### 6 Engineering Example 1

#### 6.1 Field due to point charges

In free space, point charge ${q}_{1}=10$   $n$ C (1 $n$ C $=1{0}^{-9}$ C, i.e. a nanocoulomb) is at ${P}_{1}\left(0,-4,0\right)$ and charge ${q}_{2}=20$   $n$ C is at ${P}_{2}=\left(0,0,4\right)$ .

[Note: Since the $x$ -coordinate of both charges is zero, the problem is two-dimensional in the $yz$ plane as shown in Figure 35.]

Figure 35

1. Find the field at the origin ${\underset{̲}{E}}_{1,2}$ due to ${q}_{1}$ and ${q}_{2}$ .
2. Where should a third charge ${q}_{3}=30$   $n$ C be placed in the $yz$ plane so that the total field due to ${q}_{1},{q}_{2},{q}_{3}$ is zero at the origin?
##### Solution
1. Total field at the origin ${\underset{̲}{E}}_{1,2}$ = (field at origin due to charge at ${P}_{1}$ ) + (field at origin due to charge at ${P}_{2}$ ). Therefore

$\phantom{\rule{1em}{0ex}}{\underset{̲}{E}}_{1,2}=\frac{10×1{0}^{-9}}{4\pi ×8.854×1{0}^{-12}×{4}^{2}}\underset{̲}{j}+\frac{20×1{0}^{-9}}{4\pi ×8.854×1{0}^{-12}×{4}^{2}}\left(-\underset{̲}{k}\right)=5.617\underset{̲}{j}-11.23\underset{̲}{k}$

(The negative sign in front of the second term results from the fact that the direction from ${P}_{2}$ to $O$ is in the $-z$ direction.)

2. Suppose the third charge ${q}_{3}=30$   $n$ C is placed at ${P}_{3}\left(0,a,b\right)$ . The field at the origin due to the third charge is

$\phantom{\rule{2em}{0ex}}{\underset{̲}{E}}_{3}=\frac{30×1{0}^{-9}}{4\pi ×8.854×1{0}^{-12}×\left({a}^{2}+{b}^{2}\right)}×\frac{-\left(a\underset{̲}{j}+b\underset{̲}{k}\right)}{{\left({a}^{2}+{b}^{2}\right)}^{1∕2}}$ ,

where $\frac{a\underset{̲}{j}+b\underset{̲}{k}}{{\left({a}^{2}+{b}^{2}\right)}^{1∕2}}$ is the unit vector in the direction from $O$ to ${P}_{3}$

If the position of the third charge is such that the total field at the origin is zero, then ${\underset{̲}{E}}_{3}=-{\underset{̲}{E}}_{1,2}$ . There are two unknowns ( $a$ and $b$ ). We can write down two equations by considering the $\underset{̲}{j}$ and $\underset{̲}{k}$ directions.

$\phantom{\rule{2em}{0ex}}{\underset{̲}{E}}_{3}=-269.6\left[\frac{a}{{\left({a}^{2}+{b}^{2}\right)}^{3∕2}}\underset{̲}{j}+\frac{b}{{\left({a}^{2}+{b}^{2}\right)}^{3∕2}}\underset{̲}{k}\right]\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}{\underset{̲}{E}}_{1,2}=5.617\underset{̲}{j}-11.23\underset{̲}{k}$

So

$\phantom{\rule{2em}{0ex}}5.617=269.6×\frac{a}{{\left({a}^{2}+{b}^{2}\right)}^{3∕2}}$ (1)

$\phantom{\rule{2em}{0ex}}-11.23=269.6×\frac{b}{{\left({a}^{2}+{b}^{2}\right)}^{3∕2}}$ (2)

So

$\phantom{\rule{2em}{0ex}}\frac{a}{{\left({a}^{2}+{b}^{2}\right)}^{3∕2}}=0.02083$ (3)

$\phantom{\rule{2em}{0ex}}\frac{b}{{\left({a}^{2}+{b}^{2}\right)}^{3∕2}}=-0.04165$ (4)

Squaring and adding (3) and (4) gives $\phantom{\rule{1em}{0ex}}\frac{{a}^{2}+{b}^{2}}{{\left({a}^{2}+{b}^{2}\right)}^{3}}=0.002169$

So

$\phantom{\rule{2em}{0ex}}\left({a}^{2}+{b}^{2}\right)=21.47$ (5)

Substituting back from (5) into (1) and (2) gives $a=2.07$ and $b=-4.14,$ to 3 s.f.

Eight point charges of 1 nC each are located at the corners of a cube in free space which is 1 m on each side (see Figure 36). Calculate $|\underset{̲}{E}|$ at

1. the centre of the cube
2. the centre of any face
3. the centre of any edge.

Figure 36

1. The field at the centre of the cube is zero because of the symmetrical distribution of the charges.
2. Because of the symmetrical nature of the problem it does not matter which face is chosen in order to find the magnitude of the field at the centre of a face. Suppose the chosen face has corners located at $P\left(1,1,1\right),\phantom{\rule{1em}{0ex}}T\left(1,1,0\right),\phantom{\rule{1em}{0ex}}R\left(0,1,0\right)$ and $S\left(0,1,1\right)$ then the centre ( $C$ ) of this face can be seen from the diagram to be located at $C\left(\frac{1}{2},1,\frac{1}{2}\right)$ .

The electric field at $C$ due to the charges at the corners $P,T,R$ and $S$ will then be zero since the field vectors due to equal charges located at opposite corners of the square $PTRS$ cancel one another out. The field at $C$ is then due to the equal charges located at the remaining four corners $\left(OABD\right)$ of the cube, and we note from the symmetry of the cube, that the distance of each of these corners from $C$ will be the same. In particular the distance $OC=\sqrt{{\left(\frac{1}{2}\right)}^{2}+{1}^{2}+{\left(\frac{1}{2}\right)}^{2}}=\sqrt{1.5}\phantom{\rule{1em}{0ex}}\text{m}.$ The electric field $\underset{̲}{E}$ at $C$ due to the remaining charges can then be found using $\underset{̲}{E}=\frac{1}{4\pi {\epsilon }_{0}}{\sum }_{1}^{4}\frac{{q}_{i}\cdot {\underset{̲}{r}}_{i}}{{\left|{\underset{̲}{r}}_{i}\right|}^{3}}$ where ${q}_{1}$ to ${q}_{4}$ are the equal charges ( $1{0}^{-9}$ coulombs) and ${\underset{̲}{r}}_{1}$ to ${\underset{̲}{r}}_{4}$ are the vectors directed from the four corners, where the charges are located, towards $C$ . In this case since ${q}_{1}=1{0}^{-9}$ coulombs and $\left|{\underset{̲}{r}}_{i}\right|=\sqrt{1.5}$ for $i=1$ to $i=4$ we have

$\underset{̲}{E}=\frac{1}{4\pi {\epsilon }_{0}}\frac{1{0}^{-9}}{{\left(1.5\right)}^{3∕2}}\left[{\underset{̲}{r}}_{1}+{\underset{̲}{r}}_{2}+{\underset{̲}{r}}_{3}+{\underset{̲}{r}}_{4}\right],$

where ${\underset{̲}{r}}_{1}=\underset{̲}{AC}=\left[\begin{array}{c}\hfill \frac{1}{2}\hfill \\ \hfill 1\hfill \\ \hfill \frac{1}{2}\hfill \end{array}\right]-\left[\begin{array}{c}\hfill 0\hfill \\ \hfill 0\hfill \\ \hfill 1\hfill \end{array}\right],\phantom{\rule{1em}{0ex}}{\underset{̲}{r}}_{2}=\underset{̲}{BC}\phantom{\rule{1em}{0ex}}=\left[\begin{array}{c}\hfill \frac{1}{2}\hfill \\ \hfill 1\hfill \\ \hfill \frac{1}{2}\hfill \end{array}\right]-\left[\begin{array}{c}\hfill 1\hfill \\ \hfill 0\hfill \\ \hfill 1\hfill \end{array}\right]\phantom{\rule{1em}{0ex}}\text{etc.}$

Thus $\phantom{\rule{1em}{0ex}}\underset{̲}{E}=\frac{1}{4\pi {\epsilon }_{0}}\phantom{\rule{1em}{0ex}}\frac{1{0}^{-9}}{{\left(1.5\right)}^{3∕2}}\phantom{\rule{1em}{0ex}}\left[\begin{array}{c}\hfill 0\hfill \\ \hfill 4\hfill \\ \hfill 0\hfill \end{array}\right]$

and $\left|\underset{̲}{E}\right|=\frac{1}{\pi {\epsilon }_{0}}\phantom{\rule{1em}{0ex}}\frac{1{0}^{-9}}{{\left(1.5\right)}^{3∕2}}=\frac{1{0}^{-9}}{\pi ×8.854×1{0}^{-12}{\left(1.5\right)}^{3∕2}}$ $=19.57\text{V}{\text{m}}^{-1}$

3. Suppose the chosen edge to be used connects $A\left(0,0,1\right)$ to $B\left(1,0,1\right)$ then the centre point $\left(G\right)$ will be located at $G\left(\frac{1}{2},0,1\right)$ .

By symmetry the field at $G$ due to the charges at $A$ and $B$ will be zero.

We note that the distances $DG,\phantom{\rule{1em}{0ex}}OG,\phantom{\rule{1em}{0ex}}PG$ and $SG$ are all equal. In the case of $OG$ we calculate by Pythagoras that this distance is $\sqrt{{\left(\frac{1}{2}\right)}^{2}+{0}^{2}+{1}^{2}}=\sqrt{1.25}$ .

Similarly the distances $TG$ and $RG$ are equal to $\sqrt{2.25}$ .

Using the result that $\underset{̲}{E}=\frac{1}{4\pi {\epsilon }_{0}}\phantom{\rule{1em}{0ex}}\sum \frac{{q}_{i}{\underset{̲}{r}}_{i}}{{\left|{\underset{̲}{r}}_{i}\right|}^{3}}$ gives

$\phantom{\rule{2em}{0ex}}\underset{̲}{E}=\frac{1{0}^{-9}}{4\pi {\epsilon }_{0}}\left[\frac{1}{{\left(1.25\right)}^{3∕2}}\left\{\left[\begin{array}{c}\hfill -\frac{1}{2}\hfill \\ \hfill 0\hfill \\ \hfill 1\hfill \end{array}\right]+\left[\begin{array}{c}\hfill \frac{1}{2}\hfill \\ \hfill 0\hfill \\ \hfill 1\hfill \end{array}\right]+\left[\begin{array}{c}\hfill -\frac{1}{2}\hfill \\ \hfill -1\hfill \\ \hfill 0\hfill \end{array}\right]+\left[\begin{array}{c}\hfill \frac{1}{2}\hfill \\ \hfill -1\hfill \\ \hfill 0\hfill \end{array}\right]\right\}\right$

$\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}+\frac{1}{{\left(2.25\right)}^{3∕2}}\left\{\left[\begin{array}{c}\hfill -\frac{1}{2}\hfill \\ \hfill -1\hfill \\ \hfill 1\hfill \end{array}\right]+\left[\begin{array}{c}\hfill \frac{1}{2}\hfill \\ \hfill -1\hfill \\ \hfill 1\hfill \end{array}\right]\right\}]$

$\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}=\frac{1{0}^{-9}}{4\pi {\epsilon }_{0}}\left[\frac{1}{{\left(1.25\right)}^{3∕2}}\left[\begin{array}{c}\hfill 0\hfill \\ \hfill -2\hfill \\ \hfill 2\hfill \end{array}\right]+\frac{1}{{\left(2.25\right)}^{3∕2}}\left[\begin{array}{c}\hfill 0\hfill \\ \hfill -2\hfill \\ \hfill 2\hfill \end{array}\right]\right]$

$\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}=\frac{1{0}^{-9}}{4\pi {\epsilon }_{0}}\left[\begin{array}{c}\hfill 0\hfill \\ \hfill -2.02367\hfill \\ \hfill 2.02367\hfill \end{array}\right]$

Thus $\left|\underset{̲}{E}\right|=\frac{1{0}^{-9}}{4×\pi ×8.854×1{0}^{-12}}\phantom{\rule{1em}{0ex}}\sqrt{{0}^{2}+{\left(-2.02367\right)}^{2}+{\left(2.02367\right)}^{2}}$

$\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}=25.72\text{V}{\text{m}}^{-1}$ (2 d.p.).

If $\underset{̲}{E}=-50\underset{̲}{i}-50\underset{̲}{j}+30\underset{̲}{k}\text{V}{\text{m}}^{-1}$ where $\underset{̲}{i},\phantom{\rule{1em}{0ex}}\underset{̲}{j}$ and $\underset{̲}{k}$ are unit vectors in the $x,y$ and $z$ directions respectively, find the differential amount of work done in moving a $2\mu$ C point charge a distance of 5 mm.

1. From $P\left(1,2,3\right)$ towards $Q\left(2,4,1\right)$
2. From $Q\left(2,4,1\right)$ towards $P\left(1,2,3\right)$
1. The work done in moving a $2\mu$ C charge through a distance of 5 mm towards $Q$ is $\begin{array}{rcll}W=-q\underset{̲}{E}.\underset{̲}{ds}& =& -\left(2×1{0}^{-6}\right)\left(5×1{0}^{-3}\right)\underset{̲}{E}.\frac{PQ}{|\underset{̲}{PQ}|}& \text{}\\ & & & \text{}\\ & =& -1{0}^{-8}\left(-50\underset{̲}{i}-50\underset{̲}{j}+30\underset{̲}{k}\right)\cdot \frac{\left(\underset{̲}{i}+2\underset{̲}{j}-2\underset{̲}{k}\right)}{\sqrt{{1}^{2}+{2}^{2}+{\left(-2\right)}^{2}}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& \text{}\\ & =& \frac{1{0}^{-8}\left(50+100+60\right)}{3}=7×1{0}^{-7}J& \text{}\end{array}$
2. A similar calculation yields that the work done in moving the same charge through the same distance in the direction from $Q$ to $P$ is $W=-7×1{0}^{-7}J$