}\mathbf{)}}$ is computed as follows:

$\mathbf{\nabla }\mathit{F}\mathbf{(}\mathit{x}\mathbf{,}\mathit{y}\mathbf{,}\mathit{z}\mathbf{)}\mathbf{=}\frac{\mathbf{\partial }{\mathbf{F}}_{\mathbf{1}}}{\mathbf{\partial }\mathbf{x}}\mathbf{+}\frac{\mathbf{\partial }{\mathbf{F}}_{\mathbf{2}}}{\mathbf{\partial }\mathbf{y}}\mathbf{+}\frac{\mathbf{\partial }{\mathbf{F}}_{\mathbf{3}}}{\mathbf{\partial }\mathbf{z}}$

Here,$\frac{\mathbf{\partial }{\mathbf{F}}_{\mathbf{1}}}{\mathbf{\partial }\mathbf{x}}\mathbf{+}\frac{\mathbf{\partial }{\mathbf{F}}_{\mathbf{2}}}{\mathbf{\partial }\mathbf{y}}\mathbf{+}\frac{\mathbf{\partial }{\mathbf{F}}_{\mathbf{3}}}{\mathbf{\partial }\mathbf{z}}$are the partial derivatives of function ${}^{\mathbf{F}\mathbf{ }\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{,}\mathbf{z}\mathbf{)}}$ with respect to ${}^{\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{,}\mathbf{z}\mathbf{)}}$.

Step: 3 Compute the vector v.

The position vector is ${}^{\mathrm{r}=\mathrm{x}\hat{\mathrm{x}}+\mathrm{y}\hat{\mathrm{y}}+\mathrm{z}\hat{\mathrm{z}}}$ and the scalar potential is defined as .

${}^{\mathrm{v}=\frac{\hat{\mathrm{r}}}{{\mathrm{r}}^2}}$

Find the unit position vectors.

$\hat{r}=\frac{x\hat{x}+y\hat{y}+z\hat{z}}{\sqrt{x^2+y^2+z^2}}$

Substitute$\frac{x\hat{x}+y\hat{y}+z\hat{z}}{\sqrt{x^2+y^2+z^2}}$for  $\hat{r} ,\sqrt{x^2+y^2+z^2}$, for r into the equation .

${}^{\mathrm{v}=\frac{\hat{\mathrm{r}}}{{\mathrm{r}}^2}}$$$\begin{gathered} v=\frac{\frac{x\hat{x}+y\hat{y}+z\hat{z}}{\sqrt{x^2+y^2+z^2}}}{{\left(\sqrt{x^2+y^2+z^2}\right)}^2} \\ =\frac{x\hat{x}+y\hat{y}+z\hat{z}}{{\left(x^2+y^2+z^2\right)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}} \\ =x{\left(x^2+y^2+z^2\right)}^{\raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\hat{x}+y{\left(x^2+y^2+z^2\right)}^{\raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\hat{y}+z{\left(x^2+y^2+z^2\right)}^{\raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\hat{z} \end{gathered}$$

The  $x, y, z$ components of the scalar potential v are as follows:

$$\begin{gathered} v_x=x{\left(x^2+y^2+z^2\right)}^{\raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \\ {v}_y=y{\left(x^2+y^2+z^2\right)}^{\raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \\ {v}_z=z{\left(x^2+y^2+z^2\right)}^{\raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \end{gathered}$$

The sketch of the divergence is shown as follows:

                          

                            

The divergence of the vector is outwards which means it is positive and the surface is in the form of concentric spheres.

Step 4: Compute the divergence of vector v.

The divergence of the vector ${}^{\mathrm{v}}$ is obtained as follows:,

$\nabla .v=\frac{\partial v_x}{\partial \mathrm{x}}+\frac{\partial v_y}{\partial \mathrm{y}}+\frac{\partial v_z}{\partial \mathrm{z}}$

$$\begin{gathered} =\frac{\partial \left(x{\left(x^2+y^2+z^2\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)}{\partial x}+\frac{\partial \left(y{\left(x^2+y^2+z^2\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)}{\partial y}+\frac{\partial \left(z{\left(x^2+y^2+z^2\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)}{\partial z} \\ =\left[\begin{array}{c}{\left(x^2+y^2+z^2\right)}^{\raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-x\left(-\frac{3}{2}\right){\left(x^2+y^2+z^2\right)}^{\raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left(2x\right)\\ +{\left(x^2+y^2+z^2\right)}^{-\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+y\left(-\frac{3}{2}\right){\left(x^2+y^2+z^2\right)}^{\raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left(2y\right)\\ +{\left(x^2+y^2+z^2\right)}^{\raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+z\left(-\frac{3}{2}\right){\left(x^2+y^2+z^2\right)}^{\raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left(2z\right)\end{array}\right] \\ =3{\left(x^2+y^2+z^2\right)}^{\raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-3{\left(x^2+y^2+z^2\right)}^{\raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \end{gathered}$$

Solve further as,

$\nabla .\mathrm{v}=0$

Thus the divergence of the vector function ${}^{\mathrm{v}}$ is 0, that is $\nabla .\mathrm{v}=0$