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When we use Gauss theorem/ gauss law....how do you decide what shape the Gaussian surface chosen must be for a given charge distribution(finite charge distribution)....because for a given charge distribution,I can draw a sphere , a cylinder , a cube and similarly any Gaussian surface enclosing the charge completely...but all of them would give different values of electric field right at the same given point.please explain how the same universal law giving different expressions of electric field at the same point due to a finite charge distribution makes any sense...because irrespective of shape considered,the same net charge is enclosed

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    $\begingroup$ If Gauss's law is applied correctly any chosen surface will give the correct value for the electric flux through the chosen surface. Please give an example where Gauss's law "does not work". $\endgroup$ – Farcher Aug 12 '17 at 5:41
  • $\begingroup$ Gauss's Law applies to every surface that encloses a charge. However, if you want to be able to evaluate the integral, you should choose the Gaussian surface such that the electric field strength is constant over the entire surface. Once this is done, the integral usually simplifies to a known formula. $\endgroup$ – David White Aug 12 '17 at 15:13
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Let me state Gauss law very briefly:

The surface integral of electric field over a closed surface is equal to the charge contained within the surface over $\epsilon_0$. $$ \int_{\Delta s}{\mathbf{E\cdot ds}} = \frac{q_{en}}{\epsilon_0} $$ $$or$$ The divergence of Electric field at a point of space is equal to the charge density at that point over $\epsilon_0$. $$\nabla\cdot\mathbf{E} = \frac{\rho}{\epsilon_0}$$

Unlike Coulomb's law, which directly gives an expression for the electric field, Gauss law gives a differential equation of vectors to solve the electric field from. In short, electric field is difficult to calculate from Gauss law in most circumstances.

But we still can find the electric field if we somehow solve the integral. This is practical when there exists symmetry in the situation where we want to calculate electric field.

So,

for a given charge distribution,I can draw a sphere , a cylinder , a cube and similarly any Gaussian surface enclosing the charge completely...

If you take an arbitrary gaussian surface, chances are that it is not very symmetrical and you are not able to calculate the integral on the left side of the equation. Once you know the electric field, and you now calculate the integral, you'll see that Gauss law holds. So, although you have an infinite number of choices of surfaces to verify Gauss law, you don't have as many to find the electric field from.

but all of them would give different values of electric field right at the same given point.

Let me stress again that this simply isn't true. If you are able to find the electric field from one of the surfaces, and you plug that value value of electric field in Gauss law for any arbitrary surface, then you will always find that Gauss law holds good.

Take home message: You may take a closed surface of any shape as a gaussian surface and apply Gauss law, and there will be no inconsistency.

Hope it makes sense. :)

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  • $\begingroup$ I understand what you mean but I have also seen a formula evaluated directly from gauss law which is $\endgroup$ – Adithya Eshwarla Aug 12 '17 at 9:34
  • $\begingroup$ Can you send me the formula? Or a link if you saw it online... $\endgroup$ – Shreyansh Darshan Aug 12 '17 at 9:38
  • $\begingroup$ I understand what you mean but I have also seen a formula evaluated directly from gauss law which is E=q/(e0*S).....where q is charge enclosed by Gaussian surface ,e0 is epsilon 0 and S is surface area of Gaussian surface.....if u observe this equation, it is derived directly from gauss law by evaluating the lhs integral and taking integral ds as S......now here...based on whether I consider a cube,sphere,cylinder etc.....the S will change and thereby won't I get a different field at the same point for the same charge enclosed?!i hope you understood my question $\endgroup$ – Adithya Eshwarla Aug 12 '17 at 9:40
  • $\begingroup$ Good question. But notice that as @FGSUZ said in his answer, the integral $\int \vec E \cdot \vec {ds} $ can be reduced to $E\times S$ only when E has the same value at all points of the surface and is perpendicular to it at all points. Taking an arbitrary surface will NOT change the flux, but then the integral will not be reducable to such simple forms, and we won't get a direct expression for E. In short, $E=\frac {q}{S \times \epsilon_0}$ is not correct for any arbitrary surface. Hence no contradiction arises. $\endgroup$ – Shreyansh Darshan Aug 12 '17 at 10:45
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@Shreyansh Darshan already gave a nice answer. Just in case, I'll highlight some points (I hope it's useful).

The Gauss law does not actually talk about the electric field, but the electric flux. The Gauss law is ALWAYS hold, no matter what surface you take. If you take a cylinder, you can be sure that the whole flux through that cylinder will be $q_{int}/\varepsilon_{0}$. If you take another surface, the flux will be also that value (it depends on the charge inside it). The surface must be closed, of course.

But that's the flux, and not the electric field. Knowing the flux can be okay, but it is often very useless. The Gauss law is only useful it you somehow can take $\vec{E}$ out of the integral. For which you need two conditions:

1) $\vec{E}$ makes a constant angle with the surface (preferably 0). Then the scalar product becomes the product of $E$, $dS$ and a $\cos$ that is constant.

2) The surface you chose is such that $E$ is constant along all points of that surface. Only in that case you can take $E$ outside the integral and then it's true that

$ |\vec{E}|= \dfrac{q_{int}}{\varepsilon_{0} \times \iint dS} = \dfrac{q_{int}}{\varepsilon_{0} \times S}$.

This is the basic case, and it's the first they tell us at basic levels (as pointed by @Adithya Eshwarla) But this is only a particular case.

In sum, there's no ambiguity: the flux is indeed that quantity independently of the surface. If what you want is geting the electric field with that, then you must be able to intelligently find a concrete surface in which those two conditions are hold and so you can extract the $E$ inside. As the other answer says, symmetry plays a central role in this task, but it is possible that the system doesn't have any useful symmetry and then you cannot use this law to get $E$.

Also, you can use the differential form instead, as said.

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  • $\begingroup$ So basically E can be taken out of the integral If n only if the assumed Gaussian surface is symmetrical(all points on Gaussian surface are equidistant from charge distribution) $\endgroup$ – Adithya Eshwarla Aug 12 '17 at 10:20
  • $\begingroup$ Yes, that's right. This is the central idea of this "topic" of electromagnetism, and I feel is not always highlighted enough. $\endgroup$ – FGSUZ Aug 12 '17 at 10:26

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