2 Engineering Example 1

2.1 Electrostatic potential

Introduction

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

$V=\frac{Q}{4\pi {ϵ}_{0}r}$

where ${ϵ}_0$ , the permittivity of free space, $\approx 8.85\times 10^{-12}\text{F}{\text{m}}^{-1}$ and $\pi \approx 3.14$ .

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

Problem in words

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

Mathematical statement of the problem

$V=\frac{Q}{4\pi {ϵ}_{0}r}$

Substitute for ${ϵ}_0$ and $\pi$ and use $Q=1$ , so that

$V=\frac{1}{4\times 3.14\times 8.85\times 10^{-12}r}\approx \frac{9\times 10^9}{r}=9\times 10^9{r}^{-1}$

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

Mathematical analysis

$E=-\frac{dV}{dr}=-9\times 10^9\left(-r^{-2}\right)=9\times 10^9{r}^{-2}$

When $r=5$ ,

$E=\frac{9\times 10^9}{25}=3.6\times 10^8$ ( $\text{V}{\text{m}}^{-1}$ .)

Interpretation

The electric field strength is $3.6\times 10^8\text{V}{\text{m}}^{-1}$ 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.