What is the average value of electric field?

[Petar Mali]

$$\oint_{\Delta S}\vec{E}\cdot \vec{dS}=const \int_{\Delta_V}\rho dV$$

$$\Delta S$$ surface which surround domain $$\Delta V$$.

$$\vec{E}(\vec{r},t)$$ - vector field

$$\rho(\vec{r},t)$$ - scalar field

Now in the book which I read they say we use average value theorem

and get

$$\oint_{\Delta S}\vec{E}\cdot \vec{dS}=const \overline{\rho}\Delta V$$

Can you tell me something more about

$$\int_{\Delta_V}\rho dV=\overline{\rho}\Delta V$$

 

[HallsofIvy]

Well, there really isn't much to be said. The integral of a constant over a region is just that constant times the measure (length, area, or volume) of the region. Essentially you defining the average of a function to be that constant which, when integrated over the region, gives the same value as the integral integrated over the region.

It is just an extension of the average of a finite set of numbers: if the set of numbers is $\{x_1, x_2, \cdot\cdot\cdot, x_n\}$ then adding them gives $x_1+ x_2+ \cdot\cdot\cdot+ x_n$ and their average, $\overline{x}$ is the number such that [itex]\overline{x}+ \overline{x}+ \cdot\cdot\cdot+ \overline{x}= n\overline{x}= x_1+ x_2+ \cdot\cdot\cdot+ x_n[[/tex]. That is,<br /> $\overline{x}= \frac{x_1+ x_2+ \cdot\cdot\cdot+ x_n}{n}$.[/itex]

 

[Petar Mali]

Thanks! Idea is clear to me. But I have a trouble to determine when can I do that!

For example when can I say

$$\int^a_0f(x)dx=\overline{f}a$$
?

From

$$\oint_{\Delta S}\vec{E}\cdot \vec{dS}=const \overline{\rho}\Delta V$$

$$\frac{1}{\Delta V} \oint_{\Delta S}\vec{E}\cdot \vec{dS}=const \overline{\rho}$$

$$lim_{\Delta V \rightarrow 0}\frac{1}{\Delta V} \oint_{\Delta S}\vec{E}\cdot \vec{dS}<br /> =lim_{\Delta V \rightarrow 0}const \overline{\rho}$$

$$(div\vec{E})_M=(const \overline{\rho})_M$$

where $$M$$ is some point in region which volume (measure) is $$\Delta V$$.

And if I use Gauss theorem I will get

$$div\vec{E}=const\rho$$