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The derivation of Gauss electric flux is as follows :

$$\iint{\vec{E}}\cdot{\vec{dS}}=\iint E \, dS \cos\theta \, .$$

The projection of infinitesimal area on the surface $\vec{dS}$ on the radial direction of electric field $\vec{E}$ (i.e. $dS \cos\theta$) is equal to the element area of sphere $r^2 (\sin \varphi) \, d\varphi d\phi$.

Therefore, \begin{align} \iint{\vec{E}}\cdot{\vec{dS}} &=\iint\frac{1}{4\pi\epsilon_0} \frac{q}{r^2}(r^2 (\sin\varphi) \, d\varphi \, d\phi) \\ &=\dfrac{q}{4\pi\epsilon_0}\iint(\sin\varphi) \, d\varphi \, d\phi \\ &=\dfrac{q}{4\pi\epsilon_0}4\pi \\ &=\dfrac{q}{\epsilon_0} \, . \end{align}

Now coming back to the main question:

Electric field at all points on the Gaussian surface is independent of the coordinate system. Also the angle between electric field and area vector on the Gaussian surface is independent of the coordinate system.

However, the infinitesimal area on the surface is not the same in all coordinate systems. For example, in a re-scaled orthogonal coordinate system where $\hat{i}\neq\hat{j}$, then $dx\neq\ dy$ and thus $dS'=dx\ dy$ won't be the same as $dS$ in Cartesian coordinate system. Then will $dS \cos\theta$ be equal to element area of sphere? If so, why?

Will we get a result other than $q/\epsilon_0$ in the other coordinate systems or will we get the same result ($q/\epsilon_0$)?

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  • $\begingroup$ Please look at the edits I made to this post. It will help you understand how to use better math formatting and English. In particular, note that you should put a backslash before cos and sin. $\endgroup$ – DanielSank Nov 10 '15 at 9:19
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I think we should start with the local form of Gauss's law $\nabla.\vec{E}=\rho$ Now $$ \int \nabla.\vec{E}\,dv=\int \rho\,dv$$ Using Gauss's divergence theorem we have $$\int\vec{E}.\vec{ds}=q$$ I assume $\epsilon_{0}$ to be 1 but you can always put that back into this. I think this way of looking at it does not assume any coordinate dependence.. Ofcourse if you change coordinate E and ds both will manage to give the same answer as @qftishard pointed out(because this law is not an artifact of mathematics,rather deep physical meaning is associated with it. So no matter what you do you are guranteed to get the same answer.).But the local form Of Gauss's law I think is something which is postulated on physical reasoning.

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