Q13P - Introduction to Electrodynamics

Question

Let ${}^{\mathbf{r}}$ be the separation vector from a fixed point ${}^{\mathbf{(}\mathbf{x}\mathbf{\text{'}}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\mathbf{,}\mathbf{ }\mathbf{z}\mathbf{\text{'}}\mathbf{)}}$ to the point $\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{,}\mathbf{z}\mathbf{)}$, and let r be its length. Show that

(a) ${}^{\mathbf{\nabla }\mathbf{ }\mathbf{\left(}{\mathbf{r}}^{\mathbf{2}}\mathbf{\right)}\mathbf{=}\mathbf{2}\mathbf{r}}$

(b) ${}^{\mathbf{\nabla }\mathbf{ }\mathbf{(}\mathbf{1}\mathbf{/}\mathbf{r}\mathbf{)}\mathbf{=}\mathbf{-}\mathbf{r}\mathbf{/}{\mathbf{r}}^{\mathbf{2}}}$

(c) What is the general formula for ${}^{\mathbf{\nabla }\mathbf{ }\mathbf{(}\mathbf{ }{\mathbf{r}}^{\mathbf{n}}\mathbf{ }\mathbf{)}}$

Step-by-Step Solution

Answer

(a) It is proved that ${}^{\nabla ( {\mathrm{r}}^2 )=2\overline{\mathrm{r}}}$.

(b) It is proved that${}^{\nabla \left(\frac{1}{\mathrm{r}}\right)=\frac{-1}{\left( {\mathrm{r}}^2 \right)}\left\{ \hat{\mathrm{r}} \right\}}$.

(c) The general expression is ${}^{\nabla \left( {\mathrm{r}}^{\mathrm{n}} \right)={\mathrm{nr}}^{\mathrm{n}-1} \hat{\mathrm{r}}}$

Step 1: Write the given information.

Write the divergence of the scalar function.

$\mathbf{\nabla }\mathbf{ }\mathbf{=}\frac{\mathbf{\partial }}{\mathbf{\partial }\mathbf{x}}\widehat{\mathbf{i}\mathbf{ }}\mathbf{+}\frac{\mathbf{\partial }}{\mathbf{\partial }\mathbf{y}}\hat{\mathbf{j}}\mathbf{+}\frac{\mathbf{\partial }}{\mathbf{\partial }\mathbf{z}}\hat{\mathbf{k}}$

Consider the vector function as,

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

Here, the components of the function are ${}^{F_{1,} F_{2 ,}}$ and${}^{F_3}$.

Then, the divergence of the vector function as,

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

Step 2: Derive the proof for ∇   ( 1 / r ) = - r / r 2 .

(a)

Write the expression for the separation for the separation vector.

$\stackrel{\rightharpoonup }{\mathrm{r}}=(\mathrm{x}-{\mathrm{x}}^{\text{'}}) \hat{\mathrm{x}}+(\mathrm{y}-{\mathrm{y}}^{\text{'}}) \hat{\mathrm{y}}+(\mathrm{z}-{\mathrm{z}}^{\text{'}}) \hat{\mathrm{z}}$

Since, r is the length solve further as,

Calculate the divergence of the function.

$$\begin{gathered} \nabla \left({\mathrm{r}}^2\right)=\nabla \left[{\left(\mathrm{x}-{\mathrm{x}}^{\text{'}}\right)}^2+{(\mathrm{y}-{\mathrm{y}}^{\text{'}})}^2+{(\mathrm{z}-{\mathrm{z}}^{\text{'}})}^2\right] \\ =\left\{\begin{array}{c}\frac{\partial }{\partial \mathrm{x}}\left\{{\left(\mathrm{x}-{\mathrm{x}}^{\text{'}}\right)}^2+{(\mathrm{y}-{\mathrm{y}}^{\text{'}})}^2+{(\mathrm{z}-{\mathrm{z}}^{\text{'}})}^2\right\}\hat{\mathrm{x}}+\frac{\partial }{\partial \mathrm{y}}\left\{{\left(\mathrm{x}-{\mathrm{x}}^{\text{'}}\right)}^2+{(\mathrm{y}-{\mathrm{y}}^{\text{'}})}^2+{(\mathrm{z}-{\mathrm{z}}^{\text{'}})}^2\right\}\hat{\mathrm{y}}\\ +\frac{\partial }{\partial \mathrm{z}}\left\{{\left(\mathrm{x}-{\mathrm{x}}^{\text{'}}\right)}^2+{(\mathrm{y}-{\mathrm{y}}^{\text{'}})}^2+{(\mathrm{z}-{\mathrm{z}}^{\text{'}})}^2\right\}\hat{\mathrm{z}}\end{array}\right\} \\ =2{\left(\mathrm{x}-{\mathrm{x}}^{\text{'}}\right)}^2\hat{\mathrm{x}}+2{(\mathrm{y}-{\mathrm{y}}^{\text{'}})}^2\hat{\mathrm{y}}+{2(\mathrm{z}-{\mathrm{z}}^{\text{'}})}^2\hat{\mathrm{z}} \\ =2\stackrel{\rightharpoonup }{\mathrm{r}} \end{gathered}$$

Therefore,it is proved that $\nabla \left(r^2\right)=2\overline{r}$

Step 3: Derive the proof for ∇   (   r 2   ) = 2 r .

Write the expression for the separation for the separation vector.

$\nabla \left(\frac{1}{\mathrm{r}}\right)=\nabla \left[\frac{1}{\sqrt{{( \mathrm{x}-{\mathrm{x}}^{\text{'}})}^2 + {(\mathrm{y}-{\mathrm{y}}^{\text{'}})}^2 + {(\mathrm{z}-{\mathrm{z}}^{\text{'}})}^2}}\right]$

Calculate the divergence of the function.

$$\begin{gathered} \nabla \left(\frac{1}{r}\right)=\left\{\begin{array}{c}\frac{\partial }{\partial x}\left\{\frac{1}{\sqrt{{(x-x^{\text{'}})}^2+{(y-y^{\text{'}})}^2+{(z-z^{\text{'}})}^2}}\right\}\hat{x}\\ +\frac{\partial }{\partial y}\left\{\frac{1}{\sqrt{{(x-x^{\text{'}})}^2+{(y-y^{\text{'}})}^2+{(z-z^{\text{'}})}^2}}\right\}\hat{y}\\ +\frac{\partial }{\partial z}\left\{\frac{1}{\sqrt{{(x-x^{\text{'}})}^2+{(y-y^{\text{'}})}^2+{(z-z^{\text{'}})}^2}}\right\}\hat{z}\end{array}\right\} \\ =\frac{1}{2\left\{{(x-x^{\text{'}})}^2+{(y-y^{\text{'}})}^2+{(z-z^{\text{'}})}^2\right\}}\left\{2(x-x^{\text{'}})\hat{x}+(y-y^{\text{'}})\hat{y}+(z-z^{\text{'}})\hat{z}\right\} \\ =\frac{-1}{{\left(r^2\right)}^{{\displaystyle \frac{3}{2}}}}\left\{\stackrel{\rightharpoonup }{r}\right\} \\ =\frac{-1}{\left(r^3\right)}\left\{r\hat{r}\right\} \end{gathered}$$

           

Solve further as,

$\nabla \left(\frac{1}{\mathrm{r}}\right)=\frac{-1}{\left({\mathrm{r}}^2\right)}\left\{ \hat{\mathrm{r}} \right\}$

Therefore, it is proved that ${}^{\nabla \left(\frac{1}{\mathrm{r}}\right)=\frac{-1}{\left({\mathrm{r}}^2\right)}\left\{ \hat{\mathrm{r}} \right\}}$.

Step 4: Derive the general formula for ∇   (   r n   ) .

(c)

Write the expression for the separation for the separation vector.

$r^n={\left[{\left(x-x^{\text{'}}\right)}^2+ {\left(y-y^{\text{'}}\right)}^2+ {\left(z-z^{\text{'}}\right)}^2 \right]}^{\frac{n}{2}}$

Calculate the divergence of the function.

$$\begin{gathered} \nabla \left(r^n\right)=\left\{\begin{array}{c}\frac{\partial }{\partial x}\left\{\frac{1}{\sqrt{{(x-x^{\text{'}})}^2+{(y-y^{\text{'}})}^2+{(z-z^{\text{'}})}^2}}\right\}\hat{x}\\ +\frac{\partial }{\partial y}\left\{\frac{1}{\sqrt{{(x-x^{\text{'}})}^2+{(y-y^{\text{'}})}^2+{(z-z^{\text{'}})}^2}}\right\}\hat{y}\\ +\frac{\partial }{\partial z}\left\{\frac{1}{\sqrt{{(x-x^{\text{'}})}^2+{(y-y^{\text{'}})}^2+{(z-z^{\text{'}})}^2}}\right\}\hat{z}\end{array}\right\} \\ =\left\{\begin{array}{c}\frac{n}{2}{\left[{(x-x^{\text{'}})}^2 \hat{x}+{(y-y^{\text{'}})}^2 \hat{y}+{(z-z^{\text{'}})}^2 \hat{z}\right]}^{\frac{n}{2}-1}\\ \left\{2{(x-x^{\text{'}})}^2 \hat{x}+{(y-y^{\text{'}})}^2 \hat{y}+{(z-z^{\text{'}})}^2 \hat{z}\right\}\end{array}\right\} \\ =\frac{n}{2}{\left(r^2\right)}^{\frac{n}{2}-1}\left(2\stackrel{\rightharpoonup }{r}\right) \\ =n\left(r^n\right)\left(r^{-2}\right)r\hat{r} \end{gathered}$$

               

Solve further as,

$$\begin{gathered} \nabla \left(r^n\right)=n\left(r^n\right)\left(r^{-2}\right)r \\ =n{r}^{n-1}\hat{r} \end{gathered}$$

Therefore, the general expression is ${}^{\nabla \left({\mathrm{r}}^{\mathrm{n}}\right) = \mathrm{n}{\mathrm{r}}^{\mathrm{n}-1}\hat{\mathrm{r}}}$.