### 4 Engineering Example 1

#### 4.1 Work done moving a charge in an electric field

Introduction

If a charge, $q$ , is moved through an electric field, $\underset{̲}{E}$ , from $A$ to $B$ , then the work required is given by the line integral

$\phantom{\rule{2em}{0ex}}{W}_{AB}=-q{\int }_{A}^{B}\underset{̲}{E}\cdot \underset{̲}{dl}$

Problem in words

Compare the work done in moving a charge through the electric field around a point charge in a vacuum via two different paths.

Mathematical statement of problem

An electric field $\underset{̲}{E}$ is given by

$\begin{array}{rcll}\underset{̲}{E}& =& \frac{Q}{4\pi {\epsilon }_{0}{r}^{2}}\underset{̲}{\stackrel{̂}{r}}& \text{}\\ & =& \frac{Q}{4\pi {\epsilon }_{0}\left({x}^{2}+{y}^{2}+{z}^{2}\right)}×\frac{x\underset{̲}{i}+y\underset{̲}{j}+z\underset{̲}{k}}{\sqrt{{x}^{2}+{y}^{2}+{z}^{2}}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& \text{}\\ & =& \frac{Q\left(x\underset{̲}{i}+y\underset{̲}{j}+z\underset{̲}{k}\right)}{4\pi {\epsilon }_{0}{\left({x}^{2}+{y}^{2}+{z}^{2}\right)}^{\frac{3}{2}}}& \text{}\end{array}$

where $\underset{̲}{r}$ is the position vector with magnitude $r$ and unit vector $\stackrel{̂}{\underset{̲}{r}}$ , and $\frac{1}{4\pi {ϵ}_{0}}$ is a combination of constants of proportionality, where .

Given that $Q$ = $1{0}^{-8}$ C, find the work done in bringing a charge of $q$ = $1{0}^{-10}$ C from the point $A=\left(10,10,0\right)$ to the point $B=\left(1,1,0\right)$ (where the dimensions are in metres)

1. by the direct straight line $y=x$ , $z=0$
2. by the straight line pair via $C=\left(10,1,0\right)$

Figure 4: The path comprises two straight lines from $A=\left(10,10,0\right)$ to $B=\left(1,1,0\right)$ via $C=\left(10,1,0\right)$ (see Figure 4).

Mathematical analysis

1. Here $q∕\left(4\pi {\epsilon }_{0}\right)$ = 90 so

$\phantom{\rule{2em}{0ex}}\underset{̲}{E}=\frac{90\left[x\underset{̲}{i}+y\underset{̲}{j}\right]}{{\left({x}^{2}+{y}^{2}\right)}^{\frac{3}{2}}}$

as $z=0$ over the region of interest. The work done

$\begin{array}{rcll}{W}_{AB}& =& -q{\int }_{A}^{B}\underset{̲}{E}\cdot \underset{̲}{dl}& \text{}\\ & =& -1{0}^{-10}{\int }_{A}^{B}\frac{90}{{\left({x}^{2}+{y}^{2}\right)}^{\frac{3}{2}}}\phantom{\rule{0.3em}{0ex}}\left[x\underset{̲}{i}+y\underset{̲}{j}\right]\cdot \left[dx\underset{̲}{i}+dy\underset{̲}{j}\right]\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& \text{}\end{array}$

Using $\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}y=x,\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}dy=dx$

2. The first part of the path is $A$ to $C$ where $x=10$ , $dx=0$ and $y$ goes from 10 to 1.

The second part is $C$ to $B$ , where $y=1$ , $dy=0$ and $x$ goes from 10 to 1.

The sum of the two components ${W}_{AC}$ and ${W}_{CB}$ is $5.73×1{0}^{-9}$ J. Therefore the work done over the two routes is identical.

Interpretation

In fact, the work done is independent of the route taken as the electric field $\underset{̲}{E}$ around a point charge in a vacuum is a conservative field.

##### Example 11
1. Show that $I={\int }_{\left(0,0\right)}^{\left(2,1\right)}\left\{\left(2xy+1\right)dx+\left({x}^{2}-2y\right)dy\right\}$ is independent of the path taken.
2. Find $I$ using property P1.
3. Find $I$ using property P4.
4. Find $I={\oint }_{C}\left\{\left(2xy+1\right)dx+\left({x}^{2}-2y\right)dy\right\}$ where $C$ is
1. the circle ${x}^{2}+{y}^{2}=1$
2. the square with vertices $\left(0,0\right)$ , $\left(1,0\right)$ , $\left(1,1\right)$ , $\left(0,1\right)$ .
##### Solution
1. The integral $I={\int }_{\left(0,0\right)}^{\left(2,1\right)}\left\{\left(2xy+1\right)dx+\left({x}^{2}-2y\right)dy\right\}$ may be re-written ${\int }_{C}\underset{̲}{F}\cdot \underset{̲}{dr}$ where

$\underset{̲}{F}=\left(2xy+1\right)\underset{̲}{i}+\left({x}^{2}-2y\right)\underset{̲}{j}$ .

Now $\underset{̲}{\nabla }×\underset{̲}{F}=\left|\begin{array}{ccc}\hfill \underset{̲}{i}\hfill & \hfill \underset{̲}{j}\hfill & \hfill \underset{̲}{k}\hfill \\ \hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill \frac{\partial }{\partial x}\hfill & \hfill \frac{\partial }{\partial y}\hfill & \hfill \frac{\partial }{\partial z}\hfill \\ \hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill 2xy+1\hfill & \hfill {x}^{2}-2y\hfill & \hfill 0\hfill \end{array}\right|=0\underset{̲}{i}+0\underset{̲}{j}+0\underset{̲}{k}=\underset{̲}{0}$

As $\underset{̲}{\nabla }×\underset{̲}{F}=\underset{̲}{0}$ , $\underset{̲}{F}$ is a conservative field and $I$ is independent of the path taken between $\left(0,0\right)$ and $\left(2,1\right)$ .

2. As $I$ is independent of the path taken from $\left(0,0\right)$ to $\left(2,1\right)$ , it can be evaluated along any such path. One possibility is the straight line $y=\frac{1}{2}x$ . On this line, $dy=\frac{1}{2}dx$ . The integral $I$ becomes $\begin{array}{rcll}I& =& {\int }_{\left(0,0\right)}^{\left(2,1\right)}\left\{\left(2xy+1\right)dx+\left({x}^{2}-2y\right)dy\right\}& \text{}\\ & =& {\int }_{x=0}^{2}\left\{\left(2x×\frac{1}{2}x+1\right)dx+\left({x}^{2}-4x\right)\frac{1}{2}dx\right\}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& \text{}\\ & =& {\int }_{0}^{2}\left(\frac{3}{2}{x}^{2}-\frac{1}{2}x+1\right)dx& \text{}\\ & =& {\left[\frac{1}{2}{x}^{3}-\frac{1}{4}{x}^{2}+x\right]}_{0}^{2}=4-1+2-0=5& \text{}\end{array}$
3. If $\underset{̲}{F}=\underset{̲}{\nabla }\varphi$ then

$\begin{array}{ccc}\hfill \frac{\partial \varphi }{\partial x}=2xy+1\hfill & \hfill \to \hfill & \hfill \varphi ={x}^{2}y+x+f\left(y\right)\hfill \\ \hfill \hfill \\ \hfill \frac{\partial \varphi }{\partial y}={x}^{2}-2y\hfill & \hfill \to \hfill & \hfill \varphi ={x}^{2}y-{y}^{2}+g\left(x\right)\hfill \end{array}}\to \varphi ={x}^{2}y+x-{y}^{2}+C$ .

These are consistent if $\varphi ={x}^{2}y+x-{y}^{2}$ (plus a constant which may be omitted since it cancels).

So $I=\varphi \left(2,1\right)-\varphi \left(0,0\right)=\left(4+2-1\right)-0=5$

4. As $F$ is a conservative field, all integrals around a closed contour are zero.
##### Exercises
1. Determine whether the following vector fields are conservative
1. $\underset{̲}{F}=\left(x-y\right)\underset{̲}{i}+\left(x+y\right)\underset{̲}{j}$
2. $\underset{̲}{F}=3{x}^{2}{y}^{2}\underset{̲}{i}+\left(2{x}^{3}y-1\right)\underset{̲}{j}$
3. $\underset{̲}{F}=2x\underset{̲}{i}+\left(xz-2\right)\underset{̲}{j}+xy\underset{̲}{k}$
4. $\underset{̲}{F}={x}^{2}z\underset{̲}{i}+{y}^{2}z\underset{̲}{j}+\frac{1}{3}\left({x}^{3}+{y}^{3}\right)\underset{̲}{k}$
2. Consider the integral ${\int }_{C}\underset{̲}{F}\cdot \underset{̲}{dr}$ with $\underset{̲}{F}=3{x}^{2}{y}^{2}\underset{̲}{i}+\left(2{x}^{3}y-1\right)\underset{̲}{j}$ . From Exercise 1(b) $\underset{̲}{F}$ is a conservative vector field. Find a scalar field $\varphi$ so that $\underset{̲}{\nabla }\varphi =\underset{̲}{F}$ . Hence use P4 to evaluate the integral ${\int }_{C}\underset{̲}{F}\cdot \underset{̲}{dr}$ where $C$ is an integral with start-point $\left(0,0\right)$ and end point $\left(1,4\right)$ .
3. For the following conservative vector fields $\underset{̲}{F}$ , find a scalar field $\varphi$ such that $\underset{̲}{\nabla }\varphi =\underset{̲}{F}$ and hence evaluate the $I={\int }_{C}\underset{̲}{F}\cdot \underset{̲}{dr}$ for the contours $C$ indicated.
1. $\underset{̲}{F}=\left(4{x}^{3}y-2x\right)\underset{̲}{i}+\left({x}^{4}-2y\right)\underset{̲}{j}$ ; any path from $\left(0,0\right)$ to $\left(2,1\right)$ .
2. $\underset{̲}{F}=\left({e}^{x}+{y}^{3}\right)\underset{̲}{i}+\left(3x{y}^{2}\right)\underset{̲}{j}$ ; closed path starting from any point on the circle ${x}^{2}+{y}^{2}=1$ .
3. $\underset{̲}{F}=\left({y}^{2}+sinz\right)\underset{̲}{i}+2xy\underset{̲}{j}+xcosz\underset{̲}{k}$ ; any path from $\left(1,1,0\right)$ to $\left(2,0,\pi \right)$ .
4. $\underset{̲}{F}=\frac{1}{x}\underset{̲}{i}+4{y}^{3}{z}^{2}\underset{̲}{j}+2{y}^{4}z\underset{̲}{k}$ ; any path from $\left(1,1,1\right)$ to $\left(1,2,3\right)$ .
1. No,
2. Yes,
3. No,
4. Yes
1. ${x}^{3}{y}^{2}-y+C$ , $12$
1. ${x}^{4}y-{x}^{2}-{y}^{2}$ $11$ ;
2. ${e}^{x}+x{y}^{3}$ $0$ ;
3. $x{y}^{2}+xsinz$ $-1$ ;
4. $lnx+{y}^{4}{z}^{2}$ , $143$