3 Engineering Example 6

3.1 Gauss’ law

Introduction

From Gauss’ theorem, it is possible to derive a result which can be used to gain insight into situations involved in Electrical Engineering. Knowing the electric field on a closed surface, it is possible to find the electric charge within this surface. Alternatively, in a sufficiently symmetrical situation, it is possible to find the electric field produced by a given charge distribution.

Gauss’ theorem states

$\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_S\underset{̲}{F}\cdot \underset{̲}{dS}=\int \; <!--nolimits\; -->\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_V\underset{̲}{\nabla }\cdot \underset{̲}{F}dV$

If $\underset{̲}{F}=\underset{̲}{E}$ , the electric field,

$\underset{̲}{\nabla }\cdot \underset{̲}{F}=\underset{̲}{\nabla }\cdot \underset{̲}{E}=\frac{q}{{\epsilon }_0}$

where $q$ is the amount of charge per unit volume and ${\epsilon }_0$ is the permittivity of free space: ${\epsilon }_0=10^{-9}∕36\pi \text{F}{\text{m}}^{-1}\approx 8.84\times 10^{-12}\text{F}{\text{m}}^{-1}$ . Gauss’ theorem becomes

$\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_S\underset{̲}{E}\cdot \underset{̲}{dS}=\int \; <!--nolimits\; -->\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_V\frac{q}{{\epsilon }_0}dV=\frac{1}{{\epsilon }_0}\int \; <!--nolimits\; -->\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_{V}qdV=\frac{Q}{{\epsilon }_0}$

i.e.

$\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_S\underset{̲}{E}\cdot \underset{̲}{dS}=\frac{Q}{{\epsilon }_0}$

which is known as Gauss’ law. Note: this is one of the important Maxwell’s Laws. Using this law, the charge within a given surface can be found from ${\epsilon }_0$ times the surface integral of $\underset{̲}{E}\cdot \underset{̲}{dS}$ over that surface. In certain symmetrical circumstances, it can also be used to find the electric field produced by a given charge distribution.

Problem in words

A point charge lies at the centre of a cube. Given the electric field, find the magnitude of the charge, using Gauss’ law .

Mathematical statement of problem

Consider the cube $-\frac{1}{2}\le x\le \frac{1}{2}$ , $-\frac{1}{2}\le y\le \frac{1}{2}$ , $-\frac{1}{2}\le z\le \frac{1}{2}$ where the dimensions are in metres. A point charge $Q$ lies at the centre of the cube. If the electric field on the top face $\left(z=\frac{1}{2}\right)$ is given by

$\underset{̲}{E}=10\frac{x\underset{̲}{i}+y\underset{̲}{j}+z\underset{̲}{k}}{{\left(x^2+y^2+z^2\right)}^{\frac{3}{2}}}$

find the charge $Q$ from Gauss’ law .

$\left[\text{Hint : }{\int \; <!--nolimits\; -->}_{x=-\frac{1}{2}}^{\frac{1}{2}}{\int \; <!--nolimits\; -->}_{y=-\frac{1}{2}}^{\frac{1}{2}}{\left(x^2+y^2+\frac{1}{4}\right)}^{-\frac{3}{2}}dydx=\frac{4\pi }{3}\right]$

Mathematical analysis

From Gauss’ law

$\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_S\underset{̲}{E}\cdot \underset{̲}{dS}=\frac{Q}{{\epsilon }_0}$

so

$Q={\epsilon }_0\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_S\underset{̲}{E}\cdot \underset{̲}{dS}=-6{\epsilon }_0\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_{S\left(top\right)}\underset{̲}{E}\cdot \underset{̲}{dS}$

since, using the symmetry of the six faces of the cube, it is possible to integrate over just one of them (here the top face is chosen) and multiply by 6. On the top face

$\underset{̲}{E}=10\frac{x\underset{̲}{i}+y\underset{̲}{j}+\frac{1}{2}\underset{̲}{k}}{{\left(x^2+y^2+\frac{1}{4}\right)}^{\frac{3}{2}}}$

and

So

Now

So, from Gauss’ law above,

$Q=6{\epsilon }_0\times \frac{20\pi }{3}=40\pi {\epsilon }_0\approx 10^{-9}\text{C}$

Interpretation

Gauss’ law can be used to find a charge from its effects elsewhere.

The form of $\underset{̲}{E}=10\frac{x\underset{̲}{i}+y\underset{̲}{j}+\frac{1}{2}\underset{̲}{k}}{{\left(x^2+y^2+\frac{1}{4}\right)}^{\frac{3}{2}}}$ comes from the fact that $\underset{̲}{E}$ is radial and equals $10\frac{\underset{̲}{r}}{r^3}=10\frac{\stackrel{̂}{\underset{̲}{r}}}{r^2}$

Example 36

Verify Gauss’ theorem for the vector field $\underset{̲}{F}=y^2\underset{̲}{j}-xz\underset{̲}{k}$ and the triangular prism with vertices at $\left(0,0,0\right)$ , $\left(2,0,0\right)$ , $\left(0,0,1\right)$ , $\left(0,4,0\right)$ , $\left(2,4,0\right)$ and $\left(0,4,1\right)$ (see Figure 20).

Figure 20:

Solution

As $\underset{̲}{F}=y^2\underset{̲}{j}-xz\underset{̲}{k}$ , $\underset{̲}{\nabla }\cdot \underset{̲}{F}=0+2y-x=2y-x$ .

Thus

To work out $\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_S\underset{̲}{F}\cdot \underset{̲}{dS}$ , it is necessary to consider the contributions from the five faces separately.

On the front face, $y=0$ , $\underset{̲}{F}=-xz\underset{̲}{k}$ and $\underset{̲}{dS}=-\underset{̲}{j}$ thus $\underset{̲}{F}\cdot \underset{̲}{dS}=0$ and the contribution to the integral is zero.

On the back face, $y=4$ , $\underset{̲}{F}=16\underset{̲}{j}-xz\underset{̲}{k}$ and $\underset{̲}{dS}=\underset{̲}{j}$ thus $\underset{̲}{F}\cdot \underset{̲}{dS}=16$ and the contribution to the integral is

${\int \; <!--nolimits\; -->}_{x=0}^2{\int \; <!--nolimits\; -->}_{z=0}^{1-x∕2}16dzdx={\int \; <!--nolimits\; -->}_{x=0}^2{\left[16z\right]}_{z=0}^{1-x∕2}dx={\int \; <!--nolimits\; -->}_{x=0}^{2}16\left(1-x∕2\right)dx={\left[16x-4x^2\right]}_0^2=16.$

On the left face, $x=0$ , $\underset{̲}{F}=y^2\underset{̲}{j}$ and $\underset{̲}{dS}=-\underset{̲}{i}$ thus $\underset{̲}{F}\cdot \underset{̲}{dS}=0$ and the contribution to the integral is zero.

On the bottom face, $z=0$ , $\underset{̲}{F}=y^2\underset{̲}{j}$ and $\underset{̲}{dS}=-\underset{̲}{k}$ thus $\underset{̲}{F}\cdot \underset{̲}{dS}=0$ and the contribution to the integral is zero.

On the top right face, $z=1-x∕2$ , $\underset{̲}{F}=y^2\underset{̲}{j}+\left(\frac{1}{2}{x}^2-x\right)\underset{̲}{k}$ and the unit normal $\underset{̲}{\stackrel{̂}{n}}=\frac{1}{\sqrt{5}}\underset{̲}{i}+\frac{2}{\sqrt{5}}\underset{̲}{k}$

Thus $\underset{̲}{dS}=\left[\frac{1}{\sqrt{5}}\underset{̲}{i}+\frac{2}{\sqrt{5}}\underset{̲}{k}\right]dydw$ where $dw$ measures the distance along the slope for a constant $y$ . As $dw=\frac{\sqrt{5}}{2}dx$ , $\underset{̲}{dS}=\left[\frac{1}{2}\underset{̲}{i}+\underset{̲}{k}\right]dydx$ thus $\underset{̲}{F}\cdot \underset{̲}{dS}=16$ and the contribution to the integral is

${\int \; <!--nolimits\; -->}_{x=0}^2{\int \; <!--nolimits\; -->}_{y=0}^4\left(\frac{1}{2}{x}^2-x\right)dydx={\int \; <!--nolimits\; -->}_{x=0}^2\left(2x^2-4x\right)dx={\left[\frac{2}{3}{x}^3-2x^2\right]}_0^2=-\frac{8}{3}$ .

Adding together the contributions, $\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_S\underset{̲}{F}\cdot \underset{̲}{dS}=0+16+0+0-\frac{8}{3}=\frac{40}{3}$ .



Thus $\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_S\underset{̲}{F}\cdot \underset{̲}{dS}=\int \; <!--nolimits\; -->\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_V\underset{̲}{\nabla }\cdot \underset{̲}{F}dV=\frac{40}{3}$ .



Gauss’ theorem also applies using orthogonal curvilinear coordinates.