### 2 Engineering Example 1

#### 2.1 Electrostatic potential

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Introduction
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The electrostatic potential due to a point charge $Q$ coulombs at a position $r$ (m) from the charge is given by

$\phantom{\rule{2em}{0ex}}V=\frac{Q}{4\pi {\u03f5}_{0}r}$

where ${\u03f5}_{0}$ , the permittivity of free space, $\approx 8.85\times 1{0}^{-12}\text{F}{\text{m}}^{-1}$ and $\pi \approx 3.14$ .

The field strength at position $r$ is given by $E=-\frac{dV}{dr}$ .

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Problem in words
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Find the electric field strength at a distance of 5 m from a source with a charge of 1 coulomb.

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Mathematical statement of the problem
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$\phantom{\rule{2em}{0ex}}V=\frac{Q}{4\pi {\u03f5}_{0}r}$

Substitute for ${\u03f5}_{0}$ and $\pi $ and use $Q=1$ , so that

$\phantom{\rule{2em}{0ex}}V=\frac{1}{4\times 3.14\times 8.85\times 1{0}^{-12}r}\approx \frac{9\times 1{0}^{9}}{r}=9\times 1{0}^{9}{r}^{-1}$

We need to differentiate $V$ in order to find the electric field strength from the relationship $E=-\frac{dV}{dr}$

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Mathematical analysis
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$\phantom{\rule{2em}{0ex}}E=-\frac{dV}{dr}=-9\times 1{0}^{9}\left(-{r}^{-2}\right)=9\times 1{0}^{9}{r}^{-2}$

When $r=5$ ,

$\phantom{\rule{2em}{0ex}}E=\frac{9\times 1{0}^{9}}{25}=3.6\times 1{0}^{8}$ ( $\text{V}{\text{m}}^{-1}$ .)

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Interpretation
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The electric field strength is $3.6\times 1{0}^{8}\text{V}{\text{m}}^{-1}$ at $r=5$ m.

Note that the field potential varies with the reciprocal of distance (i.e. inverse linear law with distance) whereas the field strength obeys an inverse square law with distance.