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

$W_{AB}=-q{\int \; <!--nolimits\; -->}_A^B\underset{̲}{E}\cdot \underset{̲}{dl}$

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

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 ${ϵ}_0=10^{-9}∕36\pi {\mathrm{\text{F m}}}^{-1}$ .

Given that $Q$ = $10^{-8}$ C, find the work done in bringing a charge of $q$ = $10^{-10}$ 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

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

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

   $$\begin{array}{rcll}{W}_{AB}& =& -q{\int \; <!--nolimits\; -->}_A^B\underset{̲}{E}\cdot \underset{̲}{dl}& \\ & =& -10^{-10}{\int \; <!--nolimits\; -->}_A^B\frac{90}{{\left(x^2+y^2\right)}^{\frac{3}{2}}}\left[x\underset{̲}{i}+y\underset{̲}{j}\right]\cdot \left[dx\underset{̲}{i}+dy\underset{̲}{j}\right]& \end{array}$$

   Using $y=x,dy=dx$

   $$\begin{array}{rcll}{W}_{AB}& =& -10^{-10}{\int \; <!--nolimits\; -->}_{x=10}^1\frac{90}{{\left(2x^2\right)}^{\frac{3}{2}}}\left\{xdx+xdx\right\}& \\ & =& -10^{-10}{\int \; <!--nolimits\; -->}_{10}^1\frac{90}{\left(2\sqrt{2}\right)}{x}^{-3}2xdx& \\ & =& \frac{90\times -10^{-10}}{\sqrt{2}}{\int \; <!--nolimits\; -->}_{10}^1{x}^{-2}dx& \\ & =& \frac{9\times -10^{-9}}{\sqrt{2}}{\left[-x^{-1}\right]}_{10}^1& \\ & =& \frac{9\times 10^{-9}}{\sqrt{2}}{\left[x^{-1}\right]}_{10}^1& \\ & =& \frac{9\times 10^{-9}}{\sqrt{2}}\left[1-0.1\right]& \\ & =& 5.73\times 10^{-9}\text{ J}& \end{array}$$
2. The first part of the path is $A$ to $C$ where $x=10$ , $dx=0$ and $y$ goes from 10 to 1. $$\begin{array}{rcll}{W}_{AC}& =& -Q{\int \; <!--nolimits\; -->}_A^C\underset{̲}{E}\cdot \underset{̲}{dl}& \\ & =& -10^{-10}{\int \; <!--nolimits\; -->}_{y=10}^1\frac{90}{{\left(100+y^2\right)}^{\frac{3}{2}}}\left[x\underset{̲}{i}+y\underset{̲}{j}\right]\cdot \left[0\underset{̲}{i}+dy\underset{̲}{j}\right]& \\ & =& -10^{-10}{\int \; <!--nolimits\; -->}_{10}^1\frac{90ydy}{{\left(100+y^2\right)}^{\frac{3}{2}}}& \\ & =& -10^{-10}{\int \; <!--nolimits\; -->}_{u=200}^{101}\frac{45du}{u^{\frac{3}{2}}}\text{ substituting }u=100+y^2,du=2ydy& \\ & =& -45\times 10^{-10}{\int \; <!--nolimits\; -->}_{200}^{101}{u}^{-\frac{3}{2}}du& \\ & =& -45\times 10^{-10}{\left[-2u^{-\frac{1}{2}}\right]}_{200}^{101}& \\ & =& 45\times 10^{-10}\left(\frac{2}{\sqrt{101}}-\frac{2}{\sqrt{200}}\right)=2.59\times 10^{-10}\text{J}& \end{array}$$

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

   $$\begin{array}{rcll}{W}_{CB}& =& -10^{-10}{\int \; <!--nolimits\; -->}_{x=10}^1\frac{90}{{\left(x^2+1\right)}^{\frac{3}{2}}}\left[x\underset{̲}{i}+y\underset{̲}{j}\right]\cdot \left[dx\underset{̲}{i}+0\underset{̲}{j}\right]& \\ & =& -10^{-10}{\int \; <!--nolimits\; -->}_{10}^1\frac{90xdx}{{\left(x^2+1\right)}^{\frac{3}{2}}}& \\ & =& -10^{-10}{\int \; <!--nolimits\; -->}_{u=101}^2\frac{45du}{u^{\frac{3}{2}}}\text{ substituting }u=x^2+1,du=2xdx& \\ & =& -45\times 10^{-10}{\int \; <!--nolimits\; -->}_{101}^2{u}^{-\frac{3}{2}}du& \\ & =& -45\times 10^{-10}{\left[-2u^{-\frac{1}{2}}\right]}_{101}^2& \\ & =& 45\times 10^{-10}\left(\frac{2}{\sqrt{2}}-\frac{2}{\sqrt{101}}\right)=5.468\times 10^{-9}\text{J}& \end{array}$$

   The sum of the two components $W_{AC}$ and $W_{CB}$ is $5.73\times 10^{-9}$ 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 $I={\int \; <!--nolimits\; -->}_{\left(0,0\right)}^{\left(2,1\right)}\left\{\left(2xy+1\right)dx+\left(x^2-2y\right)dy\right\}$ 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 $I={\int \; <!--nolimits\; -->}_{\left(0,0\right)}^{\left(2,1\right)}\left\{\left(2xy+1\right)dx+\left(x^2-2y\right)dy\right\}$ may be re-written ${\int \; <!--nolimits\; -->}_C\underset{̲}{F}\cdot \underset{̲}{dr}$ where
   
   $\underset{̲}{F}=\left(2xy+1\right)\underset{̲}{i}+\left(x^2-2y\right)\underset{̲}{j}$ .
   
   Now $\underset{̲}{\nabla }\times \underset{̲}{F}=\left|\begin{array}{ccc}\hfill \underset{̲}{i}\hfill & \hfill \underset{̲}{j}\hfill & \hfill \underset{̲}{k}\hfill \\ \hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill \frac{\partial }{\partial x}\hfill & \hfill \frac{\partial }{\partial y}\hfill & \hfill \frac{\partial }{\partial z}\hfill \\ \hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill 2xy+1\hfill & \hfill x^2-2y\hfill & \hfill 0\hfill \end{array}\right|=0\underset{̲}{i}+0\underset{̲}{j}+0\underset{̲}{k}=\underset{̲}{0}$

   As $\underset{̲}{\nabla }\times \underset{̲}{F}=\underset{̲}{0}$ , $\underset{̲}{F}$ 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 $$\begin{array}{rcll}I& =& {\int \; <!--nolimits\; -->}_{\left(0,0\right)}^{\left(2,1\right)}\left\{\left(2xy+1\right)dx+\left(x^2-2y\right)dy\right\}& \\ & =& {\int \; <!--nolimits\; -->}_{x=0}^2\left\{\left(2x\times \frac{1}{2}x+1\right)dx+\left(x^2-4x\right)\frac{1}{2}dx\right\}& \\ & =& {\int \; <!--nolimits\; -->}_0^2\left(\frac{3}{2}{x}^2-\frac{1}{2}x+1\right)dx& \\ & =& {\left[\frac{1}{2}{x}^3-\frac{1}{4}{x}^2+x\right]}_0^2=4-1+2-0=5& \end{array}$$
3. If $\underset{̲}{F}=\underset{̲}{\nabla }\varphi$ then

   $\left.\begin{array}{ccc}\hfill \frac{\partial \varphi }{\partial x}=2xy+1\hfill & \hfill \to \hfill & \hfill \varphi =x^{2}y+x+f\left(y\right)\hfill \\ \hfill \hfill \\ \hfill \frac{\partial \varphi }{\partial y}=x^2-2y\hfill & \hfill \to \hfill & \hfill \varphi =x^{2}y-y^2+g\left(x\right)\hfill \end{array}\right\}\to \varphi =x^{2}y+x-y^2+C$ .
   
   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. $\underset{̲}{F}=3x^2{y}^2\underset{̲}{i}+\left(2x^{3}y-1\right)\underset{̲}{j}$
   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 ${\int \; <!--nolimits\; -->}_C\underset{̲}{F}\cdot \underset{̲}{dr}$ with $\underset{̲}{F}=3x^2{y}^2\underset{̲}{i}+\left(2x^{3}y-1\right)\underset{̲}{j}$ . 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 ${\int \; <!--nolimits\; -->}_C\underset{̲}{F}\cdot \underset{̲}{dr}$ 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 $I={\int \; <!--nolimits\; -->}_C\underset{̲}{F}\cdot \underset{̲}{dr}$ for the contours $C$ indicated.
   1. $\underset{̲}{F}=\left(4x^{3}y-2x\right)\underset{̲}{i}+\left(x^4-2y\right)\underset{̲}{j}$ ; 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. $\underset{̲}{F}=\frac{1}{x}\underset{̲}{i}+4y^3{z}^2\underset{̲}{j}+2y^{4}z\underset{̲}{k}$ ; any path from $\left(1,1,1\right)$ to $\left(1,2,3\right)$ .

Answer