2
$\begingroup$

Consider this this picture. Integrating over this infinitesimal box gives the following equivalencies:

$$\int_{\Delta V} d^3r~{\rm div} \vec{E}(\vec{r}) = \int_{S(\Delta V)} d\vec{f} \cdot \vec{E}(\vec{r}),$$

which for $\Delta x \rightarrow 0$ gives $\Delta F \vec{n} \cdot (\vec{E_a} - \vec{E_i})$. Since this integral has to be equal to $\frac{1}{\epsilon_0}\sigma \Delta F$ we can deduce that $$\vec{n} \cdot (\vec{E_a}-\vec{E_i}) = \frac{\sigma}{\epsilon_0}.$$

My question is, how do I calculate this limit of $\Delta x \rightarrow 0$? I get that the integral basically computes all the scalar products of the electric field on the surface and adds them up, but how do I get the vector $ (\vec{E_a}-\vec{E_i})$? Why is the electric field given by this vector? It's probably really simple but I can't make sense of this right now, so some help would be appreciated.

$\endgroup$
1
$\begingroup$

You are using Gauss's Law. $\ Electric\ Flield's\ Flux= \frac{Q}{\epsilon_0}$

Calculate the flux through the 6 surfaces first.

When $\Delta x\rightarrow0$, Then the contribution of flux coming from the two surfaces that are shaded surfaces as well as the other two "non- radial" surfaces becomes negligibly small. You can ignore them in the limit. I hope you are following. Only the 2 big surfaces where $\ df$ is marked will contribute because the limit does nothing to their area.

Now simply calculate the flux from these two. It is

$\ df\ \vec{n}\cdot\vec{E_a}$ for one and $\ -df\ \vec{n}\cdot\vec{E_i}$ for the other. Negative sign coming in the second case because the flux is inward. Net charge lies on the surface only, so $\ Q=df \ \sigma$.

Use Gauss's Law. $\ df$ will cancel giving you what you want

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy