## Introduction to Electrodynamics 4e

Published by Pearson Education

# Chapter 1 - Section 2.2 - Differential Calculus - Problem - Page 15: 13

#### Answer

a)Proved below b)Proved below c)$\nabla({r^n})=nr^{n-1}\hat{r}$

#### Work Step by Step

Separation vector, $\vec{r}=(x-x')\hat{x}+(y-y')\hat{y}+(z-z')\hat{z}$ $|\vec{r}|=r=\sqrt{[(x-x')^2+(y-y')^2+(z-z')^2]}$ a)$\nabla({r^2})=\nabla(\vec{r}\cdot\vec{r}) $$=\frac{\partial}{\partial{x}}[(x-x')^2+(y-y')^2+(z-z')^2]\hat{x}+\frac{\partial}{\partial{y}}[(x-x')^2+(y-y')^2+(z-z')^2]\hat{y}+\frac{\partial}{\partial{z}}[(x-x')^2+(y-y')^2+(z-z')^2]\hat{z} \hspace{1.5cm}=2(x-x')\hat{x}+2(y-y')\hat{y}+2(z-z')\hat{z}=2\vec{r} Hence Proved b)\nabla\large\Big({\frac{1}{r}}\Big)$$=\large\frac{\partial}{\partial{x}}\Big(\frac{1}{\sqrt{[(x-x')^2+(y-y')^2+(z-z')^2]}}\Big)\hat{x}+\large\frac{\partial}{\partial{y}}\Big(\frac{1}{\sqrt{[(x-x')^2+(y-y')^2+(z-z')^2]}}\Big)\hat{y} +\large\frac{\partial}{\partial{z}}\Big(\frac{1}{\sqrt{[(x-x')^2+(y-y')^2+(z-z')^2]}}\Big)\hat{z} =\large-\frac{1}{2}[(x-x')^2+(y-y')^2+(z-z')^2]^{-\frac{3}{2}}\normalsize(2(x-x')\hat{x}+2(y-y')\hat{y}+2(z-z')\hat{z})$ $\hspace{1.7cm}\large=-\frac{1}{2r^3}2\vec{r}=-\frac{\hat{r}}{r^2}$ Hence Proved c)$\nabla{r^n}=\frac{\partial}{\partial{x}}r^n\hat{x}+\frac{\partial}{\partial{y}}r^n\hat{y}+\frac{\partial}{\partial{z}}r^n\hat{z}=nr^{n-1}(\frac{\partial}{\partial{x}}r\hat{x}+\frac{\partial}{\partial{y}}r\hat{y}+\frac{\partial}{\partial{z}}r\hat{z})$ $\large\frac{\partial}{\partial{x}}r=\frac{1}{2}\frac{1}{\sqrt{[(x-x')^2+(y-y')^2+(z-z')^2]}}2(x-x')$ Similarly finding $\large\frac{\partial}{\partial{y}}r$ and $\large\frac{\partial}{\partial{z}}r$ , $\Rightarrow(\frac{\partial}{\partial{x}}r\hat{x}+\frac{\partial}{\partial{y}}r\hat{y}+\frac{\partial}{\partial{z}}r\hat{z})=\frac{1}{2r}2\vec{r}=\hat{r}$ $\therefore\nabla({r^n})=nr^{n-1}\hat{r}$

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