1
$\begingroup$

I've read some answers on stack exchange for this question, but I don't find them particularly clear. I am convinced of the case for a sphere, since the area of a sphere is used in the derivation for the law.

enter image description here

So maybe relating all other areas to the area of a sphere would provide an intuitive 'proof'?

I was uncertain, however, about approximating an infinitesimal curved surface to a flat surface.

$\endgroup$
1
  • $\begingroup$ The electric flux through an open in general surface $\,\rm S\,$ in empty space due to a point electric charge $\,q\,$ is $$ \Phi_{\rm S}\boldsymbol=\dfrac{\Theta}{4\pi}\dfrac{q}{\epsilon_{0}} $$ where $\,\Theta\,$ the solid angle by which the charge $\,q\,$ $''$ sees $''$ the surface $\,\rm S$. In the case of closed surfaces, like a sphere, we have a complete solid angle $\,\Theta\boldsymbol=4\pi\,$ so $$ \Phi_{\text{closed }\rm S}\boldsymbol=\dfrac{q}{\epsilon_{0}} $$ $\endgroup$
    – Frobenius
    Sep 25 at 10:23
2
$\begingroup$

Gauss law deals with electric flux and electric flux is related to number of field lines cutting or passing through a surface. And a charge create definite number of field lines, thus if any closed surface is enclosing the charge every field lines emmited or created by the charge will pass through the closed surface. So, whatever the closed surface be, the number of field lines passing through the surface will always be the same.

$\endgroup$
2
$\begingroup$

In the picture you included, there are two distinct elements. One is Gauss’s law proper $$ \oint \vec E\cdot d\vec A=\frac{q}{\epsilon_0}\, , $$ which is always true and works for any closed surface.

The other part deals with using Gauss’s law to recover the field, i.e. the step $E=kq /r^2$. This is only possible in some circumstances where the surface and the field have high levels of symmetry.

In the example of your picture, imagine first the source charge is outside the Gaussian sphere. Then by Gauss’s law $\oint \vec E\cdot d\vec A=0$ since there is no charge inside the Gaussian surface. Of course, this does not imply there is no field. In other words, $\oint \vec E\cdot d\vec A=0$ does not imply $\vec E=0$.

Next imagine the Gaussian sphere is replaced by a long rectangle containing the charge. There is now charge inside the rectangular Gaussian rectangle but I can make the area of the rectangle whatever I want so it would seem I could get any field I want. In other words, $\oint \vec E\cdot d\vec A=q/\epsilon_0\ne E A$ always.

So while $\oint \vec E\cdot d\vec A=q/\epsilon_0$ is always true irrespective of the Gaussian surface, it is only when $\oint \vec E\cdot d\vec A=E A$ (which only holds for specific surfaces placed in certain ways around the source) that Gauss’s law is useful in recovering the field.

$\endgroup$
1
$\begingroup$

This is because every shape can be changed into other more symmetric shape. We can convert it to sphere and thus derive Gauss' law.

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.